Chapter 6 Normalization of Database Tables

Chapter 6 Normalization of Database Tables
DATABASE S YSTEMS
DESIGN, IMPLEMENTATION, AND MANAGEMENT
CARLOS CORONEL • STEVEN MORRIS • PETER ROB
Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
Database Systems: Design, Implementation,
and Management, Ninth Edition
Carlos Coronel, Steven Morris, and Peter Rob
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edication
To the treasures in my life:To Victoria, for 20 wonderful years.Thank you for your unending
support, for being my angel, my sweetie and most importantly, my best friend. To Carlos
Anthony whose intelligence and wit is matched only by your good looks; you show us the way.
Thank you for your words of wisdom, contagious happiness and for bringing us shining days.
You are still young; your best times are still to come.To Gabriela Victoria who is the image of
brilliance, beauty, and faithfulness. Thank you for being the sunshine in my cloudy days. To
Christian Javier whose endless energy and delightful smiles keep us going every day. Thank you
for being the youthful reminder of life’s simple beauties.To my parents, Sarah and Carlos, thank
you for your sacrifice and example.To all of you, you are all my inspiration.“TQTATA.”
Carlos Coronel
To Pamela, from high school sweetheart through 20 years of marriage, the beautiful love of
my life who has supported, encouraged, and inspired me. More than anyone else, you are
responsible for whatever successes I have achieved. To my son, Alexander Logan, whose
depth of character is without measure.You are my pride and joy. To my daughter, Lauren
Elizabeth, whose beauty and intensity take my breath away. You are my heart and soul.
Thank you all for the sacrifices that you have made that enabled me to pursue this dream.
I love you so much more than I can express.To my mother, Florence Maryann, and to the
memory of my father, Alton Lamar, who together instilled in me the desire to learn and the
passion to achieve.To my mother-in-law, Connie Duke, and to the memory of my father-inlaw, Wayne Duke, who taught me to find joy in all things. To all of you, with all my love,
I dedicate this book.
Steven Morris
To Anne, who remains my best friend after 48 years of marriage.To our son, Peter William,
who turned out to be the man we hoped he would be and who proved his wisdom by
making Sheena our treasured daughter-in-law. To Sheena, who stole our hearts so many
years ago.To our grandsons, Adam Lee and Alan Henri, who are growing up to be the fine
human beings their parents are.To my mother-in-law, Nini Fontein, and to the memory of
my father-in-law, Henri Fontein—their life experiences in Europe and Southeast Asia would
fill a history book and they enriched my life by entrusting me with their daughter’s future.
To the memory of my parents, Hendrik and Hermine Rob, who rebuilt their lives after
World War II’s horrors, who did it again after a failed insurgency in Indonesia, and who
finally found their promised land in these United States. And to the memory of Heinz, who
taught me daily lessons about loyalty, uncritical acceptance, and boundless understanding.
I dedicate this book to you, with love.
Peter Rob
D E D I C A T I O N
D
BRIEF CONTENTS
PART I:
Database Concepts
Chapter 1: Database Systems
Chapter 2: Data Models
PART II:
Design Concepts
Chapter 3: The Relational Database Model
Chapter 4: Entity Relationship (ER) Modeling
Chapter 5: Advanced Data Modeling
Chapter 6: Normalization of Database Tables
PART III:
Advanced Design and Implementation
Chapter 7: Introduction to Structured Query Language (SQL)
Chapter 8: Advanced SQL
Chapter 9: Database Design
PART IV:
Advanced Database Concepts
Chapter 10: Transaction Management and Concurrency Control
Chapter 11: Database Performance Tuning and Query Optimization
Chapter 12: Distributed Database Management Systems
Chapter 13: Business Intelligence and Data Warehouses
PART V:
Databases and the Internet
Chapter 14: Database Connectivity and Web Technologies
PART VI:
Database Administration
Chapter 15: Database Administration and Security
GLOSSARY
INDEX
IV
BRIEF CONTENTS
The following appendixes and answers to selected questions and problems are included in the Premium Website for
this text, found at cengage.com/mis/coronel.
Appendix A:
Designing Databases with Visio Professional: A Tutorial
Appendix B:
The University Lab: Conceptual Design
Appendix C:
The University Lab: Conceptual Design Verification, Logical
Design, and Implementation
Appendix D:
Converting an ER Model into a Database Structure
Appendix E:
Comparison of ER Model Notations
Appendix F:
Client/Server Systems
Appendix G:
Object-Oriented Databases
Appendix H:
Unified Modeling Language (UML)
Appendix I:
Databases in Electronic Commerce
Appendix J:
Web Database Development with ColdFusion
Appendix K:
The Hierarchical Database Model
Appendix L:
The Network Database Model
Appendix M:
Microsoft ® Access ® Tutorial
Appendix N:
Creating a New Database Using Oracle 11g
Answers to Selected Questions and Problems
V
TABLE OF CONTENTS
PA R T I D ATA B A S E C O N C E P T S
Business Vignette: The Relational Revolution
3
Chapter 1
4
Database Systems
1.1 Why Databases?
1.2 Data vs. Information
1.3 Introducing the Database
1.3.1
Role and Advantages of the DBMS
1.3.2
Types of Databases
1.4 Why Database Design is Important
1.5 Evolution of File System Data Processing
1.5.1
Manual File Systems
1.5.2
Computerized File Systems
1.5.3
File System Redux: Modern End-User Productivity Tools
1.6 Problems with File System Data Processing
1.6.1
Structural and Data Dependence
1.6.2
Data Redundancy
1.6.3
Lack of Design and Data-Modeling Skills
1.7 Database Systems
1.7.1
The Database System Environment
1.7.2
DBMS Functions
1.7.3
Managing the Database System: A Shift in Focus
Summary
Key Terms
Review Questions
Problems
Chapter 2
Data Models
2.1 Data Modeling and Data Models
2.2 The Importance of Data Models
2.3 Data Model Basic Building Blocks
2.4 Business Rules
2.4.1
Discovering Business Rules
2.4.2
Translating Business Rules into Data Model Components
2.4.3
Naming Conventions
2.5 The Evolution of Data Models
2.5.1
Hierarchical and Network Models
2.5.2
The Relational Model
2.5.3
The Entity Relationship Model
2.5.4
The Object-Oriented (OO) Model
2.5.5
Newer Data Models: Object/Relational and XML
2.5.6
The Future of Data Models
2.5.7
Data Models: A Summary
2.6 Degrees of Data Abstraction
2.6.1
The External Model
2.6.2
The Conceptual Model
2.6.3
The Internal Model
2.6.4
The Physical Model
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Summary
Key Terms
Review Questions
Problems
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PA R T I I D E S I G N C O N C E P T S
Business Vignette: BP’s Data Modeling Initiative
Chapter 3
The Relational Database Model
3.1 A Logical View of Data
3.1.1
Tables and Their Characteristics
3.2 Keys
3.3 Integrity Rules
3.4 Relational Set Operators
3.5 The Data Dictionary and the System Catalog
3.6 Relationships within the Relational Database
3.6.1
The 1:M Relationship
3.6.2
The 1:1 Relationship
3.6.3
The M:N Relationship
3.7 Data Redundancy Revisited
3.8 Indexes
3.9 Codd’s Relational Database Rules
Summary
Key Terms
Review Questions
Problems
Chapter 4
Entity Relationship (ER) Modeling
4.1 The Entity Relationship Model (ERM)
4.1.1
Entities
4.1.2
Attributes
4.1.3
Relationships
4.1.4
Connectivity and Cardinality
4.1.5
Existence Dependence
4.1.6
Relationship Strength
4.1.7
Weak Entities
4.1.8
Relationship Participation
4.1.9
Relationship Degree
4.1.10
Recursive Relationships
4.1.11
Associative (Composite) Entities
4.2 Developing an ER Diagram
4.3 Database Design Challenges: Conflicting Goals
Summary
Key Terms
Review Questions
Problems
Cases
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Chapter 5
Advanced Data Modeling
5.1 The Extended Entity Relationship Model
5.1.1
Entity Supertypes and Subtypes
5.1.2
Specialization Hierarchy
5.1.3
Inheritance
5.1.4
Subtype Discriminator
5.1.5
Disjoint and Overlapping Constraints
5.1.6
Completeness Constraint
5.1.7
Specialization and Generalization
5.2 Entity Clustering
5.3 Entity Integrity: Selecting Primary Keys
5.3.1
Natural Keys and Primary Keys
5.3.2
Primary Key Guidelines
5.3.3
When to Use Composite Primary Keys
5.3.4
When to Use Surrogate Primary Keys
5.4 Design Cases: Learning Flexible Database Design
5.4.1
Design Case #1: Implementing 1:1 Relationships
5.4.2
Design Case #2: Maintaining History of Time-Variant Data
5.4.3
Design Case #3: Fan Traps
5.4.4
Design Case #4: Redundant Relationships
Summary
Key Terms
Review Questions
Problems
Cases
Chapter 6
Normalization of Database Tables
6.1 Database Tables and Normalization
6.2 The Need for Normalization
6.3 The Normalization Process
6.3.1
Conversion to First Normal Form
6.3.2
Conversion to Second Normal Form
6.3.3
Conversion to Third Normal Form
6.4 Improving the Design
6.5 Surrogate Key Considerations
6.6 Higher-Level Normal Forms
6.6.1
The Boyce-Codd Normal Form (BCNF)
6.6.2
Fourth Normal Form (4NF)
6.7 Normalization and Database Design
6.8 Denormalization
6.9 Data-Modeling Checklist
Summary
Key Terms
Review Questions
Problems
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PA R T I I I A D VA N C E D D E S I G N A N D I M P L E M E N TAT I O N
Business Vignette: The Many Benefits of BI
Chapter 7
Introduction to Structured Query Language (SQL)
7.1 Introduction to SQL
7.2 Data Definition Commands
7.2.1
The Database Model
7.2.2
Creating the Database
7.2.3
The Database Schema
7.2.4
Data Types
7.2.5
Creating Table Structures
7.2.6
SQL Constraints
7.2.7
SQL Indexes
7.3 Data Manipulation Commands
7.3.1
Adding Table Rows
7.3.2
Saving Table Changes
7.3.3
Listing Table Rows
7.3.4
Updating Table Rows
7.3.5
Restoring Table Contents
7.3.6
Deleting Table Rows
7.3.7
Inserting Table Rows with a Select Subquery
7.4 SELECT Queries
7.4.1
Selecting Rows with Conditional Restrictions
7.4.2
Arithmetic Operators:The Rule of Precedence
7.4.3
Logical Operators: AND, OR, and NOT
7.4.4
Special Operators
7.5 Additional Data Definition Commands
7.5.1
Changing a Column’s Data Type
7.5.2
Changing a Column’s Data Characteristics
7.5.3
Adding a Column
7.5.4
Dropping a Column
7.5.5
Advanced Data Updates
7.5.6
Copying Parts of Tables
7.5.7
Adding Primary and Foreign Key Designations
7.5.8
Deleting a Table from the Database
7.6 Additional SELECT Query Keywords
7.6.1
Ordering a Listing
7.6.2
Listing Unique Values
7.6.3
Aggregate Functions
7.6.4
Grouping Data
7.7 Virtual Tables: Creating a View
7.8 Joining Database Tables
7.8.1
Joining Tables with an Alias
7.8.2
Recursive Joins
7.8.3
Outer Joins
Summary
Key Terms
Review Questions
Problems
Cases
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Chapter 8
Advanced SQL
8.1 Relational Set Operators
8.1.1
UNION
8.1.2
UNION ALL
8.1.3
INTERSECT
8.1.4
MINUS
8.1.5
Syntax Alternatives
8.2 SQL Join Operators
8.2.1
Cross Join
8.2.2
Natural Join
8.2.3
Join USING Clause
8.2.4
JOIN ON Clause
8.2.5
Outer Joins
8.3 Subqueries and Correlated Queries
8.3.1
WHERE Subqueries
8.3.2
IN Subqueries
8.3.3
HAVING Subqueries
8.3.4
Multirow Subquery Operators: ANY and ALL
8.3.5
FROM Subqueries
8.3.6
Attribute List Subqueries
8.3.7
Correlated Subqueries
8.4 SQL Functions
8.4.1
Date and Time Functions
8.4.2
Numeric Functions
8.4.3
String Functions
8.4.4
Conversion Functions
8.5 Oracle Sequences
8.6 Updatable Views
8.7 Procedural SQL
8.7.1
Triggers
8.7.2
Stored Procedures
8.7.3
PL/SQL Processing with Cursors
8.7.4
PL/SQL Stored Functions
8.8 Embedded SQL
Summary
Key Terms
Review Questions
Problems
Cases
Chapter 9
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Database Design
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9.1 The Information System
9.2 The Systems Development Life Cycle (SDLC)
9.2.1
Planning
9.2.2
Analysis
9.2.3
Detailed Systems Design
9.2.4
Implementation
9.2.5
Maintenance
9.3 The Database Life Cycle (DBLC)
9.3.1
The Database Initial Study
9.3.2
Database Design
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9.3.3
Implementation and Loading
9.3.4
Testing and Evaluation
9.3.5
Operation
9.3.6
Maintenance and Evolution
9.4 Conceptual Design
9.4.1
Data Analysis and Requirements
9.4.2
Entity Relationship Modeling and Normalization
9.4.3
Data Model Verification
9.4.4
Distributed Database Design
9.5 DBMS Software Selection
9.6 Logical Design
9.6.1
Map the Conceptual Model to the Logical Model
9.6.2
Validate the Logical Model Using Normalization
9.6.3
Validate Logical Model Integrity Constraints
9.6.4
Validate the Logical Model against User Requirements
9.7 Physical Design
9.7.1
Define Data Storage Organization
9.7.2
Define Integrity and Security Measures
9.7.3
Determine Performance Measures
9.8 Database Design Strategies
9.9 Centralized vs. Decentralized Design
Summary
Key Terms
Review Questions
Problems
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PA R T I V A D VA N C E D D ATA B A S E C O N C E P T S
Business Vignette: Combating Data Explosion
Chapter 10
Transaction Management and Concurrency Control
10.1 What Is a Transaction?
10.1.1
Evaluating Transaction Results
10.1.2
Transaction Properties
10.1.3
Transaction Management with SQL
10.1.4
The Transaction Log
10.2 Concurrency Control
10.2.1
Lost Updates
10.2.2
Uncommitted Data
10.2.3
Inconsistent Retrievals
10.2.4
The Scheduler
10.3 Concurrency Control with Locking Methods
10.3.1
Lock Granularity
10.3.2
Lock Types
10.3.3
Two-Phase Locking to Ensure Serializability
10.3.4
Deadlocks
10.4 Concurrency Control with Time Stamping Methods
10.4.1
Wait/Die and Wound/Wait Schemes
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10.5 Concurrency Control with Optimistic Methods
10.6 Database Recovery Management
10.6.1
Transaction Recovery
Summary
Key Terms
Review Questions
Problems
Chapter 11
Database Performance Tuning and Query Optimization
11.1 Database Performance-Tuning Concepts
11.1.1
Performance Tuning: Client and Server
11.1.2
DBMS Architecture
11.1.3
Database Statistics
11.2 Query Processing
11.2.1
SQL Parsing Phase
11.2.2
SQL Execution Phase
11.2.3
SQL Fetching Phase
11.2.4
Query Processing Bottlenecks
11.3 Indexes and Query Optimization
11.4 Optimizer Choices
11.4.1
Using Hints to Affect Optimizer Choices
11.5 SQL Performance Tuning
11.5.1
Index Selectivity
11.5.2
Conditional Expressions
11.6 Query Formulation
11.7 DBMS Performance Tuning
11.8 Query Optimization Example
Summary
Key Terms
Review Questions
Problems
Chapter 12
Distributed Database Management Systems
12.1 The Evolution of Distributed Database Management Systems
12.2 DDBMS Advantages and Disadvantages
12.3 Distributed Processing and Distributed Databases
12.4 Characteristics of Distributed Database Management Systems
12.5 DDBMS Components
12.6 Levels of Data and Process Distribution
12.6.1
Single-Site Processing, Single-Site Data (SPSD)
12.6.2
Multiple-Site Processing, Single-Site Data (MPSD)
12.6.3
Multiple-Site Processing, Multiple-Site Data (MPMD)
12.7 Distributed Database Transparency Features
12.8 Distribution Transparency
12.9 Transaction Transparency
12.9.1
Distributed Requests and Distributed Transactions
12.9.2
Distributed Concurrency Control
12.9.3
Two-Phase Commit Protocol
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12.10 Performance Transparency and Query Optimization
12.11 Distributed Database Design
12.11.1 Data Fragmentation
12.11.2 Data Replication
12.11.3 Data Allocation
12.12 Client/Server vs. DDBMS
12.13 C. J. Date’s Twelve Commandments for Distributed Databases
Summary
Key Terms
Review Questions
Problems
Chapter 13 Business Intelligence and Data Warehouses
13.1 The Need for Data Analysis
13.2 Business Intelligence
13.3 Business Intelligence Architecture
13.4 Decision Support Data
13.4.1
Operational Data vs. Decision Support Data
13.4.2
Decision Support Database Requirements
13.5 The Data Warehouse
13.5.1
Twelve Rules that Define a Data Warehouse
13.5.2
Decision Support Architectural Styles
13.6 Online Analytical Processing
13.6.1
Multidimensional Data Analysis Techniques
13.6.2
Advanced Database Support
13.6.3
Easy-to-Use End-User Interface
13.6.4
Client/Server Architecture
13.6.5
OLAP Architecture
13.6.6
Relational OLAP
13.6.7
Multidimensional OLAP
13.6.8
Relational vs. Multidimensional OLAP
13.7 Star Schemas
13.7.1
Facts
13.7.2
Dimensions
13.7.3
Attributes
13.7.4
Attribute Hierarchies
13.7.5
Star Schema Representation
13.7.6
Performance-Improving Techniques for the Star Schema
13.8 Implementing a Data Warehouse
13.8.1
The Data Warehouse as an Active Decision Support Framework
13.8.2
A Company-Wide Effort that Requires User Involvement
13.8.3
Satisfy the Trilogy: Data,Analysis, and Users
13.8.4
Apply Database Design Procedures
13.9 Data Mining
13.10 SQL Extensions for OLAP
13.10.1 The ROLLUP Extension
13.10.2 The CUBE Extension
13.10.3 Materialized Views
Summary
Key Terms
Review Questions
Problems
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PA R T V D ATA B A S E S A N D T H E I N T E R N E T
Business Vignette: KBB Transforms with Innovative Web Services
Chapter 14 Database Connectivity and Web Technologies
14.1 Database Connectivity
14.1.1
Native SQL Connectivity
14.1.2
ODBC, DAO, and RDO
14.1.3
OLE-DB
14.1.4
ADO.NET
14.1.5
Java Database Connectivity (JDBC)
14.2 Internet Databases
14.2.1
Web-to-Database Middleware: Server-Side Extensions
14.2.2
Web Server Interfaces
14.2.3
The Web Browser
14.2.4
Client-Side Extensions
14.2.5
Web Application Servers
14.3 Extensible Markup Language (XML)
14.3.1
Document Type Definitions (DTD) and XML Schemas
14.3.2
XML Presentation
14.3.3
XML Applications
14.4 SQL Data Services
Summary
Key Terms
Review Questions
Problems
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PA R T V I D ATA B A S E A D M I N I S T R AT I O N
Business Vignette: The Rising SQL Injection Threat
Chapter 15
Database Administration and Security
15.1 Data as a Corporate Asset
15.2 The Need for and Role of a Database in an Organization
15.3 Introduction of a Database: Special Considerations
15.4 The Evolution of the Database Administration Function
15.5 The Database Environment’s Human Component
15.5.1
The DBA’s Managerial Role
15.5.2
The DBA’s Technical Role
15.6 Security
15.6.1
Security Policies
15.6.2
Security Vulnerabilities
15.6.3
Database Security
15.7 Database Administration Tools
15.7.1
The Data Dictionary
15.7.2
CASE Tools
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612
613
616
618
623
629
629
630
631
633
633
635
TABLE OF CONTENTS
15.8 Developing a Data Administration Strategy
15.9 The DBA at Work: Using Oracle for Database Administration
15.9.1
Oracle Database Administration Tools
15.9.2
The Default Login
15.9.3
Ensuring an Automatic RDBMS Start
15.9.4
Creating Tablespaces and Datafiles
15.9.5
Managing the Database Objects:Tables,Views,Triggers, and Procedures
15.9.6
Managing Users and Establishing Security
15.9.7
Customizing the Database Initialization Parameters
Summary
Key Terms
Review Questions
637
639
640
640
641
642
643
644
647
648
649
649
Glossary
653
Index
672
IN THE PREMIUM WEBSITE
The Premium Website can be found at cengage.com/mis/coronel. Locate your premium access card in the front of each new
book purchase, and click “Create My Account” to begin the registration process. If you’ve purchased a used book, please
search for Database Systems, Ninth Edition at www.ichapters.com where you can purchase instant access.
Appendix A
Designing Databases with Visio Professional: A Tutorial
Appendix B
The University Lab: Conceptual Design
Appendix C
The University Lab: Conceptual Design Verification, Logical Design,
and Implementation
Appendix D
Converting an ER Model into a Database Structure
Appendix E
Comparison of ER Model Notations
Appendix F
Client/Server Systems
Appendix G
Object-Oriented Databases
Appendix H
Unified Modeling Language (UML)
Appendix I
Databases in Electronic Commerce
Appendix J
Web Database Development with ColdFusion
Appendix K
The Hierarchical Database Model
Appendix L
The Network Database Model
Appendix M
Microsoft® Access® Tutorial
Appendix N
Creating a New Database with Oracle 11g
Answers to Selected Questions and Problems
XV
PREFACE
This database systems book has been successful through eight editions because the authors, editors, and the publisher
paid attention to the impact of technology and to adopter questions and suggestions.We believe that this ninth edition
successfully reflects the same attention to such stimuli. Furthermore this ninth edition marks the addition of a new
co-author, Steven Morris. Steven brings his wealth of knowledge, teaching experience, and expertise to this work.
In many respects, rewriting a book is more difficult than writing it the first time. If the book is successful, as this one is,
a major concern is that the updates, inserts, and deletions will adversely affect writing style and continuity of coverage.
Fortunately, this edition benefits from the incorporation of a new co-author with fresh ideas and perspectives balanced
by the experience of the original authors to ensure continuity of writing style and quality of presentation. In addition, the
efforts of a combination of superb reviewers and editors, plus a wealth of feedback from adopters and students of the
previous editions, helped provide the guidance that make this new edition the best yet.
XVI
PREFACE
CHANGES TO THE NINTH EDITION
In this ninth edition, we have added some new features and we have reorganized some of the coverage to provide a
better flow of material. Aside from enhancing the already strong database design coverage, we have made other
improvements in the topical coverage. Here are a few of the highlights:
• Updated Business Vignettes showing the impact of database technologies in the real world.
• Strengthened the database design contents by more clearly differentiating among the conceptual, logical, and physical design stages.
• Streamlined and modernized the coverage of database evolution and the importance of database design
skills.
• Enhanced the coverage of data models by shifting the focus from a historical perspective to emerging
data technologies.
• Expanded end-of-chapter review questions and problems and introduced a new Cases section to
selected chapters.
• Formalized and improved consistency of normalization concepts.
• Improved readability and overall visual appeal of the book.
• Created a database design process guide and a data modeling checklist as cover inserts.
This ninth edition continues to provide a solid and practical foundation for the design, implementation, and management of database systems. This foundation is built on the notion that, while databases are very practical things, their
successful creation depends on understanding the important concepts that define them. It’s not easy to come up with
the proper mix of theory and practice, but we are grateful that the previously mentioned feedback suggests that we
largely succeeded in our quest to maintain the proper balance.
THE APPROACH: A CONTINUED EMPHASIS ON DESIGN
As the title suggests, Database Systems: Design, Implementation, and Management covers three broad aspects of
database systems. However, for several important reasons, special attention is given to database design.
• The availability of excellent database software enables even database-inexperienced people to create
databases and database applications. Unfortunately, the “create without design” approach usually
paves the road to any number of database disasters. In our experience, many, if not most, database system failures are traceable to poor design and cannot be solved with the help of even the best programmers and managers. Nor is better DBMS software likely to overcome problems created or magnified by
poor design. Using an analogy, even the best bricklayers and carpenters can’t create a good building
from a bad blueprint.
• Most of the vexing database system management problems seem to be triggered by poorly designed databases. It hardly seems worthwhile to use scarce resources to develop excellent and extensive database system management skills in order to exercise them on crises induced by poorly designed databases.
• Design provides an excellent means of communication. Clients are more likely to get what they need
when database system design is approached carefully and thoughtfully. In fact, clients may discover
how their organizations really function once a good database design is completed.
XVII
PREFACE
• Familiarity with database design techniques promotes one’s understanding of current database technologies. For example, because data warehouses derive much of their data from operational databases,
data warehouse concepts, structures, and procedures make more sense when the operational database’s structure and implementation are understood.
Because the practical aspects of database design are stressed, we have covered design concepts and procedures in
detail, making sure that the numerous end-of-chapter problems and cases are sufficiently challenging so students can
develop real and useful design skills. We also make sure that students understand the potential and actual conflicts
between database design elegance, information requirements, and transaction processing speed. For example, it
makes little sense to design databases that meet design elegance standards while they fail to meet end-user information
requirements. Therefore, we explore the use of carefully defined trade-offs to ensure that the databases are capable of
meeting end-user requirements while conforming to high design standards.
TOPICAL COVERAGE
The Systems View
The book’s title begins with Database Systems. Therefore, we examine the
database and design concepts covered in Chapters 1–6 as part of a larger
whole by placing them within the systems analysis framework of Chapter 9.
We believe that database designers who fail to understand that the database
is part of a larger system are likely to overlook important database design
requirements. In fact, Chapter 9, Database Design, provides the map for
the advanced database design developed in Appendixes B and C. Within
the larger systems framework, we can also explore issues such as transaction management and concurrency control (Chapter 10), distributed database management systems (Chapter 12), business intelligence and data
warehouses (Chapter 13), database connectivity and Web technologies
(Chapter 14), and database administration and security (Chapter 15).
Database Design
The first item in the
book’s subtitle is Design,
and our examination of
database design is comprehensive. For example, Chapters 1 and 2 examine the development
and future of databases and data models and illustrate the need for
design. Chapter 3 examines the details of the relational database
model; Chapter 4 provides extensive, in-depth, and practical database
design coverage; and Chapter 5 explores advanced database design
topics. Chapter 6 is devoted to critical normalization issues that affect
database efficiency and effectiveness. Chapter 9 examines database
design within the systems framework and maps the activities required
to successfully design and implement the complex real-world database
XVIII
PREFACE
developed in Appendixes B and C. Appendix A, Designing Databases with Visio Professional: A Tutorial, provides a
good introductory tutorial for the use of a database design tool.
Because database design is affected by real-world transactions, the way data are distributed, and ever-increasing
information requirements, we examine major database features that must be supported in current-generation databases and models. For example, Chapter 10, Transaction Management and Concurrency Control, focuses on the
characteristics of database transactions and how they affect database integrity and consistency. Chapter 11,
Database Performance Tuning and Query Optimization, illustrates the need for query efficiency in a real world that
routinely generates and uses terabyte-sized databases and tables with millions of records. Chapter 12, Distributed
Database Management Systems, focuses on data distribution, replication, and allocation. In Chapter 13, Business
Intelligence and Data Warehouses, we explore the characteristics of the databases that are used in decision support
and online analytical processing. Chapter 14, Database Connectivity and Web Technologies, covers the basic database connectivity issues encountered in a Web-based data world, and it shows the development of Web-based database front ends.
Implementation
The second portion of the subtitle is Implementation. We use Structured
Query Language (SQL) in Chapters 7 and 8 to show how databases are
implemented and managed. Appendix M, Microsoft Access Tutorial, provides a quick but comprehensive guide to MS Access database implementation. Appendixes B and C demonstrate the design of a database that was
fully implemented and they illustrate a wide range of implementation
issues. We had to deal with conflicting design goals: design elegance, information requirements, and operational speed. Therefore, we carefully
audited the initial design (Appendix B) to check its ability to meet end-user
needs and to establish appropriate implementation protocols. The result of
this audit yielded the final, implementable design developed in Appendix
C. The special issues encountered in an Internet database environment are
addressed in Chapter 14, Database Connectivity and Web Technologies,
and in Appendix J, Web
Database
Development
with ColdFusion.
Management
The final portion of the subtitle is Management. We deal with database
management issues in Chapter 10, Transaction Management and
Concurrency Control; Chapter 12, Distributed Database Management
Systems; and Chapter 15, Database Administration and Security.
Chapter 11, Database Performance Tuning and Query Optimization, is
a valuable resource that illustrates how a DBMS manages the data
retrieval operations. In addition, Appendix N, Creating a New Database
Using Oracle 11g, walks you through the process of setting up a new
database.
XIX
PREFACE
TEACHING DATABASE: A MATTER OF FOCUS
Given the wealth of detailed coverage, instructors can “mix and match” chapters to produce the desired coverage.
Depending on where database courses fit into the curriculum, instructors may choose to emphasize database design or
database management. (See Figure 1.)
The hands-on nature of database design lends itself particularly well to class projects for which students use instructorselected software to prototype a student-designed system for the end user. Several of the end-of-chapter problems are
sufficiently complex to serve as projects, or an instructor may work with local businesses to give students hands-on
experience. Note that some elements of the database design track are also found in the database management track.
This is because it is difficult to manage database technologies that are not understood.
The options shown in Figure 1 serve only as a starting point. Naturally, instructors will tailor their coverage based on
their specific course requirements. For example, an instructor may decide to make Appendix I an outside reading
assignment and Appendix A a self-taught tutorial, then use that time to cover client/server systems or object-oriented
databases. The latter choice would serve as a gateway to UML coverage.
FIGURE
1
Core Coverage
(1) Database Systems
(2) Data Models
(3) The Relational Database Model
(4) Entity Relationship (ER) Modeling
(6) Normalization of Database Tables
(7) Introduction to Structured Query Language (SQL)
XX
Database Design and Implementation Focus
Database Management Focus
(5) Advanced Data Modeling
(8) Advanced SQL
(9) Database Design
(D) Converting an ER Model into a Database Structure
(E) Comparison of ER Model Notations
(H) Unified Modeling Language (UML)
(11) Database Performance Tuning and Query Optimization
(14) Database Connectivity and Web Development
(J) Web Database Development with ColdFusion
(10) Transaction Management and Concurrency Control
(11) Database Performance Tuning and Query Optimization
(12) Distributed Database Management Systems
(13) The Data Warehouse
(15) Database Administration
(F) Client/Server Systems
(G) Object-Oriented Databases
(I) Databases in Electronic Commerce
Supplementary Reading
Supplementary Reading
(A) Designing Databases with Visio Professional: A Tutorial
(B) The University Lab: Conceptual Design
(C) The University Lab: Conceptual Design Verification,
Logical Design, and Implementation
(F) Client/Server Systems
(L) The Network Database Model
(M) Microsoft Access Tutorial
(9) Database Design
(A) Designing Databases with Visio Professional: A Tutorial
(D) Converting an ER Model into a Database Structure
(E) Comparison of ER Model Notations
(K) The Hierarchical Database Model
(L) The Network Database Model
(N) Creating a New Database Using Oracle 11g
TEXT FEATURES
The Relational Revolution
Today, we take for granted the benefits brought to us by relational databases: the ability
to store, access, and change data quickly and easily on low-cost computers.Yet, until the
late 1970s, databases stored large amounts of data in a hierarchical structure that was
difficult to navigate and inflexible. Programmers needed to know what clients wanted to
Business Vignettes
highlight part topics in
a real-life setting.
do with the data before the database was designed. Adding or changing the way the data
was analyzed was a time-consuming and expensive process. As a result, you searched
B
V
usiness
ignette
through huge card catalogs to find a library book, you used road maps that didnt show
changes made in the last year, and you had to buy a newspaper to find information on
stock prices.
In 1970, Edgar “Ted” Codd, a mathematician employed by IBM, wrote an article that
would change all that. At the time, nobody realized that Codd’s obscure theories would
Online Content
Online Content boxes
draw attention to material
in the Premium Website
for this text and provide
ideas for incorporating
this content into the course.
The databases used in each chapter are available in the Premium Website for this book. Throughout the book,
Online Content boxes highlight material related to chapter content located in the Premium Website.
Note
Data that display data inconsistency are also referred to as data that lack data integrity. Data integrity is defined
as the condition in which all of the data in the database are consistent with the real-world events and conditions.
In other words, data integrity means that:
• Data are accurate—there are no data inconsistencies.
• Data are verifiable—the data will always yield consistent results.
Notes highlight important facts about the concepts introduced in the
chapter.
FIGURE
Illustrating data storage management with Oracle
1.9
Database Name: ORALAB.MTSU.EDU
A variety of four-color
figures, including ER
models and implementations, tables, and illustrations, clearly illustrate
difficult concepts.
The ORALAB database is
actually stored in nine
datafiles located on the C:
drive of the database server
computer.
The Oracle DBA Studio
Management interface also
shows the amount of space
used by each of the datafiles
that constitute the single
logical database.
The Oracle DBA Studio Administrator GUI shows the data storage
management characteristics for the ORALAB database.
XXI
TEXT FEATURES
S u m m a r y
The ERM uses ERDs to represent the conceptual database as viewed by the end user. The ERM’s main components
are entities, relationships, and attributes. The ERD also includes connectivity and cardinality notations. An ERD can
also show relationship strength, relationship participation (optional or mandatory), and degree of relationship
(unary, binary, ternary, etc.).
Connectivity describes the relationship classification (1:1, 1:M, or M:N). Cardinality expresses the specific number
of entity occurrences associated with an occurrence of a related entity. Connectivities and cardinalities are usually
based on business rules.
In the ERM, a M:N relationship is valid at the conceptual level. However, when implementing the ERM in a
relational database, the M:N relationship must be mapped to a set of 1:M relationships through a composite entity.
K e y
T e r m s
binary relationship, 118
identifying relationship, 112
relationship degree, 118
cardinality, 109
iterative process, 124
simple attribute, 106
composite attribute, 105
mandatory participation, 116
single-valued attribute, 106
composite key, 105
multivalued attribute, 106
strong relationship, 112
derived attribute, 107
non-identifying relationship, 111
ternary relationship, 118
existence-dependent, 110
optional participation, 116
unary relationship, 118
R e v i e w
A robust Summary at
the end of each chapter ties together the
major concepts and
serves as a quick
review for students.
An alphabetic list of
Key Terms points to
the pages where terms
are first explained.
Q u e s t i o n s
1. What two conditions must be met before an entity can be classified as a weak entity? Give an example of a
weak entity.
2. What is a strong (or identifying) relationship, and how is it depicted in a Crow’s Foot ERD?
3. Given the business rule “an employee may have many degrees,” discuss its effect on attributes, entities, and
relationships. (Hint: Remember what a multivalued attribute is and how it might be implemented.)
4. What is a composite entity, and when is it used?
Review Questions
challenge students to
apply the skills learned
in each chapter.
5. Suppose you are working within the framework of the conceptual model in Figure Q4.5.
P r o b l e m s
1. Given the following business rules, create the appropriate Crow’s Foot ERD.
a. A company operates many departments.
b. Each department employs one or more employees.
c. Each of the employees may or may not have one or more dependents.
d. Each employee may or may not have an employment history.
2. The Hudson Engineering Group (HEG) has contacted you to create a conceptual model whose application will
meet the expected database requirements for the company’s training program. The HEG administrator gives you
XXII
Problems become
progressively more
complex as students
draw on the lessons
learned from the completion of preceding
problems.
ADDITIONAL FEATURES
PREMIUM WEBSITE
Single Sign On (SSO) provides a central location from which you can access Cengage Learning’s online learning solutions with convenience and flexibility. You can:
• Gain access to online resources including robust Premium Website.
• Simplify your coursework by reducing human error and the need to keep track of multiple passwords.
See the insert card at the front of this book for instructions on how to access this text’s SSO site.
This Web resource, which you will find referenced throughout the book in the Online Content boxes, includes the following features:
Appendixes
Fourteen appendixes provide additional material on a variety of important areas, such as using Microsoft® Visio® and
Microsoft® Access®, ER model notations, UML, object-oriented databases, databases and electronic commerce, and
Adobe® ColdFusion®.
Answers to Selected Questions and Problems
The authors have provided answers to selected Review Questions and Problems from each chapter to help students
check their comprehension of chapter content and database skills.
Database, SQL Script, and ColdFusion Files
The Premium Website for this book includes all of the database structures and table contents used in the text. For students
using Oracle® and Microsoft SQL Server™, the SQL scripts to create and load all tables used in the SQL chapters (7 and 8)
are included. In addition, all ColdFusion scripts used to develop the Web interfaces shown Appendix J are included.
Video Tutorials
Custom-made video tutorials by Peter Rob and Carlos Coronel, exclusive to this textbook, provide clear explanations
of the essential concepts presented in the book. These unique tutorials will help the user gain a better understanding of
topics such as SQL, Oracle, ERDs, and ColdFusion.
Test Yourself on Database Systems
Brand new quizzes, created specifically for this site, allow users to test themselves on the content of each chapter and
immediately see what answers they answered right and wrong. For each question answered incorrectly, users are provided with the correct answer and the page in the text where that information is covered. Special testing software randomly compiles a selection of questions from a large database, so students can take quizzes multiple times on a given
chapter, with some new questions each time.
Microsoft PowerPoint® Slides
Direct access is offered to the book’s PowerPoint presentations that cover the key points from each chapter. These
presentations are a useful study tool.
Useful Web Links
Students can access a chapter-by-chapter repository of helpful and relevant links for further research.
Glossary of Key Terms
Students can view a PDF file of the glossary from the book. They can also search for keywords and terms in this file;
it’s quick and easy to use!
Q&A
Helpful question-and-answer documents are available for download. Here you will find supporting material in areas
such as Data Dependency/Structural Dependency and Weak Entities/Strong Relationships.
XXIII
ADDITIONAL FEATURES
INSTRUCTOR RESOURCES
Database Systems: Design, Implementation, and Management, Ninth Edition, includes teaching tools to support
instructors in the classroom. The ancillaries that accompany the textbook are listed below. Most of the teaching tools
available with this book are provided to the instructor on a single CD-ROM; they are also available on the Web at
www.cengage.com/mis/coronel.
Instructor’s Manual
The authors have created this manual to help instructors make their classes informative and interesting. Because the
authors tackle so many problems in depth, instructors will find the Instructor’s Manual especially useful. The details of
the design solution process are shown in detail in the Instructor’s Manual as well as notes about alternative
approaches that may be used to solve a particular problem. Finally, the book’s questions and problems together with
their answers and solutions are included.
SQL Script Files for Instructors
The authors have provided teacher’s SQL script files to create and delete users. They have also provided SQL scripts
to let instructors cut and paste the SQL code into the SQL windows. (Scripts are provided for Oracle as well as for MS
SQL Server.) The SQL scripts, which have all been tested by Course Technology, are a major convenience for instructors. You won’t have to type in the SQL commands and the use of the scripts eliminates errors due to “typos” that are
sometimes difficult to trace.
ColdFusion Files for Instructors
The ColdFusion Web development solutions are provided. Instructors have access to a menu-driven system that lets
teachers show the code as well as the execution of that code.
Databases
Microsoft® Access® Instructor databases are available for many chapters that include features not found in the student
databases. For example, the databases that accompany Chapters 7 and 8 include many of the queries that produce the
problem solutions. Other Access databases, such as the ones accompanying Chapters 3, 4, 5, and 6, include the
implementation of the design problem solutions to let instructors illustrate the effect of design decisions. All the MS
Access files are in the Access 2000 format so that students can use them regardless of what version they have installed
on their computers. In addition, instructors have access to all the script files for both Oracle and MS SQL Server so that
all the databases and their tables can be converted easily and precisely.
XXIV
ADDITIONAL FEATURES
Solutions
Instructors are provided with solutions to all Review Questions and Problems. Intermediate solution steps for the more
complex problems are shown to make the instructor’s job easier. Solutions may also be found on the Course
Technology Web site at www.cengage.com/mis/coronel. The solutions are password-protected.
ExamView®
This objective-based test generator lets the instructor create paper, LAN, or Web-based tests from test banks designed
specifically for this Course Technology text. Instructors can use the QuickTest Wizard to create tests in fewer than five
minutes by taking advantage of Course Technology’s question banks, or instructors can create customized exams.
PowerPoint® Presentations
Microsoft PowerPoint slides are included for each chapter. Instructors can use the slides in a variety of ways; for example, as teaching aids during classroom presentations or as printed handouts for classroom distribution. Instructors can
modify the slides provided or include slides of their own for additional topics introduced to the class.
Figure Files
Figure files for solutions presented in the Instructor’s Manual allow instructors to create their own presentations.
Instructors can also manipulate these files to meet their particular needs.
WebTutor™
Whether you want to Web-enable your class or teach entirely online, WebTutor provides customizable, rich, text-specific content that can be used with both WebCT® and BlackBoard®. WebTutor allows instructors to easily blend, add,
edit, reorganize, or delete content. Each WebTutor product provides media assets, quizzing, Web links, discussion topics, and more.
XXV
ACKNOWLEDGMENTS
Regardless of how many editions of this book are published, they will always rest on the solid foundation created by the first
edition. We remain convinced that our work has become successful because that first edition was guided by Frank Ruggirello,
a former Wadsworth senior editor and publisher. Aside from guiding the book’s development, Frank also managed to solicit
the great Peter Keen’s evaluation (thankfully favorable) and subsequently convinced PK to write the foreword for the first edition. Although we sometimes found Frank to be an especially demanding taskmaster, we also found him to be a superb professional and a fine friend. We suspect Frank will still see his fingerprints all over our current work. Many thanks.
A difficult task in rewriting a book is deciding what new approaches, topical coverage, and depth of coverage changes
can or cannot fit into a book that has successfully weathered the test of the marketplace. The comments and suggestions made by the book’s adopters, students, and reviewers play a major role in deciding what coverage is desirable and
how that coverage is to be treated.
Some adopters became extraordinary reviewers, providing incredibly detailed and well-reasoned critiques even as they
praised the book’s coverage and style. Dr. David Hatherly, a superb database professional who is a senior lecturer in the
School of Information Technology, Charles Sturt University–Mitchell, Bathhurst, Australia, made sure that we knew precisely
what issues led to his critiques. Even better for us, he provided the suggestions that made it much easier for us to improve
the topical coverage in earlier editions. Dr. Hatherly’s recommendations continue to be reflected in this ninth edition. All of
his help was given freely and without prompting on our part. His efforts are much appreciated, and our thanks are heartfelt.
We also owe a debt of gratitude to Professor Emil T. Cipolla, who teaches at St. Mary College. Professor Cipolla’s wealth
of IBM experience turned out to be a valuable resource when we tackled the embedded SQL coverage in Chapter 8.
Every technical book receives careful scrutiny by several groups of reviewers selected by the publisher. We were fortunate to face the scrutiny of reviewers who were superbly qualified to offer their critiques, comments, and suggestions—
many of which were used to strengthen this edition. While holding them blameless for any remaining shortcomings, we
owe these reviewers many thanks for their contributions:
Amita G. Chin, Virginia Commonwealth University
Samuel Conn, Regis University
Bill Hochstettler, Franklin University
Lionel M. Holguin, Jr., Athens State University
Larry Molloy, Oakland Community College
Bruce Myers, Austin Peay State University
Steven Robinett, Allegany College of Maryland
Ioulia Rytikova, George Mason University
Samuel Sambasivam, Azusa Pacific University
Kevin Scheibe, Iowa State University
Ganesan Shankaranarayanan, Boston University
Xingzhong (Frank) Shi, New Jersey Institute of Technology
Yingbing Yu, Austin Peay State Univeristy
XXVI
ACKNOWLEDGMENTS
Because this ninth edition is build solidly on the foundation of the previous editions, we would also like to thank the
following reviewers for their efforts in helping to make the previous editions successful: Dr. Reza Barkhi, Pamplin
College of Business, Virginia Polytechnic Institute and State University; Dr. Vance Cooney, Xavier University; Harpal
S. Dillion, Southwestern Oklahoma State University; Janusz Szczypula, Carnegie Mellon University; Dr. Ahmad
Abuhejleh, University of Wisconsin, River Falls; Dr. Terence M. Baron, University of Toledo; Dr. Juan Estava, Eastern
Michigan University; Dr. Kevin Gorman, University of North Carolina, Charlotte; Dr. Jeff Hedrington, University of
Wisconsin, Eau Claire; Dr. Herman P. Hoplin, Syracuse University; Dr. Sophie Lee, University of Massachusetts,
Boston; Dr. Michael Mannino, University of Washington; Dr. Carol Chrisman, Illinois State University; Dr. Timothy
Heintz, Marquette University; Dr. Herman Hoplin, Syracuse University; Dr. Dean James, Embry-Riddle University;
Dr. Constance Knapp, Pace University; Dr. Mary Ann Robbert, Bentley College; Dr. Francis J. Van Wetering,
University of Nebraska; Dr. Joseph Walls, University of Southern California; Dr. Stephen C. Solosky, Nassau
Community College; Dr. Robert Chiang, Syracuse University; Dr. Crist Costa, Rhode Island College; Dr. Sudesh M.
Duggal, Northern Kentucky University; Dr. Chang Koh, University of North Carolina, Greensboro; Paul A. Seibert,
North Greenville College; Neil Dunlop, Vista Community College; Ylber Ramadani, George Brown College; Samuel
Sambasivam, Azusa Pacific University; Arjan Sadhwani, San Jose State University; Genard Catalano, Columbia
College; Craig Shaw, Central Community College; Lei-da Chen, Creighton University; Linda K. Lau, Longwood
University; Anita Lee-Post, University of Kentucky; Lenore Horowitz, Schenectady County Community College;
Dr. Scott L. Schneberger, Georgia State University; Tony Pollard, University of Western Sydney; Lejla Vrazalic,
University of Wollongong; and David Witzany, Parkland College.
In some respects, writing books resembles building construction: When 90 percent of the work seems done, 90 percent of the work remains to be done. Fortunately for us, we had a great team on our side.
• How can we possibly pay sufficient homage to Deb Kaufmann’s many contributions? Even our best
superlatives don’t begin to paint a proper picture of our professional relationship with Deb
Kaufmann, our developmental editor since the fifth edition. Deb has that magic combination of good
judgment, intelligence, technical skill, and the rare ability to organize and sharpen an author’s writing without affecting its intent or its flow. And she does it all with style, class, and humor. She is the
best of the best.
• After writing so many books and eight editions of this book, we know just how difficult it can be to
transform the authors’ work into an attractive book. The production team, both at Course
Technology (Matt Hutchinson) and GEX Publishing Services (Louise Capulli and Marisa Taylor),
have done an excellent job.
• We also owe Kate Mason, our product manager, special thanks for her ability to guide this book to a
successful conclusion. Kate’s work touched all of the publication bases, and her managerial skills protected us from those publishing gremlins that might have become a major nuisance. Not to mention the
fact that her skills in dealing with occasionally cranky authors far exceed those of any diplomat we can
think of. And did we mention that Kate is, quite simply, a delightful person?
• Many thanks to Andrea Schein, our copyeditor. Given her ability to spot even the smallest discrepancies, we suspect that her middle name is “Thorough.” We can only imagine the level of mental discipline required to perform her job and we salute her.
XXVII
ACKNOWLEDGMENTS
We also thank our students for their comments and suggestions. They are the reason for writing this book in the first
place. One comment stands out in particular: “I majored in systems for four years, and I finally discovered why when I
took your course.” And one of our favorite comments by a former student was triggered by a question about the challenges created by a real-world information systems job: “Doc, it’s just like class, only easier. You really prepared me
well. Thanks!”
Last, and certainly not least, we thank our families for the solid home support. They graciously accepted the fact that
during more than a year’s worth of rewriting, there would be no free weekends, rare free nights, and even rarer free
days. We owe you much, and the dedication we wrote to you is but a small reflection of the important space you
occupy in our hearts.
Carlos Coronel, Steven Morris, and Peter Rob
XXVIII
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PART
I
DATABASE CONCEPTS
Database Systems
1
Data Models
2
The Relational Revolution
Today, we take for granted the benefits brought to us by relational databases: the ability
to store, access, and change data quickly and easily on low-cost computers.Yet, until the
late 1970s, databases stored large amounts of data in a hierarchical structure that was
difficult to navigate and inflexible. Programmers needed to know what clients wanted to
do with the data before the database was designed. Adding or changing the way the data
was analyzed was a time-consuming and expensive process. As a result, you searched
through huge card catalogs to find a library book, you used road maps that didn’t show
changes made in the last year, and you had to buy a newspaper to find information on
stock prices.
In 1970, Edgar “Ted” Codd, a mathematician employed by IBM, wrote an article that
would change all that. At the time, nobody realized that Codd’s obscure theories would
spark a technological revolution on par with the development of personal computers and
the Internet. Don Chamberlin, coinventor of SQL, the most popular computer language
used by database systems today, explains, “There was this guy Ted Codd who had some
kind of strange mathematical notation, but nobody took it very seriously.” Then Ted Codd
organized a symposium, and Chamberlin listened as Codd reduced complicated five-page
programs to one line. “And I said, ‘Wow,’” Chamberlin recalls.
The symposium convinced IBM to fund System R, a research project that built a
prototype of a relational database and that would eventually lead to the creation of SQL
and DB2. IBM, however, kept System R on the back burner for a number of crucial years.
The company had a vested interest in IMS, a reliable, high-end database system that had
come out in 1968. Unaware of the market potential of this research, IBM allowed its staff
to publish these papers publicly.
Among those reading these papers was Larry Ellison, who had just founded a small
company. Recruiting programmers from System R and the University of California, Ellison
was able to market the first SQL-based relational database in 1979, well before IBM. By
1983, the company had released a portable version of the database, grossed over
$5,000,000 annually, and changed its name to Oracle. Spurred on by competition, IBM
finally released SQL/DS, its first relational database, in 1980.
IBM has yet to catch up. By 2007, global sales of relational database management systems
rose to $18.8 billion. Oracle captured 48.6% of the market share, more than its two
closest competitors, IBM and Microsoft, combined.
B
V
usiness
ignette
O N E
1
Database Systems
In this chapter, you will learn:
쐍 The difference between data and information
쐍 What a database is, the various types of databases, and why they are valuable assets for
decision making
쐍 The importance of database design
쐍 How modern databases evolved from file systems
쐍 About flaws in file system data management
쐍 The main components of the database system
쐍 The main functions of a database management system (DBMS)
Good decisions require good information that is derived from raw facts.These raw facts are
known as data. Data are likely to be managed most efficiently when they are stored in a
database. In this chapter, you will learn what a database is, what it does, and why it yields
better results than other data management methods.You will also learn about various types
of databases and why database design is so important.
Databases evolved from computer file systems. Although file system data management is
now largely outmoded, understanding the characteristics of file systems is important
because file systems are the source of serious data management limitations. In this chapter,
you will also learn how the database system approach helps eliminate most of the
shortcomings of file system data management.
P
review
D A T A B A S E
S Y S T E M S
1.1 WHY DATABASES?
Imagine trying to operate a business without knowing who your customers are, what products you are selling, who is
working for you, who owes you money, and whom you owe money. All businesses have to keep this type of data and
much more; and just as importantly, they must have those data available to decision makers when they need them. It
can be argued that the ultimate purpose of all business information systems is to help businesses use information as
an organizational resource. At the heart of all of these systems are the collection, storage, aggregation, manipulation,
dissemination, and management of data.
Depending on the type of information system and the characteristics of the business, these data could vary from a few
megabytes on just one or two topics to terabytes covering hundreds of topics within the business’s internal and external
environment. Telecommunications companies such as Sprint and AT&T are known to have systems that keep data on
trillions of phone calls, with new data being added to the system at speeds up to 70,000 calls per second!1 Not only
do these companies have to store and manage these immense collections of data, they have to be able to find any given
fact in that data quickly. Consider the case of Internet search staple Google. While Google is reluctant to disclose many
details about its data storage specifications, it is estimated that the company responds to over 91 million searches per
day across a collection of data that is several terabytes in size. Impressively, the results of these searches are available
nearly instantly.
How can these businesses process this much data? How can they store it all, and then quickly retrieve just the facts
that decision makers want to know, just when they want to know it? The answer is that they use databases. Databases,
as explained in detail throughout this book, are specialized structures that allow computer-based systems to store,
manage, and retrieve data very quickly. Virtually all modern business systems rely on databases; therefore, a good
understanding of how these structures are created and their proper use is vital for any information systems
professional. Even if your career does not take you down the amazing path of database design and development,
databases will be a key component underpinning the systems that you work with. In any case, it is very likely that, in
your career, you will be making decisions based on information generated from data. Thus, it is important that you
know the difference between data and information.
1.2 DATA VS. INFORMATION
To understand what drives database design, you must understand the difference between data and information. Data
are raw facts. The word raw indicates that the facts have not yet been processed to reveal their meaning. For example,
suppose that you want to know what the users of a computer lab think of its services. Typically, you would begin by
surveying users to assess the computer lab’s performance. Figure 1.1, Panel A, shows the Web survey form that
enables users to respond to your questions. When the survey form has been completed, the form’s raw data are saved
to a data repository, such as the one shown in Figure 1.1, Panel B. Although you now have the facts in hand, they
are not particularly useful in this format—reading page after page of zeros and ones is not likely to provide much
insight. Therefore, you transform the raw data into a data summary like the one shown in Figure 1.1, Panel C. Now
it’s possible to get quick answers to questions such as “What is the composition of our lab’s customer base?” In this
case, you can quickly determine that most of your customers are juniors (24.59%) and seniors (53.01%). Because
graphics can enhance your ability to quickly extract meaning from data, you show the data summary bar graph in
Figure 1.1, Panel D.
Information is the result of processing raw data to reveal its meaning. Data processing can be as simple as organizing
data to reveal patterns or as complex as making forecasts or drawing inferences using statistical modeling. To reveal
meaning, information requires context. For example, an average temperature reading of 105 degrees does not mean
1“Top Ten Largest Databases in the World,” Business Intelligence Lowdown, February 15, 2007, http://www.businessintelligencelowdown.com/
2007/02/top_10_largest_.html
5
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C H A P T E R
FIGURE
1
Transforming raw data into information
1.1
a) Initial Survey Screen
c) Information in Summary Format
b) Raw Data
d) Information in Graphic Format
much unless you also know its context: Is this in degrees Fahrenheit or Celsius? Is this a machine temperature, a body
temperature, or an outside air temperature? Information can be used as the foundation for decision making. For
example, the data summary for each question on the survey form can point out the lab’s strengths and weaknesses,
helping you to make informed decisions to better meet the needs of lab customers.
Keep in mind that raw data must be properly formatted for storage, processing, and presentation. For example, in
Panel C of Figure 1.1, the student classification is formatted to show the results based on the classifications Freshman,
Sophomore, Junior, Senior, and Graduate Student. The respondents’ yes/no responses might need to be converted
to a Y/N format for data storage. More complex formatting is required when working with complex data types, such
as sounds, videos, or images.
In this “information age,” production of accurate, relevant, and timely information is the key to good decision making.
In turn, good decision making is the key to business survival in a global market. We are now said to be entering the
D A T A B A S E
S Y S T E M S
“knowledge age.”2 Data are the foundation of information, which is the bedrock of knowledge—that is, the body of
information and facts about a specific subject. Knowledge implies familiarity, awareness, and understanding of
information as it applies to an environment. A key characteristic of knowledge is that “new” knowledge can be derived
from “old” knowledge.
Let’s summarize some key points:
쐌
Data constitute the building blocks of information.
쐌
Information is produced by processing data.
쐌
Information is used to reveal the meaning of data.
쐌
Accurate, relevant, and timely information is the key to good decision making.
쐌
Good decision making is the key to organizational survival in a global environment.
Timely and useful information requires accurate data. Such data must be properly generated and stored in a format
that is easy to access and process. And, like any basic resource, the data environment must be managed carefully. Data
management is a discipline that focuses on the proper generation, storage, and retrieval of data. Given the crucial
role that data play, it should not surprise you that data management is a core activity for any business, government
agency, service organization, or charity.
1.3 INTRODUCING THE DATABASE
Efficient data management typically requires the use of a computer database. A database is a shared, integrated
computer structure that stores a collection of:
쐌
End-user data, that is, raw facts of interest to the end user.
쐌
Metadata, or data about data, through which the end-user data are integrated and managed.
The metadata provide a description of the data characteristics and the set of relationships that links the data found
within the database. For example, the metadata component stores information such as the name of each data element,
the type of values (numeric, dates, or text) stored on each data element, whether or not the data element can be left
empty, and so on. The metadata provide information that complements and expands the value and use of the data.
In short, metadata present a more complete picture of the data in the database. Given the characteristics of metadata,
you might hear a database described as a “collection of self-describing data.”
A database management system (DBMS) is a collection of programs that manages the database structure and
controls access to the data stored in the database. In a sense, a database resembles a very well-organized electronic
filing cabinet in which powerful software, known as a database management system, helps manage the cabinet’s
contents.
1.3.1 Role and Advantages of the DBMS
The DBMS serves as the intermediary between the user and the database. The database structure itself is stored as a
collection of files, and the only way to access the data in those files is through the DBMS. Figure 1.2 emphasizes the
point that the DBMS presents the end user (or application program) with a single, integrated view of the data in the
database. The DBMS receives all application requests and translates them into the complex operations required to fulfill
those requests. The DBMS hides much of the database’s internal complexity from the application programs and users.
The application program might be written by a programmer using a programming language such as Visual Basic.NET,
Java, or C#, or it might be created through a DBMS utility program.
2Peter Drucker coined the phrase “knowledge worker” in 1959 in his book Landmarks of Tomorrow. In 1994, Ms. Esther Dyson, Mr. George
Keyworth, and Dr. Alvin Toffler introduced the concept of the “knowledge age.”
7
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C H A P T E R
1
FIGURE
The DBMS manages the interaction between the end user and the database
1.2
End users
Database structure
Data
Metadata
Application
request
Customers
DBMS
Database
Management System
http://
Single
View of Data
Invoices
Integrated
End-user
data
End users
Application
request
Products
Data
Having a DBMS between the end user’s applications and the database offers some important advantages. First, the
DBMS enables the data in the database to be shared among multiple applications or users. Second, the DBMS
integrates the many different users’ views of the data into a single all-encompassing data repository.
Because data are the crucial raw material from which information is derived, you must have a good method to manage
such data. As you will discover in this book, the DBMS helps make data management more efficient and effective. In
particular, a DBMS provides advantages such as:
쐌
Improved data sharing. The DBMS helps create an environment in which end users have better access to
more and better-managed data. Such access makes it possible for end users to respond quickly to changes in
their environment.
쐌
Improved data security. The more users access the data, the greater the risks of data security breaches.
Corporations invest considerable amounts of time, effort, and money to ensure that corporate data are used
properly. A DBMS provides a framework for better enforcement of data privacy and security policies.
쐌
Better data integration. Wider access to well-managed data promotes an integrated view of the organization’s
operations and a clearer view of the big picture. It becomes much easier to see how actions in one segment
of the company affect other segments.
쐌
Minimized data inconsistency. Data inconsistency exists when different versions of the same data appear
in different places. For example, data inconsistency exists when a company’s sales department stores a sales
representative’s name as “Bill Brown” and the company’s personnel department stores that same person’s
name as “William G. Brown,” or when the company’s regional sales office shows the price of a product as
$45.95 and its national sales office shows the same product’s price as $43.95. The probability of data
inconsistency is greatly reduced in a properly designed database.
쐌
Improved data access. The DBMS makes it possible to produce quick answers to ad hoc queries. From a
database perspective, a query is a specific request issued to the DBMS for data manipulation—for example,
to read or update the data. Simply put, a query is a question, and an ad hoc query is a spur-of-the-moment
question. The DBMS sends back an answer (called the query result set) to the application. For example, end
D A T A B A S E
S Y S T E M S
users, when dealing with large amounts of sales data, might want quick answers to questions (ad hoc queries)
such as:
-
What was the dollar volume of sales by product during the past six months?
-
What is the sales bonus figure for each of our salespeople during the past three months?
-
How many of our customers have credit balances of $3,000 or more?
쐌
Improved decision making. Better-managed data and improved data access make it possible to generate
better-quality information, on which better decisions are based. The quality of the information generated
depends on the quality of the underlying data. Data quality is a comprehensive approach to promoting the
accuracy, validity, and timeliness of the data. While the DBMS does not guarantee data quality, it provides a
framework to facilitate data quality initiatives. Data quality concepts will be covered in more detail in Chapter
15, Database Administration and Security.
쐌
Increased end-user productivity. The availability of data, combined with the tools that transform data into
usable information, empowers end users to make quick, informed decisions that can make the difference
between success and failure in the global economy.
The advantages of using a DBMS are not limited to the few just listed. In fact, you will discover many more advantages
as you learn more about the technical details of databases and their proper design.
1.3.2 Types of Databases
A DBMS can support many different types of databases. Databases can be classified according to the number of users,
the database location(s), and the expected type and extent of use.
The number of users determines whether the database is classified as single-user or multiuser. A single-user
database supports only one user at a time. In other words, if user A is using the database, users B and C must wait
until user A is done. A single-user database that runs on a personal computer is called a desktop database. In
contrast, a multiuser database supports multiple users at the same time. When the multiuser database supports a
relatively small number of users (usually fewer than 50) or a specific department within an organization, it is called a
workgroup database. When the database is used by the entire organization and supports many users (more than 50,
usually hundreds) across many departments, the database is known as an enterprise database.
Location might also be used to classify the database. For example, a database that supports data located at a single
site is called a centralized database. A database that supports data distributed across several different sites is called
a distributed database. The extent to which a database can be distributed and the way in which such distribution
is managed are addressed in detail in Chapter 12, Distributed Database Management Systems.
The most popular way of classifying databases today, however, is based on how they will be used and on the time
sensitivity of the information gathered from them. For example, transactions such as product or service sales,
payments, and supply purchases reflect critical day-to-day operations. Such transactions must be recorded accurately
and immediately. A database that is designed primarily to support a company’s day-to-day operations is classified as
an operational database (sometimes referred to as a transactional or production database). In contrast, a data
warehouse focuses primarily on storing data used to generate information required to make tactical or strategic
decisions. Such decisions typically require extensive “data massaging” (data manipulation) to extract information to
formulate pricing decisions, sales forecasts, market positioning, and so on. Most decision support data are based on
data obtained from operational databases over time and stored in data warehouses. Additionally, the data warehouse
can store data derived from many sources. To make it easier to retrieve such data, the data warehouse structure is quite
different from that of an operational or transactional database. The design, implementation, and use of data
warehouses are covered in detail in Chapter 13, Business Intelligence and Data Warehouses.
Databases can also be classified to reflect the degree to which the data are structured. Unstructured data are data
that exist in their original (raw) state, that is, in the format in which they were collected. Therefore, unstructured data
exist in a format that does not lend itself to the processing that yields information. Structured data are the result of
9
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1
taking unstructured data and formatting (structuring) such data to facilitate storage, use, and the generation of
information. You apply structure (format) based on the type of processing that you intend to perform on the data.
Some data might not be ready (unstructured) for some types of processing, but they might be ready (structured) for
other types of processing. For example, the data value 37890 might refer to a zip code, a sales value, or a product
code. If this value represents a zip code or a product code and is stored as text, you cannot perform mathematical
computations with it. On the other hand, if this value represents a sales transaction, it is necessary to format it as
numeric.
To further illustrate the structure concept, imagine a stack of printed paper invoices. If you want to merely store these
invoices as images for future retrieval and display, you can scan them and save them in a graphic format. On the other
hand, if you want to derive information such as monthly totals and average sales, such graphic storage would not be
useful. Instead, you could store the invoice data in a (structured) spreadsheet format so that you can perform the
requisite computations. Actually, most data you encounter are best classified as semistructured. Semistructured data
are data that have already been processed to some extent. For example, if you look at a typical Web page, the data
are presented to you in a prearranged format to convey some information.
The database types mentioned thus far focus on the storage and management of highly structured data. However,
corporations are not limited to the use of structured data. They also use semistructured and unstructured data. Just
think of the very valuable information that can be found on company e-mails, memos, documents such as procedures
and rules, Web pages, and so on. Unstructured and semistructured data storage and management needs are being
addressed through a new generation of databases known as XML databases. Extensible Markup Language (XML)
is a special language used to represent and manipulate data elements in a textual format. An XML database supports
the storage and management of semistructured XML data.
Table 1.1 compares the features of several well-known database management systems.
TABLE
Types of Databases
1.1
NUMBER OF USERS
DATA LOCATION
PRODUCT
SINGLE
USER
WORKGROUP
MS Access
X
X
MS SQL
Server
X3
X
X
IBM DB2
X3
X
MySQL
X
X
Oracle
RDBMS
X3
X
DATA USAGE
MULTIUSER
ENTERPRISE
CENTRALIZED
DISTRIBUTED
X
OPERATIONAL
XML
DATA
WAREHOUSE
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X*
X
X
X
X
* Supports XML functions only. XML data are stored in large text objects.
1.4 WHY DATABASE DESIGN IS IMPORTANT
Database design refers to the activities that focus on the design of the database structure that will be used to store
and manage end-user data. A database that meets all user requirements does not just happen; its structure must be
designed carefully. In fact, database design is such a crucial aspect of working with databases that most of this book
is dedicated to the development of good database design techniques. Even a good DBMS will perform poorly with a
badly designed database.
Proper database design requires the designer to identify precisely the database’s expected use. Designing a
transactional database emphasizes accurate and consistent data and operational speed. Designing a data warehouse
database emphasizes the use of historical and aggregated data. Designing a database to be used in a centralized,
3Vendor offers single-user/personal DBMS version.
D A T A B A S E
S Y S T E M S
single-user environment requires a different approach from that used in the design of a distributed, multiuser database.
This book emphasizes the design of transactional, centralized, single-user, and multiuser databases. Chapters 12 and
13 also examine critical issues confronting the designer of distributed and data warehouse databases.
Designing appropriate data repositories of integrated information using the two-dimensional table structures found in
most databases is a process of decomposition. The integrated data must be decomposed properly into its constituent
parts, with each part stored in its own table. Further, the relationships between these tables must be carefully
considered and implemented so that the integrated view of the data can be re-created later as information for the end
user. A well-designed database facilitates data management and generates accurate and valuable information. A poorly
designed database is likely to become a breeding ground for difficult-to-trace errors that may lead to bad decision
making—and bad decision making can lead to the failure of an organization. Database design is simply too important
to be left to luck. That’s why college students study database design, why organizations of all types and sizes send
personnel to database design seminars, and why database design consultants often make an excellent living.
1.5 EVOLUTION OF FILE SYSTEM DATA PROCESSING
Understanding what a database is, what it does, and the proper way to use it can be clarified by considering what a
database is not. A brief explanation of the evolution of file system data processing can be helpful in understanding the
data access limitations that databases attempt to overcome. Understanding these limitations is relevant to database
designers and developers because database technologies do not make these problems magically disappear—database
technologies simply make it easier to create solutions that avoid these problems. Creating database designs that avoid
the pitfalls of earlier systems requires that the designer understand what the problems of the earlier systems were and
how to avoid them, or else the database technologies are no better (potentially even worse!) than the technologies and
techniques that they have replaced.
1.5.1 Manual File Systems
In order to be successful, an organization must come up with systems for handling core business tasks. Historically,
such systems were often manual, paper-and-pencil systems. The papers within these systems were organized in order
to facilitate the expected use of the data. Typically, this was accomplished through a system of file folders and filing
cabinets. As long as a data collection was relatively small and an organization’s business users had few reporting
requirements, the manual system served its role well as a data repository. However, as organizations grew and as
reporting requirements became more complex, keeping track of data in a manual file system became more difficult.
Therefore, companies looked to computer technology for help.
1.5.2 Computerized File Systems
Generating reports from manual file systems was slow and cumbersome. In fact, some business managers faced
government-imposed reporting requirements that required weeks of intensive effort each quarter, even when a
well-designed manual system was used. Therefore, a data processing (DP) specialist was hired to create a
computer-based system that would track data and produce required reports.
Initially, the computer files within the file system were similar to the manual files. A simple example of a customer data file
for a small insurance company is shown in Figure 1.3. (You will discover later that the file structure shown in Figure 1.3,
although typically found in early file systems, is unsatisfactory for a database.)
The description of computer files requires a specialized vocabulary. Every discipline develops its own jargon to enable
its practitioners to communicate clearly. The basic file vocabulary shown in Table 1.2 will help you to understand
subsequent discussions more easily.
11
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C H A P T E R
FIGURE
1
Contents of the CUSTOMER file
1.3
C_NAME
C_PHONE
C_ADDRESS
C_ZIP
= Customer name
= Customer phone
= Customer address
= Customer zip code
A_NAME
A_PHONE
TP
AMT
REN
= Agent name
= Agent phone
= Insurance type
= Insurance policy amount, in thousands of $
= Insurance renewal date
Online Content
The databases used in each chapter are available in the Premium Website for this book. Throughout the book,
Online Content boxes highlight material related to chapter content located in the Premium Website.
TABLE
1.2
TERM
Data
Field
Record
File
Basic File Terminology
DEFINITION
“Raw” facts, such as a telephone number, a birth date, a customer name, and a year-to-date (YTD)
sales value. Data have little meaning unless they have been organized in some logical manner.
A character or group of characters (alphabetic or numeric) that has a specific meaning. A field is
used to define and store data.
A logically connected set of one or more fields that describes a person, place, or thing. For example,
the fields that constitute a record for a customer might consist of the customer’s name, address,
phone number, date of birth, credit limit, and unpaid balance.
A collection of related records. For example, a file might contain data about the students currently
enrolled at Gigantic University.
Using the proper file terminology given in Table 1.2, you can identify the file components shown in Figure 1.3. The
CUSTOMER file shown in Figure 1.3 contains 10 records. Each record is composed of nine fields: C_NAME,
C_PHONE, C_ADDRESS, C_ZlP, A_NAME, A_PHONE, TP, AMT, and REN. The 10 records are stored in a named
file. Because the file in Figure 1.3 contains customer data for the insurance company, its filename is CUSTOMER.
When business users wanted data from the computerized file, they sent requests for the data to the DP specialist. For each
request, the DP specialist had to create programs to retrieve the data from the file, manipulate it in whatever manner the
user had requested, and present it as a printed report. If a request was for a report that had been previously run, the DP
specialist could rerun the existing program and provide the printed results to the user. As other business users saw the
new and innovative ways that the customer data were being reported, they wanted to be able to view their data in similar
fashions. This generated more requests for the DP specialist to create more computerized files of other business data,
which in turn meant that more data management programs had to be created, and more requests for reports. For
example, the sales department at the insurance company created a file named SALES, which helped track daily sales
D A T A B A S E
S Y S T E M S
efforts. The sales department’s success was so obvious that the personnel department manager demanded access to the
DP specialist to automate payroll processing and other personnel functions. Consequently, the DP specialist was asked
to create the AGENT file shown in Figure 1.4. The data in the AGENT file were used to write checks, keep track of taxes
paid, and summarize insurance coverage, among other tasks.
FIGURE
Contents of the AGENT file
1.4
A_NAME
A_PHONE
A_ADDRESS
ZIP
HIRED
= Agent name
= Agent phone
= Agent address
= Agent zip code
= Agent date of hire
YTD_PAY
YTD_FIT
YTD_FICA
YTD_SLS
DEP
= Year-to-date pay
= Year-to-date federal income tax paid
= Year-to-date Social Security taxes paid
= Year-to-date sales
= Number of dependents
As more and more computerized files were developed, the problems with this type of file system became apparent.
While these problems are explored in detail in the next section, briefly, the problems centered on having lots of data
files that contained related, often overlapping, data with no means of controlling or managing the data consistently
across all of the files. As shown in Figure 1.5, each file in the system used its own application program to store,
retrieve, and modify data. And each file was owned by the individual or the department that commissioned its creation.
FIGURE
A simple file system
1.5
Sales department
Personnel department
SALES
file
File
Report
Programs
File
Management
Programs
CUSTOMER
file
File
Report
Programs
File
Management
Programs
AGENT
file
The advent of computer files to store company data was significant; it not only established a landmark in the use of
computer technologies but also represented a huge step forward in a business’s ability to process data. Previously, users
had direct, hands-on access to all of the business data. But they didn’t have the tools to convert those data into the
information that they needed. The creation of computerized file systems gave them improved tools for manipulating
the company data that allowed them to create new information. However, it had the additional effect of introducing
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1
a schism between the end users and their data. The desire to close the gap between the end users and the data
influenced the development of all types of computer technologies, system designs, and uses (and misuse) of many
technologies and techniques. However, such developments also created a split between the ways DP specialists and
end users viewed the data.
쐌
From the DP specialist’s perspective, the computer files within the file system were created to be similar to the
manual files. Data management programs were created to add to, update, and delete data from the file.
쐌
From the end user’s perspective, the systems separated the users from the data. As the users’ competitive
environment pushed them to make more and more decisions in less and less time, the delay from when the
users conceived of a new way to create information from the data to when the DP specialist could create the
programs to generate that information was a source of great frustration.
1.5.3 File System Redux: Modern End-User Productivity Tools
The users’ desire for direct, hands-on access to the data helped to fuel the adoption of personal computers for business
use. Although not directly related to file system evolution, the ubiquitous use of personal productivity tools can
introduce the same problems as the old file systems.
Personal computer spreadsheet programs such as Microsoft Excel are widely used by business users, and allow the user
to enter data in a series of rows and columns so that the data can be manipulated using a wide range of functions. The
popularity of spreadsheet applications has enabled users to conduct sophisticated analysis of data that has greatly
enhanced their ability to understand the data and make better decisions. Unfortunately, as in the old adage “When the
only tool you have is a hammer, every problem looks like a nail,” users have become so adept at working with
spreadsheets, they tend to use them to complete tasks for which spreadsheets are not appropriate.
One of the common misuses of spreadsheets is as a substitute for a database. Interestingly, end users often take the
limited data to which they have direct access and place it in a spreadsheet in a format similar to that of the traditional,
manual data storage systems—which is precisely what the early DP specialists did when creating computerized data
files. Due to the large number of users with spreadsheets, each making separate copies of the data, the resulting “file
system” of spreadsheets suffers from the same problems as the file systems created by the early DP specialists, which
are outlined in the next section.
1.6 PROBLEMS WITH FILE SYSTEM DATA PROCESSING
The file system method of organizing and managing data was a definite improvement over the manual system, and the
file system served a useful purpose in data management for over two decades—a very long time in the computer era.
Nonetheless, many problems and limitations became evident in this approach. A critique of the file system method
serves two major purposes:
쐌
Understanding the shortcomings of the file system enables you to understand the development of modern
databases.
쐌
Many of the problems are not unique to file systems. Failure to understand such problems is likely to lead to
their duplication in a database environment, even though database technology makes it easy to avoid them.
The following problems associated with file systems, whether created by DP specialists or through a series of
spreadsheets, severely challenge the types of information that can be created from the data as well as the accuracy of
the information:
쐌
Lengthy development times. The first and most glaring problem with the file system approach is that even
the simplest data-retrieval task requires extensive programming. With the older file systems, programmers had
to specify what must be done and how it was to be done. As you will learn in upcoming chapters, modern
databases use a nonprocedural data manipulation language that allows the user to specify what must be done
without specifying how it must be done.
D A T A B A S E
S Y S T E M S
쐌
Difficulty of getting quick answers. The need to write programs to produce even the simplest reports makes
ad hoc queries impossible. Harried DP specialists who work with mature file systems often receive numerous
requests for new reports. They are often forced to say that the report will be ready “next week” or even “next
month.” If you need the information now, getting it next week or next month will not serve your information
needs.
쐌
Complex system administration. System administration becomes more difficult as the number of files in the
system expands. Even a simple file system with a few files requires creating and maintaining several file
management programs (each file must have its own file management programs that allow the user to add,
modify, and delete records; to list the file contents; and to generate reports). Because ad hoc queries are not
possible, the file reporting programs can multiply quickly. The problem is compounded by the fact that each
department in the organization “owns” its data by creating its own files.
쐌
Lack of security and limited data sharing. Another fault of a file system data repository is a lack of security
and limited data sharing. Data sharing and security are closely related. Sharing data among multiple
geographically dispersed users introduces a lot of security risks. In terms of spreadsheet data, while many
spreadsheet programs provide rudimentary security options, they are not always used, and even when they are
used, they are insufficient for robust data sharing among users. In terms of the creation of data management
and reporting programs, security and data-sharing features are difficult to program and are, therefore, often
omitted in a file system environment. Such features include effective password protection, the ability to lock out
parts of files or parts of the system itself, and other measures designed to safeguard data confidentiality. Even
when an attempt is made to improve system and data security, the security devices tend to be limited in scope
and effectiveness.
쐌
Extensive programming. Making changes to an existing file structure can be difficult in a file system environment. For example, changing just one field in the original CUSTOMER file would require a program that:
1.
Reads a record from the original file.
2.
Transforms the original data to conform to the new structure’s storage requirements.
3.
Writes the transformed data into the new file structure.
4.
Repeats steps 2 to 4 for each record in the original file.
In fact, any change to a file structure, no matter how minor, forces modifications in all of the programs that use the
data in that file. Modifications are likely to produce errors (bugs), and additional time is spent using a debugging process
to find those errors. Those limitations, in turn, lead to problems of structural and data dependence.
1.6.1 Structural and Data Dependence
A file system exhibits structural dependence, which means that access to a file is dependent on its structure. For
example, adding a customer date-of-birth field to the CUSTOMER file shown in Figure 1.3 would require the four steps
described in the previous section. Given this change, none of the previous programs will work with the new
CUSTOMER file structure. Therefore, all of the file system programs must be modified to conform to the new file
structure. In short, because the file system application programs are affected by change in the file structure, they exhibit
structural dependence. Conversely, structural independence exists when it is possible to make changes in the file
structure without affecting the application program’s ability to access the data.
Even changes in the characteristics of data, such as changing a field from integer to decimal, require changes in all the
programs that access the file. Because all data access programs are subject to change when any of the file’s data
storage characteristics change (that is, changing the data type), the file system is said to exhibit data dependence.
Conversely, data independence exists when it is possible to make changes in the data storage characteristics without
affecting the application program’s ability to access the data.
The practical significance of data dependence is the difference between the logical data format (how the human
being views the data) and the physical data format (how the computer must work with the data). Any program that
accesses a file system’s file must tell the computer not only what to do but also how to do it. Consequently, each
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program must contain lines that specify the opening of a specific file type, its record specification, and its field
definitions. Data dependence makes the file system extremely cumbersome from the point of view of a programmer
and database manager.
1.6.2 Data Redundancy
The file system’s structure makes it difficult to combine data from multiple sources, and its lack of security renders the
file system vulnerable to security breaches. The organizational structure promotes the storage of the same basic data
in different locations. (Database professionals use the term islands of information for such scattered data locations.)
The dispersion of data is exacerbated by the use of spreadsheets to store data. In a file system, the entire sales
department would share access to the SALES data file through the data management and reporting programs created
by the DP specialist. With the use of spreadsheets, it is possible for each member of the sales department to create
his or her own copy of the sales data. Because it is unlikely that data stored in different locations will always be updated
consistently, the islands of information often contain different versions of the same data. For example, in Figures 1.3
and 1.4, the agent names and phone numbers occur in both the CUSTOMER and the AGENT files. You only need
one correct copy of the agent names and phone numbers. Having them occur in more than one place produces data
redundancy. Data redundancy exists when the same data are stored unnecessarily at different places.
Uncontrolled data redundancy sets the stage for:
쐌
Poor data security. Having multiple copies of data increases the chances for a copy of the data to be
susceptible to unauthorized access. Chapter 15, Database Administration and Security, explores the issues and
techniques associated with securing data.
쐌
Data inconsistency. Data inconsistency exists when different and conflicting versions of the same data appear
in different places. For example, suppose you change an agent’s phone number or address in the AGENT file.
If you forget to make corresponding changes in the CUSTOMER file, the files contain different data for the
same agent. Reports will yield inconsistent results that depend on which version of the data is used.
Note
Data that display data inconsistency are also referred to as data that lack data integrity. Data integrity is defined
as the condition in which all of the data in the database are consistent with the real-world events and conditions.
In other words, data integrity means that:
• Data are accurate—there are no data inconsistencies.
• Data are verifiable—the data will always yield consistent results.
Data entry errors are more likely to occur when complex entries (such as 10-digit phone numbers) are made
in several different files and/or recur frequently in one or more files. In fact, the CUSTOMER file shown in
Figure 1.3 contains just such an entry error: the third record in the CUSTOMER file has a transposed digit in
the agent’s phone number (615-882-2144 rather than 615-882-1244).
It is possible to enter a nonexistent sales agent’s name and phone number into the CUSTOMER file, but
customers are not likely to be impressed if the insurance agency supplies the name and phone number of an
agent who does not exist. Should the personnel manager allow a nonexistent agent to accrue bonuses and
benefits? In fact, a data entry error such as an incorrectly spelled name or an incorrect phone number yields
the same kind of data integrity problems.
쐌
Data anomalies. The dictionary defines anomaly as “an abnormality.” Ideally, a field value change should be
made in only a single place. Data redundancy, however, fosters an abnormal condition by forcing field value
changes in many different locations. Look at the CUSTOMER file in Figure 1.3. If agent Leah F. Hahn decides
to get married and move, the agent name, address, and phone number are likely to change. Instead of making
just a single name and/or phone/address change in a single file (AGENT), you must also make the change each
time that agent’s name, phone number, and address occur in the CUSTOMER file. You could be faced with
D A T A B A S E
S Y S T E M S
the prospect of making hundreds of corrections, one for each of the customers served by that agent! The same
problem occurs when an agent decides to quit. Each customer served by that agent must be assigned a new
agent. Any change in any field value must be correctly made in many places to maintain data integrity. A data
anomaly develops when not all of the required changes in the redundant data are made successfully. The data
anomalies found in Figure 1.3 are commonly defined as follows:
-
Update anomalies. If agent Leah F. Hahn has a new phone number, that number must be entered in each
of the CUSTOMER file records in which Ms. Hahn’s phone number is shown. In this case, only three
changes must be made. In a large file system, such a change might occur in hundreds or even thousands
of records. Clearly, the potential for data inconsistencies is great.
-
Insertion anomalies. If only the CUSTOMER file existed, to add a new agent, you would also add a dummy
customer data entry to reflect the new agent’s addition. Again, the potential for creating data inconsistencies
would be great.
-
Deletion anomalies. If you delete the customers Amy B. O’Brian, George Williams, and Olette K. Smith,
you will also delete John T. Okon’s agent data. Clearly, this is not desirable.
1.6.3 LACK OF DESIGN AND DATA-MODELING SKILLS
A new problem that has evolved with the use of personal productivity tools (such as spreadsheet and desktop databases)
is that users typically lack the knowledge of proper design and data-modeling skills. People naturally have an integrated
view of the data in their environment. For example, consider a student’s class schedule. The schedule probably contains
the student’s identification number and name, class code, class description, class credit hours, the name of the instructor
teaching the class, the class meeting days and times, and the class room number. In the mind of the student, these various
data items compose a single unit. If a student organization wanted to keep a record of the schedules of all of the
organization members, an end user might make a spreadsheet to store the schedule information. Even if the student makes
a foray into the realm of desktop databases, he or she is likely to create a structure composed of a single table that mimics
the structure of the schedule. As you will learn in the coming chapters, forcing this type of integrated data into a single
two-dimensional table structure is a poor data design that leads to a large degree of redundancy for several data items.
Data-modeling skills are also a vital part of the design process. It is important that the design that is created be properly
documented. Design documentation is necessary to facilitate communication among the database designer, the end
user, and the developer. Data modeling, as introduced later in this text, is the most common method of documenting
database designs. Using a standardized data-modeling technique ensures that the data model fulfills its role in facilitating
communication among the designer, user, and developer. The data model also provides an invaluable resource when
maintaining or modifying a database as business requirements change. The data designs created by end users are rarely
documented and never with an appropriate standardized data-modeling technique. On a positive note, however, if you
are reading this book, then you are engaged in the type of training that is necessary to develop the skills in database
design and data modeling that it takes to successfully design a database that ensures consistency of the data, enforces
integrity, and provides a stable and flexible platform for providing users with timely, accurate information.
1.7 DATABASE SYSTEMS
The problems inherent in file systems make using a database system very desirable. Unlike the file system, with its
many separate and unrelated files, the database system consists of logically related data stored in a single logical data
repository. (The “logical” label reflects the fact that, although the data repository appears to be a single unit to the end
user, its contents may actually be physically distributed among multiple data storage facilities and/or locations.) Because
the database’s data repository is a single logical unit, the database represents a major change in the way end-user data
are stored, accessed, and managed. The database’s DBMS, shown in Figure 1.6, provides numerous advantages over
file system management, shown in Figure 1.5, by making it possible to eliminate most of the file system’s data
inconsistency, data anomaly, data dependence, and structural dependence problems. Better yet, the current generation
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of DBMS software stores not only the data structures, but also the relationships between those structures and the
access paths to those structures—all in a central location. The current generation of DBMS software also takes care
of defining, storing, and managing all required access paths to those components.
FIGURE
Contrasting database and file systems
1.6
A Database System
Database
Personnel dept.
DBMS
Sales dept.
Employees
Customers
Sales
Inventory
Accounts
Accounting dept.
A File System
Personnel dept.
Employees
Sales dept.
Accounting dept.
Sales
Accounts
Customers
Remember that the DBMS is just one of several crucial components of a database system. The DBMS may even be
referred to as the database system’s heart. However, just as it takes more than a heart to make a human being function,
it takes more than a DBMS to make a database system function. In the sections that follow, you’ll learn what a database
system is, what its components are, and how the DBMS fits into the database system picture.
1.7.1 The Database System Environment
The term database system refers to an organization of components that define and regulate the collection, storage,
management, and use of data within a database environment. From a general management point of view, the database
system is composed of the five major parts shown in Figure 1.7: hardware, software, people, procedures, and data.
Let’s take a closer look at the five components shown in Figure 1.7:
쐌
Hardware. Hardware refers to all of the system’s physical devices; for example, computers (PCs, workstations,
servers, and supercomputers), storage devices, printers, network devices (hubs, switches, routers, fiber optics),
and other devices (automated teller machines, ID readers, and so on).
D A T A B A S E
FIGURE
S Y S T E M S
The database system environment
1.7
Procedures
and standards
writes
and
enforces
Analysts
supervises
Database
System
Database administrator administrator
manages
designer
designs
use
Hardware
Programmers
End users
Application
programs
write
DBMS utilities
DBMS
access
Data
쐌
쐌
Software. Although the most readily identified software is the DBMS itself, to make the database system
function fully, three types of software are needed: operating system software, DBMS software, and application
programs and utilities.
-
Operating system software manages all hardware components and makes it possible for all other software
to run on the computers. Examples of operating system software include Microsoft Windows, Linux, Mac
OS, UNIX, and MVS.
-
DBMS software manages the database within the database system. Some examples of DBMS software
include Microsoft’s SQL Server, Oracle Corporation’s Oracle, Sun’s MySQL, and IBM’s DB2.
-
Application programs and utility software are used to access and manipulate data in the DBMS and to
manage the computer environment in which data access and manipulation take place. Application
programs are most commonly used to access data found within the database to generate reports,
tabulations, and other information to facilitate decision making. Utilities are the software tools used to help
manage the database system’s computer components. For example, all of the major DBMS vendors now
provide graphical user interfaces (GUIs) to help create database structures, control database access, and
monitor database operations.
People. This component includes all users of the database system. On the basis of primary job functions, five
types of users can be identified in a database system: system administrators, database administrators, database
designers, system analysts and programmers, and end users. Each user type, described below, performs both
unique and complementary functions.
-
System administrators oversee the database system’s general operations.
-
Database administrators, also known as DBAs, manage the DBMS and ensure that the database is
functioning properly. The DBA’s role is sufficiently important to warrant a detailed exploration in Chapter
15, Database Administration and Security.
-
Database designers design the database structure. They are, in effect, the database architects. If the
database design is poor, even the best application programmers and the most dedicated DBAs cannot
produce a useful database environment. Because organizations strive to optimize their data resources, the
database designer’s job description has expanded to cover new dimensions and growing responsibilities.
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System analysts and programmers design and implement the application programs. They design and
create the data entry screens, reports, and procedures through which end users access and manipulate the
database’s data.
-
End users are the people who use the application programs to run the organization’s daily operations. For
example, salesclerks, supervisors, managers, and directors are all classified as end users. High-level end
users employ the information obtained from the database to make tactical and strategic business decisions.
쐌
Procedures. Procedures are the instructions and rules that govern the design and use of the database system.
Procedures are a critical, although occasionally forgotten, component of the system. Procedures play an
important role in a company because they enforce the standards by which business is conducted within the
organization and with customers. Procedures are also used to ensure that there is an organized way to monitor
and audit both the data that enter the database and the information that is generated through the use of
those data.
쐌
Data. The word data covers the collection of facts stored in the database. Because data are the raw material
from which information is generated, the determination of what data are to be entered into the database and
how those data are to be organized is a vital part of the database designer’s job.
A database system adds a new dimension to an organization’s management structure. Just how complex this
managerial structure is depends on the organization’s size, its functions, and its corporate culture. Therefore, database
systems can be created and managed at different levels of complexity and with varying adherence to precise standards.
For example, compare a local movie rental system with a national insurance claims system. The movie rental system
may be managed by two people, the hardware used is probably a single PC, the procedures are probably simple, and
the data volume tends to be low. The national insurance claims system is likely to have at least one systems
administrator, several full-time DBAs, and many designers and programmers; the hardware probably includes several
servers at multiple locations throughout the United States; the procedures are likely to be numerous, complex, and
rigorous; and the data volume tends to be high.
In addition to the different levels of database system complexity, managers must also take another important fact into
account: database solutions must be cost-effective as well as tactically and strategically effective. Producing a
million-dollar solution to a thousand-dollar problem is hardly an example of good database system selection or of good
database design and management. Finally, the database technology already in use is likely to affect the selection of a
database system.
1.7.2 DBMS Functions
A DBMS performs several important functions that guarantee the integrity and consistency of the data in the database.
Most of those functions are transparent to end users, and most can be achieved only through the use of a DBMS. They
include data dictionary management, data storage management, data transformation and presentation, security
management, multiuser access control, backup and recovery management, data integrity management, database
access languages and application programming interfaces, and database communication interfaces. Each of these
functions is explained below.
쐌
Data dictionary management. The DBMS stores definitions of the data elements and their relationships
(metadata) in a data dictionary. In turn, all programs that access the data in the database work through the
DBMS. The DBMS uses the data dictionary to look up the required data component structures and
relationships, thus relieving you from having to code such complex relationships in each program. Additionally,
any changes made in a database structure are automatically recorded in the data dictionary, thereby freeing you
from having to modify all of the programs that access the changed structure. In other words, the DBMS
provides data abstraction, and it removes structural and data dependence from the system. For example,
Figure 1.8 shows how Microsoft SQL Server Express presents the data definition for the CUSTOMER table.
쐌
Data storage management. The DBMS creates and manages the complex structures required for data storage,
thus relieving you from the difficult task of defining and programming the physical data characteristics. A
D A T A B A S E
FIGURE
S Y S T E M S
Illustrating metadata with Microsoft SQL Server Express
1.8
Metadata
modern DBMS provides storage not only for the data, but also for related data entry forms or screen
definitions, report definitions, data validation rules, procedural code, structures to handle video and picture
formats, and so on. Data storage management is also important for database performance tuning.
Performance tuning relates to the activities that make the database perform more efficiently in terms of
storage and access speed. Although the user sees the database as a single data storage unit, the DBMS actually
stores the database in multiple physical data files. (See Figure 1.9.) Such data files may even be stored on
different storage media. Therefore, the DBMS doesn’t have to wait for one disk request to finish before the
next one starts. In other words, the DBMS can fulfill database requests concurrently. Data storage management
and performance tuning issues are addressed in Chapter 11, Database Performance Tuning and Query
Optimization.
쐌
Data transformation and presentation. The DBMS transforms entered data to conform to required data
structures. The DBMS relieves you of the chore of making a distinction between the logical data format and
the physical data format. That is, the DBMS formats the physically retrieved data to make it conform to the
user’s logical expectations. For example, imagine an enterprise database used by a multinational company. An
end user in England would expect to enter data such as July 11, 2010, as “11/07/2010.” In contrast, the
same date would be entered in the United States as “07/11/2010.” Regardless of the data presentation
format, the DBMS must manage the date in the proper format for each country.
쐌
Security management. The DBMS creates a security system that enforces user security and data privacy.
Security rules determine which users can access the database, which data items each user can access, and
which data operations (read, add, delete, or modify) the user can perform. This is especially important in
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FIGURE
1
Illustrating data storage management with Oracle
1.9
Database Name: ORALAB.MTSU.EDU
The ORALAB database is
actually stored in nine
datafiles located on the C:
drive of the database server
computer.
The Oracle DBA Studio
Management interface also
shows the amount of space
used by each of the datafiles
that constitute the single
logical database.
The Oracle DBA Studio Administrator GUI shows the data storage
management characteristics for the ORALAB database.
multiuser database systems. Chapter 15, Database Administration and Security, examines data security and
privacy issues in greater detail. All database users may be authenticated to the DBMS through a username and
password or through biometric authentication such as a fingerprint scan. The DBMS uses this information to
assign access privileges to various database components such as queries and reports.
쐌
Multiuser access control. To provide data integrity and data consistency, the DBMS uses sophisticated
algorithms to ensure that multiple users can access the database concurrently without compromising the
integrity of the database. Chapter 10, Transaction Management and Concurrency Control, covers the details
of the multiuser access control.
쐌
Backup and recovery management. The DBMS provides backup and data recovery to ensure data safety and
integrity. Current DBMS systems provide special utilities that allow the DBA to perform routine and special
backup and restore procedures. Recovery management deals with the recovery of the database after a failure,
such as a bad sector in the disk or a power failure. Such capability is critical to preserving the database’s
integrity. Chapter 15 covers backup and recovery issues.
쐌
Data integrity management. The DBMS promotes and enforces integrity rules, thus minimizing data
redundancy and maximizing data consistency. The data relationships stored in the data dictionary are used to
enforce data integrity. Ensuring data integrity is especially important in transaction-oriented database systems.
Data integrity and transaction management issues are addressed in Chapter 7, Introduction to Structured
Query Language (SQL), and Chapter 10.
쐌
Database access languages and application programming interfaces. The DBMS provides data access through
a query language. A query language is a nonprocedural language—one that lets the user specify what must
be done without having to specify how it is to be done. Structured Query Language (SQL) is the de facto
D A T A B A S E
S Y S T E M S
query language and data access standard supported by the majority of DBMS vendors. Chapter 7, Introduction
to Structured Query Language (SQL), and Chapter 8, Advanced SQL, address the use of SQL. The DBMS also
provides application programming interfaces to procedural languages such as COBOL, C, Java, Visual
Basic.NET, and C#. In addition, the DBMS provides administrative utilities used by the DBA and the database
designer to create, implement, monitor, and maintain the database.
쐌
Database communication interfaces. Current-generation DBMSs accept end-user requests via multiple,
different network environments. For example, the DBMS might provide access to the database via the Internet
through the use of Web browsers such as Mozilla Firefox or Microsoft Internet Explorer. In this environment,
communications can be accomplished in several ways:
-
End users can generate answers to queries by filling in screen forms through their preferred Web browser.
-
The DBMS can automatically publish predefined reports on a Website.
-
The DBMS can connect to third-party systems to distribute information via e-mail or other productivity
applications.
Database communication interfaces are examined in greater detail in Chapter 12, Distributed Database Management
Systems, in Chapter 14, Database Connectivity and Web Technologies, and in Appendix I, Databases in Electronic
Commerce. (Appendixes are found in the Premium Website.)
NOTE
Why a Spreadsheet Is Not a Database
While a spreadsheet allows for the creation of multiple tables, it does not support even the most basic database
functionality such as support for self-documentation through metadata, enforcement of data types or domains
to ensure consistency of data within a column, defined relationships among tables, or constraints to ensure
consistency of data across related tables. Most users lack the necessary training to recognize the limitations of
spreadsheets for these types of tasks.
1.7.3 Managing the Database System: A Shift in Focus
The introduction of a database system over the file system provides a framework in which strict procedures and
standards can be enforced. Consequently, the role of the human component changes from an emphasis on
programming (in the file system) to a focus on the broader aspects of managing the organization’s data resources and
on the administration of the complex database software itself.
The database system makes it possible to tackle far more sophisticated uses of the data resources, as long as the database
is designed to make use of that available power. The kinds of data structures created within the database and the extent
of the relationships among them play a powerful role in determining the effectiveness of the database system.
Although the database system yields considerable advantages over previous data management approaches, database
systems do carry significant disadvantages. For example:
쐌
Increased costs. Database systems require sophisticated hardware and software and highly skilled personnel.
The cost of maintaining the hardware, software, and personnel required to operate and manage a database
system can be substantial. Training, licensing, and regulation compliance costs are often overlooked when
database systems are implemented.
쐌
Management complexity. Database systems interface with many different technologies and have a significant
impact on a company’s resources and culture. The changes introduced by the adoption of a database system
must be properly managed to ensure that they help advance the company’s objectives. Given the fact that
database systems hold crucial company data that are accessed from multiple sources, security issues must be
assessed constantly.
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쐌
Maintaining currency. To maximize the efficiency of the database system, you must keep your system current.
Therefore, you must perform frequent updates and apply the latest patches and security measures to all
components. Because database technology advances rapidly, personnel training costs tend to be significant.
쐌
Vendor dependence. Given the heavy investment in technology and personnel training, companies might be
reluctant to change database vendors. As a consequence, vendors are less likely to offer pricing point
advantages to existing customers, and those customers might be limited in their choice of database system
components.
쐌
Frequent upgrade/replacement cycles. DBMS vendors frequently upgrade their products by adding new
functionality. Such new features often come bundled in new upgrade versions of the software. Some of these
versions require hardware upgrades. Not only do the upgrades themselves cost money, but it also costs money
to train database users and administrators to properly use and manage the new features.
Now that we have considered what a database and DBMS are, and why they are necessary, it is natural for our
thoughts to turn to developing the skills of database design. However, before we can create a design, we must know
what tools are at our disposal. Throughout this chapter, we have generalized the discussion of database technology
such that it appears that there is a single, common approach to database design. As a database designer and developer,
however, you need to understand that there are different approaches, and you need to know how these approaches
influence the designs that you can create and how those designs are modeled.
D A T A B A S E
S Y S T E M S
S u m m a r y
◗
◗
◗
◗
◗
◗
Data are raw facts. Information is the result of processing data to reveal its meaning. Accurate, relevant, and timely
information is the key to good decision making, and good decision making is the key to organizational survival in
a global environment.
Data are usually stored in a database. To implement a database and to manage its contents, you need a database
management system (DBMS). The DBMS serves as the intermediary between the user and the database. The
database contains the data you have collected and “data about data,” known as metadata.
Database design defines the database structure. A well-designed database facilitates data management and
generates accurate and valuable information. A poorly designed database can lead to bad decision making, and bad
decision making can lead to the failure of an organization.
Databases evolved from manual and then computerized file systems. In a file system, data are stored in independent
files, each requiring its own data management programs. Although this method of data management is largely
outmoded, understanding its characteristics makes database design easier to comprehend.
Some limitations of file system data management are that it requires extensive programming, system administration
can be complex and difficult, making changes to existing structures is difficult, and security features are likely to be
inadequate. Also, independent files tend to contain redundant data, leading to problems of structural and data
dependence.
Database management systems were developed to address the file system’s inherent weaknesses. Rather than
depositing data in independent files, a DBMS presents the database to the end user as a single data repository. This
arrangement promotes data sharing, thus eliminating the potential problem of islands of information. In addition,
the DBMS enforces data integrity, eliminates redundancy, and promotes data security.
K e y
T e r m s
ad hoc query, 8
database system, 18
query, 8
centralized database, 9
desktop database, 9
query language, 22
data, 5
distributed database, 9
query result set, 8
data anomaly, 17
enterprise database, 9
record, 12
data dependence, 15
Extensible Markup
Language (XML), 10
semistructured data, 10
data dictionary, 20
data inconsistency, 8
field, 12
structural dependence, 15
data independence, 15
file, 12
structural independence, 15
data integrity, 16
information, 5
structured data, 9
data management, 7
islands of information, 16
data processing (DP) specialist, 11
knowledge, 7
Structured Query
Language (SQL), 22
data quality, 9
logical data format, 15
transactional database, 9
data redundancy, 16
metadata, 7
unstructured data, 9
data warehouse, 9
multiuser database, 9
workgroup database, 9
database, 7
operational database, 9
XML database, 10
database design, 10
performance tuning, 21
database management
system (DBMS), 7
physical data format, 15
production database, 9
single-user database, 9
25
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1
Online Content
Answers to selected Review Questions and Problems for this chapter are contained in the Premium Website for
this book.
R e v i e w
Q u e s t i o n s
1. Define each of the following terms:
a. data
b. field
c. record
d. file
2. What is data redundancy, and which characteristics of the file system can lead to it?
3. What is data independence, and why is it lacking in file systems?
4. What is a DBMS, and what are its functions?
5. What is structural independence, and why is it important?
6. Explain the difference between data and information.
7. What is the role of a DBMS, and what are its advantages? What are its disadvantages?
8. List and describe the different types of databases.
9. What are the main components of a database system?
10. What are metadata?
11. Explain why database design is important.
12. What are the potential costs of implementing a database system?
13. Use examples to compare and contrast unstructured and structured data. Which type is more prevalent in a
typical business environment?
14. What are some basic database functions that a spreadsheet cannot perform?
15. What common problems does a collection of spreadsheets created by end users share with the typical file system?
16. Explain the significance of the loss of direct, hands-on access to business data that end users experienced with
the advent of computerized data repositories.
P r o b l e m s
Online Content
The file structures you see in this problem set are simulated in a Microsoft Access database named
Ch01_Problems, available in the Premium Website for this book.
D A T A B A S E
FIGURE
S Y S T E M S
The file structure for Problems 1–4
P1.1
Given the file structure shown in Figure P1.1, answer Problems 1−4.
1. How many records does the file contain? How many fields are there per record?
2. What problem would you encounter if you wanted to produce a listing by city? How would you solve this problem
by altering the file structure?
3. If you wanted to produce a listing of the file contents by last name, area code, city, state, or zip code, how would
you alter the file structure?
4. What data redundancies do you detect? How could those redundancies lead to anomalies?
FIGURE
The file structure for Problems 5–8
P1.5
5. Identify and discuss the serious data redundancy problems exhibited by the file structure shown in Figure P1.5.
6. Looking at the EMP_NAME and EMP_PHONE contents in Figure P1.5, what change(s) would you recommend?
7. Identify the various data sources in the file you examined in Problem 5.
8. Given your answer to Problem 7, what new files should you create to help eliminate the data redundancies found
in the file shown in Figure P1.5?
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C H A P T E R
FIGURE
1
The file structure for Problems 9–10
P1.9
9. Identify and discuss the serious data redundancy problems exhibited by the file structure shown in Figure P1.9.
(The file is meant to be used as a teacher class assignment schedule. One of the many problems with data
redundancy is the likely occurrence of data inconsistencies—two different initials have been entered for the
teacher named Maria Cordoza.)
10. Given the file structure shown in Figure P1.9, what problem(s) might you encounter if building KOM were deleted?
In this chapter, you will learn:
쐍 About data modeling and why data models are important
쐍 About the basic data-modeling building blocks
쐍 What business rules are and how they influence database design
쐍 How the major data models evolved
쐍 How data models can be classified by level of abstraction
This chapter examines data modeling. Data modeling is the first step in the database design
journey, serving as a bridge between real-world objects and the database that resides in the
computer.
P
review
One of the most vexing problems of database design is that designers, programmers, and
end users see data in different ways. Consequently, different views of the same data can lead
to database designs that do not reflect an organization’s actual operation, thus failing to
meet end-user needs and data efficiency requirements. To avoid such failures, database
designers must obtain a precise description of the nature of the data and of the many uses
of that data within the organization. Communication among database designers, programmers, and end users should be frequent and clear. Data modeling clarifies such communication by reducing the complexities of database design to more easily understood
abstractions that define entities and the relations among them.
First, you will learn what some of the basic data-modeling concepts are and how current
data models developed from earlier models. Tracing the development of those database
models will help you understand the database design and implementation issues that are
addressed in the rest of this book. Second, you will be introduced to the entity
relationship diagram (ERD) as a data-modeling tool. ER diagrams can be drawn using a
variety of notations. Within this chapter, you will be introduced to the traditional Chen
notation, the more current Crow’s Foot notation, and the newer class diagram notation,
which is part of the Unified Modeling Language (UML). Finally, you will learn how various
degrees of data abstraction help reconcile varying views of the same data.
2
T W O
Data Models
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2
2.1 DATA MODELING AND DATA MODELS
Database design focuses on how the database structure will be used to store and manage end-user data. Data modeling,
the first step in designing a database, refers to the process of creating a specific data model for a determined problem
domain. (A problem domain is a clearly defined area within the real-world environment, with well-defined scope and
boundaries, that is to be systematically addressed.) A data model is a relatively simple representation, usually
graphical, of more complex real-world data structures. In general terms, a model is an abstraction of a more complex
real-world object or event. A model’s main function is to help you understand the complexities of the real-world
environment. Within the database environment, a data model represents data structures and their characteristics,
relations, constraints, transformations, and other constructs with the purpose of supporting a specific problem domain.
Note
The terms data model and database model are often used interchangeably. In this book, the term database
model is used to refer to the implementation of a data model in a specific database system.
Data modeling is an iterative, progressive process. You start with a simple understanding of the problem domain, and
as your understanding of the problem domain increases, so does the level of detail of the data model. Done properly,
the final data model is in effect a “blueprint” containing all the instructions to build a database that will meet all end-user
requirements. This blueprint is narrative and graphical in nature, meaning that it contains both text descriptions in
plain, unambiguous language and clear, useful diagrams depicting the main data elements.
Note
An implementation-ready data model should contain at least the following components:
• A description of the data structure that will store the end-user data.
• A set of enforceable rules to guarantee the integrity of the data.
• A data manipulation methodology to support the real-world data transformations.
Traditionally, database designers relied on good judgment to help them develop a good data model. Unfortunately,
good judgment is often in the eye of the beholder, and it often develops after much trial and error. For example, if each
of the students in this class has to create a data model for a video store, it’s very likely that each of them will come
up with a different model. Which one would be the correct one? The simple answer is “the one that meets all the
end-user requirements,” and there may be more than one correct solution! Fortunately, database designers make use
of existing data-modeling constructs and powerful database design tools that substantially diminish the potential for
errors in database modeling. In the following sections, you will learn how existing data models are used to represent
real-world data and how the different degrees of data abstraction facilitate data modeling. For example, if each student
in a class has to create a data model for a video store, it’s very likely that each will come up with a different model.
2.2 THE IMPORTANCE OF DATA MODELS
Data models can facilitate interaction among the designer, the applications programmer, and the end user. A
well-developed data model can even foster improved understanding of the organization for which the database design
is developed. In short, data models are a communication tool. This important aspect of data modeling was summed
up neatly by a client whose reaction was as follows: “I created this business, I worked with this business for years, and
this is the first time I’ve really understood how all the pieces really fit together.”
D A T A
M O D E L S
The importance of data modeling cannot be overstated. Data constitute the most basic information units employed by
a system. Applications are created to manage data and to help transform data into information. But data are viewed
in different ways by different people. For example, contrast the (data) view of a company manager with that of a
company clerk. Although the manager and the clerk both work for the same company, the manager is more likely to
have an enterprise-wide view of company data than the clerk.
Even different managers view data differently. For example, a company president is likely to take a universal view of
the data because he or she must be able to tie the company’s divisions to a common (database) vision. A purchasing
manager in the same company is likely to have a more restricted view of the data, as is the company’s inventory
manager. In effect, each department manager works with a subset of the company’s data. The inventory manager is
more concerned about inventory levels, while the purchasing manager is more concerned about the cost of items and
about personal/business relationships with the suppliers of those items.
Applications programmers have yet another view of data, being more concerned with data location, formatting, and
specific reporting requirements. Basically, applications programmers translate company policies and procedures from
a variety of sources into appropriate interfaces, reports, and query screens.
The different users and producers of data and information often reflect the “blind people and the elephant” analogy: the
blind person who felt the elephant’s trunk had quite a different view of the elephant from the one who felt the elephant’s
leg or tail. What is needed is a view of the whole elephant. Similarly, a house is not a random collection of rooms; if
someone is going to build a house, he or she should first have the overall view that is provided by blueprints. Likewise, a
sound data environment requires an overall database blueprint based on an appropriate data model.
When a good database blueprint is available, it does not matter that an applications programmer’s view of the data is
different from that of the manager and/or the end user. Conversely, when a good database blueprint is not available,
problems are likely to ensue. For instance, an inventory management program and an order entry system may use
conflicting product-numbering schemes, thereby costing the company thousands (or even millions) of dollars.
Keep in mind that a house blueprint is an abstraction; you cannot live in the blueprint. Similarly, the data model is an
abstraction; you cannot draw the required data out of the data model. Just as you are not likely to build a good house
without a blueprint, you are equally unlikely to create a good database without first creating an appropriate data model.
2.3 DATA MODEL BASIC BUILDING BLOCKS
The basic building blocks of all data models are entities, attributes, relationships, and constraints. An entity is anything
(a person, a place, a thing, or an event) about which data are to be collected and stored. An entity represents a
particular type of object in the real world. Because an entity represents a particular type of object, entities are
“distinguishable”—that is, each entity occurrence is unique and distinct. For example, a CUSTOMER entity would have
many distinguishable customer occurrences, such as John Smith, Pedro Dinamita, Tom Strickland, etc. Entities may
be physical objects, such as customers or products, but entities may also be abstractions, such as flight routes or musical
concerts.
An attribute is a characteristic of an entity. For example, a CUSTOMER entity would be described by attributes such
as customer last name, customer first name, customer phone, customer address, and customer credit limit. Attributes
are the equivalent of fields in file systems.
A relationship describes an association among entities. For example, a relationship exists between customers and
agents that can be described as follows: an agent can serve many customers, and each customer may be served by one
agent. Data models use three types of relationships: one-to-many, many-to-many, and one-to-one. Database designers
usually use the shorthand notations 1:M or 1..*, M:N or *..*, and 1:1 or 1..1, respectively. (Although the M:N notation
is a standard label for the many-to-many relationship, the label M:M may also be used.) The following examples
illustrate the distinctions among the three.
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2
쐌
One-to-many (1:M or 1..*) relationship. A painter paints many different paintings, but each one of them
is painted by only one painter. Thus, the painter (the “one”) is related to the paintings (the “many”). Therefore,
database designers label the relationship “PAINTER paints PAINTING” as 1:M. (Note that entity names are
often capitalized as a convention, so they are easily identified.) Similarly, a customer (the “one”) may generate
many invoices, but each invoice (the “many”) is generated by only a single customer. The “CUSTOMER
generates INVOICE” relationship would also be labeled 1:M.
쐌
Many-to-many (M:N or *..*) relationship. An employee may learn many job skills, and each job skill may
be learned by many employees. Database designers label the relationship “EMPLOYEE learns SKILL” as M:N.
Similarly, a student can take many classes and each class can be taken by many students, thus yielding the M:N
relationship label for the relationship expressed by “STUDENT takes CLASS.”
쐌
One-to-one (1:1 or 1..1) relationship. A retail company’s management structure may require that each of
its stores be managed by a single employee. In turn, each store manager, who is an employee, manages only
a single store. Therefore, the relationship “EMPLOYEE manages STORE” is labeled 1:1.
The preceding discussion identified each relationship in both directions; that is, relationships are bidirectional:
쐌
One CUSTOMER can generate many INVOICEs.
쐌
Each of the many INVOICEs is generated by only one CUSTOMER.
A constraint is a restriction placed on the data. Constraints are important because they help to ensure data integrity.
Constraints are normally expressed in the form of rules. For example:
쐌
An employee’s salary must have values that are between 6,000 and 350,000.
쐌
A student’s GPA must be between 0.00 and 4.00.
쐌
Each class must have one and only one teacher.
How do you properly identify entities, attributes, relationships, and constraints? The first step is to clearly identify the
business rules for the problem domain you are modeling.
2.4 BUSINESS RULES
When database designers go about selecting or determining the entities, attributes, and relationships that will be used
to build a data model, they might start by gaining a thorough understanding of what types of data are in an
organization, how the data are used, and in what time frames they are used. But such data and information do not,
by themselves, yield the required understanding of the total business. From a database point of view, the collection of
data becomes meaningful only when it reflects properly defined business rules. A business rule is a brief, precise,
and unambiguous description of a policy, procedure, or principle within a specific organization. In a sense, business
rules are misnamed: they apply to any organization, large or small—a business, a government unit, a religious group,
or a research laboratory—that stores and uses data to generate information.
Business rules, derived from a detailed description of an organization’s operations, help to create and enforce actions
within that organization’s environment. Business rules must be rendered in writing and updated to reflect any change
in the organization’s operational environment.
Properly written business rules are used to define entities, attributes, relationships, and constraints. Any time you see
relationship statements such as “an agent can serve many customers, and each customer can be served by only one
agent,” you are seeing business rules at work. You will see the application of business rules throughout this book,
especially in the chapters devoted to data modeling and database design.
D A T A
M O D E L S
To be effective, business rules must be easy to understand and widely disseminated, to ensure that every person in the
organization shares a common interpretation of the rules. Business rules describe, in simple language, the main and
distinguishing characteristics of the data as viewed by the company. Examples of business rules are as follows:
쐌
A customer may generate many invoices.
쐌
An invoice is generated by only one customer.
쐌
A training session cannot be scheduled for fewer than 10 employees or for more than 30 employees.
Note that those business rules establish entities, relationships, and constraints. For example, the first two business rules
establish two entities (CUSTOMER and INVOICE) and a 1:M relationship between those two entities. The third
business rule establishes a constraint (no fewer than 10 people and no more than 30 people), two entities (EMPLOYEE
and TRAINING), and a relationship between EMPLOYEE and TRAINING.
2.4.1 Discovering Business Rules
The main sources of business rules are company managers, policy makers, department managers, and written
documentation such as a company’s procedures, standards, and operations manuals. A faster and more direct source
of business rules is direct interviews with end users. Unfortunately, because perceptions differ, end users are sometimes
a less reliable source when it comes to specifying business rules. For example, a maintenance department mechanic
might believe that any mechanic can initiate a maintenance procedure, when actually only mechanics with inspection
authorization can perform such a task. Such a distinction might seem trivial, but it can have major legal consequences.
Although end users are crucial contributors to the development of business rules, it pays to verify end-user
perceptions. Too often, interviews with several people who perform the same job yield very different perceptions of
what the job components are. While such a discovery may point to “management problems,” that general diagnosis
does not help the database designer. The database designer’s job is to reconcile such differences and verify the results
of the reconciliation to ensure that the business rules are appropriate and accurate.
The process of identifying and documenting business rules is essential to database design for several reasons:
쐌
They help to standardize the company’s view of data.
쐌
They can be a communications tool between users and designers.
쐌
They allow the designer to understand the nature, role, and scope of the data.
쐌
They allow the designer to understand business processes.
쐌
They allow the designer to develop appropriate relationship participation rules and constraints and to create
an accurate data model.
Of course, not all business rules can be modeled. For example, a business rule that specifies that “no pilot can fly more
than 10 hours within any 24-hour period” cannot be modeled. However, such a business rule can be enforced by
application software.
2.4.2 Translating Business Rules into Data Model Components
Business rules set the stage for the proper identification of entities, attributes, relationships, and constraints. In the real
world, names are used to identify objects. If the business environment wants to keep track of the objects, there will be
specific business rules for them. As a general rule, a noun in a business rule will translate into an entity in the model,
and a verb (active or passive) associating nouns will translate into a relationship among the entities. For example, the
business rule “a customer may generate many invoices” contains two nouns (customer and invoices) and a verb
( generate) that associates the nouns. From this business rule, you could deduce that:
쐌
Customer and invoice are objects of interest for the environment and should be represented by their respective
entities.
쐌
There is a “generate” relationship between customer and invoice.
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2
To properly identify the type of relationship, you should consider that relationships are bidirectional; that is, they go
both ways. For example, the business rule “a customer may generate many invoices” is complemented by the business
rule “an invoice is generated by only one customer.” In that case, the relationship is one-to-many (1:M). Customer is
the “1” side, and invoice is the “many” side.
As a general rule, to properly identify the relationship type, you should ask two questions:
쐌
How many instances of B are related to one instance of A?
쐌
How many instances of A are related to one instance of B?
For example, you can assess the relationship between student and class by asking two questions:
쐌
In how many classes can one student enroll? Answer: many classes.
쐌
How many students can enroll in one class? Answer: many students.
Therefore, the relationship between student and class is many-to-many (M:N). You will have many opportunities to
determine the relationships between entities as you proceed through this book, and soon the process will become
second nature.
2.4.3 Naming Conventions
During the translation of business rules to data model components, you identify entities, attributes, relationships, and
constraints. This identification process includes naming the object in a way that makes the object unique and
distinguishable from other objects in the problem domain. Therefore, it is important that you pay special attention to
how you name the objects you are discovering.
Entity names should be descriptive of the objects in the business environment, and use terminology that is familiar to
the users. An attribute name should also be descriptive of the data represented by that attribute. It is also a good
practice to prefix the name of an attribute with the name of the entity (or an abbreviation of the entity name) in which
it occurs. For example, in the CUSTOMER entity, the customer’s credit limit may be called CUS_CREDIT_LIMIT. The
CUS indicates that the attribute is descriptive of the CUSTOMER entity, while CREDIT_LIMIT makes it easy to
recognize the data that will be contained in the attribute. This will become increasingly important in later chapters when
we discuss the need to use common attributes to specify relationships between entities. The use of a proper naming
convention will improve the data model’s ability to facilitate communication among the designer, application
programmer, and the end user. In fact, a proper naming convention can go a long way toward making your model
self-documenting.
2.5 THE EVOLUTION OF DATA MODELS
The quest for better data management has led to several models that attempt to resolve the file system’s critical
shortcomings. These models represent schools of thought as to what a database is, what it should do, the types of
structures that it should employ, and the technology that would be used to implement these structures. Perhaps
confusingly, these models are called data models just as are the graphical data models that we have been discussing.
This section gives an overview of the major data models in roughly chronological order. You will discover that many
of the “new” database concepts and structures bear a remarkable resemblance to some of the “old” data model
concepts and structures. Table 2.1 traces the evolution of the major data models.
D A T A
TABLE
M O D E L S
Evolution of Major Data Models
2.1
GENERATION
First
TIME
1960s−1970s
DATA MODEL
File system
EXAMPLES
VMS/VSAM
Second
1970s
Hierarchical and
network
Third
Mid-1970s to
present
Relational
Fourth
Mid-1980s to
present
Object-oriented
Object/relational
(O/R)
IMS
ADABAS
IDS-II
DB2
Oracle
MS SQL-Server
MySQL
Versant
Objectivity/DB
DB/2 UDB
Oracle 11g
Next
generation
Present to
future
XML
Hybrid DBMS
dbXML
Tamino
DB2 UDB
Oracle 11g
MS SQL Server
COMMENTS
Used mainly on IBM mainframe
systems
Managed records, not relationships
Early database systems
Navigational access
Conceptual simplicity
Entity relationship (ER) modeling and
support for relational data modeling
Object/relational supports object
data types
Star Schema support for data
warehousing
Web databases become common
Unstructured data support
O/R model supports XML documents
Hybrid DBMS adds an object front
end to relational databases
Online Content
The hierarchical and network models are largely of historical interest, yet they do contain some elements and
features that interest current database professionals. The technical details of those two models are discussed in
detail in Appendixes K and L, respectively, in the Premium Website for this book. Appendix G is devoted to the
object-oriented (OO) model. However, given the dominant market presence of the relational model, most of
the book focuses on that model.
2.5.1 Hierarchical and Network Models
The hierarchical model was developed in the 1960s to manage large amounts of data for complex manufacturing
projects such as the Apollo rocket that landed on the moon in 1969. Its basic logical structure is represented by an
upside-down tree. The hierarchical structure contains levels, or segments. A segment is the equivalent of a file
system’s record type. Within the hierarchy, a higher layer is perceived as the parent of the segment directly beneath
it, which is called the child. The hierarchical model depicts a set of one-to-many (1:M) relationships between a parent
and its children segments. (Each parent can have many children, but each child has only one parent.)
The network model was created to represent complex data relationships more effectively than the hierarchical
model, to improve database performance, and to impose a database standard. In the network model, the user
perceives the network database as a collection of records in 1:M relationships. However, unlike the hierarchical model,
the network model allows a record to have more than one parent. While the network database model is generally not
used today, the definitions of standard database concepts that emerged with the network model are still used by
modern data models. Some important concepts that were defined at this time are:
쐌
The schema, which is the conceptual organization of the entire database as viewed by the database
administrator.
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쐌
The subschema, which defines the portion of the database “seen” by the application programs that actually
produce the desired information from the data contained within the database.
쐌
A data management language (DML), which defines the environment in which data can be managed and
to work with the data in the database.
쐌
A schema data definition language (DDL), which enables the database administrator to define the schema
components.
As information needs grew and as more sophisticated databases and applications were required, the network model
became too cumbersome. The lack of ad hoc query capability put heavy pressure on programmers to generate the
code required to produce even the simplest reports. And although the existing databases provided limited data
independence, any structural change in the database could still produce havoc in all application programs that drew
data from the database. Because of the disadvantages of the hierarchical and network models, they were largely
replaced by the relational data model in the 1980s.
2.5.2 The Relational Model
The relational model was introduced in 1970 by E. F. Codd (of IBM) in his landmark paper “A Relational Model of
Data for Large Shared Databanks” (Communications of the ACM, June 1970, pp. 377−387). The relational model
represented a major breakthrough for both users and designers. To use an analogy, the relational model produced an
“automatic transmission” database to replace the “standard transmission” databases that preceded it. Its conceptual
simplicity set the stage for a genuine database revolution.
Note
The relational database model presented in this chapter is an introduction and an overview. A more detailed
discussion is in Chapter 3, The Relational Database Model. In fact, the relational model is so important that
it will serve as the basis for discussions in most of the remaining chapters.
The relational model foundation is a mathematical concept known as a relation. To avoid the complexity of abstract
mathematical theory, you can think of a relation (sometimes called a table) as a matrix composed of intersecting rows
and columns. Each row in a relation is called a tuple. Each column represents an attribute. The relational model also
describes a precise set of data manipulation constructs based on advanced mathematical concepts.
In 1970, Codd’s work was considered ingenious but impractical. The relational model’s conceptual simplicity was
bought at the expense of computer overhead; computers at that time lacked the power to implement the relational
model. Fortunately, computer power grew exponentially, as did operating system efficiency. Better yet, the cost of
computers diminished rapidly as their power grew. Today even PCs, costing a fraction of what their mainframe
ancestors did, can run sophisticated relational database software such as Oracle, DB2, Microsoft SQL Server, MySQL,
and other mainframe relational software.
The relational data model is implemented through a very sophisticated relational database management system
(RDBMS). The RDBMS performs the same basic functions provided by the hierarchical and network DBMS systems,
in addition to a host of other functions that make the relational data model easier to understand and implement.
Arguably the most important advantage of the RDBMS is its ability to hide the complexities of the relational model
from the user. The RDBMS manages all of the physical details, while the user sees the relational database as a
collection of tables in which data are stored. The user can manipulate and query the data in a way that seems intuitive
and logical.
Tables are related to each other through the sharing of a common attribute (value in a column). For example, the
CUSTOMER table in Figure 2.1 might contain a sales agent’s number that is also contained in the AGENT table.
D A T A
FIGURE
M O D E L S
Linking relational tables
2.1
Table name: AGENT (first six attributes)
Database name: Ch02_InsureCo
Link through AGENT_CODE
Table name: CUSTOMER
Online Content
This chapter’s databases can be found in the Premium Website. For example, the contents of the AGENT and
CUSTOMER tables shown in Figure 2.1 are found in the database named Ch02_InsureCo.
The common link between the CUSTOMER and AGENT tables enables you to match the customer to his or her sales
agent, even though the customer data are stored in one table and the sales representative data are stored in another
table. For example, you can easily determine that customer Dunne’s agent is Alex Alby because for customer Dunne,
the CUSTOMER table’s AGENT_CODE is 501, which matches the AGENT table’s AGENT_CODE for Alex Alby.
Although the tables are independent of one another, you can easily associate the data between tables. The relational
model provides a minimum level of controlled redundancy to eliminate most of the redundancies commonly found in
file systems.
FIGURE
2.2
A relational diagram
The relationship type (1:1, 1:M, or M:N) is often shown
in a relational schema, an example of which is shown in
Figure 2.2. A relational diagram is a representation of
the relational database’s entities, the attributes within
those entities, and the relationships between those
entities.
In Figure 2.2, the relational diagram shows the connecting
fields (in this case, AGENT_CODE) and the relationship
type, 1:M. Microsoft Access, the database software application used to generate Figure 2.2, employs the ⬁ (infinity)
symbol to indicate the “many” side. In this example, the
CUSTOMER represents the “many” side because an
AGENT can have many CUSTOMERs. The AGENT represents the “1” side because each CUSTOMER has only
one AGENT.
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2
A relational table stores a collection of related entities. In this respect, the relational database table resembles a file. But
there is one crucial difference between a table and a file: A table yields complete data and structural independence
because it is a purely logical structure. How the data are physically stored in the database is of no concern to the user
or the designer; the perception is what counts. And this property of the relational data model, explored in depth in
the next chapter, became the source of a real database revolution.
Another reason for the relational data model’s rise to dominance is its powerful and flexible query language. For most
relational database software, the query language is Structured Query Language (SQL), which allows the user to specify
what must be done without specifying how it must be done. The RDBMS uses SQL to translate user queries into
instructions for retrieving the requested data. SQL makes it possible to retrieve data with far less effort than any other
database or file environment.
From an end-user perspective, any SQL-based relational database application involves three parts: a user interface, a
set of tables stored in the database, and the SQL “engine.” Each of these parts is explained below.
쐌
The end-user interface. Basically, the interface allows the end user to interact with the data (by auto-generating
SQL code). Each interface is a product of the software vendor’s idea of meaningful interaction with the data.
You can also design your own customized interface with the help of application generators that are now
standard fare in the database software arena.
쐌
A collection of tables stored in the database. In a relational database, all data are perceived to be stored in
tables. The tables simply “present” the data to the end user in a way that is easy to understand. Each table is
independent. Rows in different tables are related by common values in common attributes.
쐌
SQL engine. Largely hidden from the end user, the SQL engine executes all queries, or data requests. Keep
in mind that the SQL engine is part of the DBMS software. The end user uses SQL to create table structures
and to perform data access and table maintenance. The SQL engine processes all user requests—largely
behind the scenes and without the end user’s knowledge. Hence, it’s said that SQL is a declarative language
that tells what must be done but not how it must be done. (You will learn more about the SQL engine in
Chapter 11, Database Performance Tuning and Query Optimization.)
Because the RDBMS performs the behind-the-scenes tasks, it is not necessary to focus on the physical aspects of the
database. Instead, the chapters that follow concentrate on the logical portion of the relational database and its design.
Furthermore, SQL is covered in detail in Chapter 7, Introduction to Structured Query Language (SQL), and in Chapter 8,
Advanced SQL.
2.5.3 The Entity Relationship Model
The conceptual simplicity of relational database technology triggered the demand for RDBMSs. In turn, the rapidly
increasing requirements for transaction and information created the need for more complex database implementation
structures, thus creating the need for more effective database design tools. (Building a skyscraper requires more detailed
design activities than building a doghouse, for example.)
Complex design activities require conceptual simplicity to yield successful results. Although the relational model was a
vast improvement over the hierarchical and network models, it still lacked the features that would make it an effective
database design tool. Because it is easier to examine structures graphically than to describe them in text, database
designers prefer to use a graphical tool in which entities and their relationships are pictured. Thus, the entity
relationship (ER) model, or ERM, has become a widely accepted standard for data modeling.
Peter Chen first introduced the ER data model in 1976; it was the graphical representation of entities and their
relationships in a database structure that quickly became popular because it complemented the relational data model
concepts. The relational data model and ERM combined to provide the foundation for tightly structured database
design. ER models are normally represented in an entity relationship diagram (ERD), which uses graphical
representations to model database components.
D A T A
M O D E L S
Note
Because this chapter’s objective is to introduce data-modeling concepts, a simplified ERD is discussed in this
section. You will learn how to use ERDs to design databases in Chapter 4, Entity Relationship (ER) Modeling.
The ER model is based on the following components:
쐌
Entity. Earlier in this chapter, an entity was defined as anything about which data are to be collected and
stored. An entity is represented in the ERD by a rectangle, also known as an entity box. The name of the entity,
a noun, is written in the center of the rectangle. The entity name is generally written in capital letters and is
written in the singular form: PAINTER rather than PAINTERS, and EMPLOYEE rather than EMPLOYEES.
Usually, when applying the ERD to the relational model, an entity is mapped to a relational table. Each row
in the relational table is known as an entity instance or entity occurrence in the ER model.
Note
A collection of like entities is known as an entity set. For example, you can think of the AGENT file in Figure 2.1
as a collection of three agents (entities) in the AGENT entity set. Technically speaking, the ERD depicts entity
sets. Unfortunately, ERD designers use the word entity as a substitute for entity set, and this book will conform
to that established practice when discussing any ERD and its components.
Each entity is described by a set of attributes that describes particular characteristics of the entity. For example,
the entity EMPLOYEE will have attributes such as a Social Security number, a last name, and a first name.
(Chapter 4 explains how attributes are included in the ERD.)
쐌
Relationships. Relationships describe associations among data. Most relationships describe associations
between two entities. When the basic data model components were introduced, three types of relationships
among data were illustrated: one-to-many (1:M), many-to-many (M:N), and one-to-one (1:1). The ER model
uses the term connectivity to label the relationship types. The name of the relationship is usually an active
or passive verb. For example, a PAINTER paints many PAINTINGs; an EMPLOYEE learns many SKILLs; an
EMPLOYEE manages a STORE.
Figure 2.3 shows the different types of relationships using two ER notations: the original Chen notation and the more
current Crow’s Foot notation.
The left side of the ER diagram shows the Chen notation, based on Peter Chen’s landmark paper. In this notation, the
connectivities are written next to each entity box. Relationships are represented by a diamond connected to the related
entities through a relationship line. The relationship name is written inside the diamond.
The right side of Figure 2.3 illustrates the Crow’s Foot notation. The name “Crow’s Foot” is derived from the
three-pronged symbol used to represent the “many” side of the relationship. As you examine the basic Crow’s Foot
ERD in Figure 2.3, note that the connectivities are represented by symbols. For example, the “1” is represented by
a short line segment, and the “M” is represented by the three-pronged “crow’s foot.” In this example, the relationship
name is written above the relationship line.
In Figure 2.3, entities and relationships are shown in a horizontal format, but they may also be oriented vertically. The
entity location and the order in which the entities are presented are immaterial; just remember to read a 1:M
relationship from the “1” side to the “M” side.
The Crow’s Foot notation is used as the design standard in this book. However, the Chen notation is used to illustrate
some of the ER modeling concepts whenever necessary. Most data modeling tools let you select the Crow’s Foot
notation. Microsoft Visio Professional software was used to generate the Crow’s Foot designs you will see in
subsequent chapters.
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C H A P T E R
FIGURE
2
The Chen and Crow’s Foot notations
2.3
Note
Many-to-many (M:N) relationships exist at a conceptual level, and you should know how to recognize them.
However, you will learn in Chapter 3 that M:N relationships are not appropriate in a relational model. For that
reason, Microsoft Visio does not support the M:N relationship directly. Therefore, to illustrate the existence of
a M:N relationship using Visio, you have to change the line style of the connector (see Appendix A, Designing
Databases with Visio Professional: A Tutorial, in the Premium Website).
Online Content
Aside from the Chen and Crow’s Foot notations, there are other ER model notations. For a summary of the
symbols used by several additional ER model notations, see Appendix D, Comparison of ER Model Notations,
in the Premium Website.
Its exceptional visual simplicity makes the ER model the dominant database modeling and design tool. Nevertheless,
the search for better data-modeling tools continues as the data environment continues to evolve.
2.5.4 The Object-Oriented (OO) Model
Increasingly complex real-world problems demonstrated a need for a data model that more closely represented the real
world. In the object-oriented data model (OODM), both data and their relationships are contained in a single
structure known as an object. In turn, the OODM is the basis for the object-oriented database management
system (OODBMS).
D A T A
M O D E L S
Online Content
This chapter introduces only basic OO concepts. You’ll have a chance to examine object-orientation concepts
and principles in detail in Appendix G, Object-Oriented Databases, in the Premium Website.
An OODM reflects a very different way to define and use entities. Like the relational model’s entity, an object is
described by its factual content. But quite unlike an entity, an object includes information about relationships between
the facts within the object, as well as information about its relationships with other objects. Therefore, the facts within
the object are given greater meaning. The OODM is said to be a semantic data model because semantic indicates
meaning.
Subsequent OODM development has allowed an object to also contain all operations that can be performed on it, such
as changing its data values, finding a specific data value, and printing data values. Because objects include data, various
types of relationships, and operational procedures, the object becomes self-contained, thus making the object—at least
potentially—a basic building block for autonomous structures.
The OO data model is based on the following components:
쐌
An object is an abstraction of a real-world entity. In general terms, an object may be considered equivalent to
an ER model’s entity. More precisely, an object represents only one occurrence of an entity. (The object’s
semantic content is defined through several of the items in this list.)
쐌
Attributes describe the properties of an object. For example, a PERSON object includes the attributes Name,
Social Security Number, and Date of Birth.
쐌
Objects that share similar characteristics are grouped in classes. A class is a collection of similar objects with
shared structure (attributes) and behavior (methods). In a general sense, a class resembles the ER model’s entity
set. However, a class is different from an entity set in that it contains a set of procedures known as methods.
A class’s method represents a real-world action such as finding a selected PERSON’s name, changing a
PERSON’s name, or printing a PERSON’s address. In other words, methods are the equivalent of procedures
in traditional programming languages. In OO terms, methods define an object’s behavior.
쐌
Classes are organized in a class hierarchy. The class hierarchy resembles an upside-down tree in which each
class has only one parent. For example, the CUSTOMER class and the EMPLOYEE class share a parent
PERSON class. (Note the similarity to the hierarchical data model in this respect.)
쐌
Inheritance is the ability of an object within the class hierarchy to inherit the attributes and methods of the
classes above it. For example, two classes, CUSTOMER and EMPLOYEE, can be created as subclasses from
the class PERSON. In this case, CUSTOMER and EMPLOYEE will inherit all attributes and methods from
PERSON.
Object-oriented data models are typically depicted using Unified Modeling Language (UML) class diagrams. Unified
Modeling Language (UML) is a language based on OO concepts that describes a set of diagrams and symbols that
can be used to graphically model a system. UML class diagrams are used to represent data and their relationships
within the larger UML object-oriented system’s modeling language. For a more complete description of UML see
Appendix H, Unified Modeling Language (UML).
To illustrate the main concepts of the object-oriented data model, let’s use a simple invoicing problem. In this case,
invoices are generated by customers, each invoice references one or more lines, and each line represents an item
purchased by a customer. Figure 2.4 illustrates the object representation for this simple invoicing problem, as well as
the equivalent UML class diagram and ER model. The object representation is a simple way to visualize a single object
occurrence.
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2
FIGURE
A comparison of OO, UML, and ER models
2.4
Object Representation
+generates
CUSTOMER
INVOICE
1..1
INV_DATE
INV_NUMBER
INV_SHIP_DATE
INV_TOTAL
1
CUSTOMER
LINE
ER Model
UML Class Diagram
M
+belongs to
INVOICE
+INV_NUMBER : Integer
0..* +INV_DATE : Date
+INV_SHIP_DATE : Date
+INV_TOTAL : Double
1..1
+generates
1..*
+belongs to
LINE
As you examine Figure 2.4, note that:
쐌
The object representation of the INVOICE includes all related objects within the same object box. Note that
the connectivities (1 and M) indicate the relationship of the related objects to the INVOICE. For example, the
1 next to the CUSTOMER object indicates that each INVOICE is related to only one CUSTOMER. The M next
to the LINE object indicates that each INVOICE contains many LINEs.
쐌
The UML class diagram uses three separate object classes (CUSTOMER, INVOICE, and LINE) and two
relationships to represent this simple invoicing problem. Note that the relationship connectivities are
represented by the 1..1, 0..*, and 1..* symbols and that the relationships are named in both ends to represent
the different “roles” that the objects play in the relationship.
쐌
The ER model also uses three separate entities and two relationships to represent this simple invoice problem.
The OODM advances were felt in many areas, from system modeling to programming. The added semantics of the
OODM allowed for a richer representation of complex objects. This in turn enabled applications to support
increasingly complex objects in innovative ways. As you will see in the next section, such evolutionary advances also
affected the relational model.
2.5.5 Newer Data Models: Object/Relational and XML
Facing the demand to support more complex data representations, the relational model’s main vendors evolved the
model further and created the extended relational data model (ERDM). The ERDM adds many of the OO model’s
features within the inherently simpler relational database structure. The ERDM gave birth to a new generation of
relational databases supporting OO features such as objects (encapsulated data and methods), extensible data types
based on classes, and inheritance. That’s why a DBMS based on the ERDM is often described as an object/relational
database management system (O/R DBMS).
The use of complex objects received a boost with the Internet revolution. When organizations integrated their business
models with the Internet, they realized the potential of the Internet to access, distribute, and exchange critical business
information. This resulted in the widespread adoption of the Internet as a business communication tool. It is in this
environment that Extensible Markup Language (XML) emerged as the de facto standard for the efficient and effective
exchange of structured, semistructured, and unstructured data. Organizations using XML data soon realized there was a
need to manage the large amounts of unstructured data such as word-processing documents, Web pages, e-mails,
diagrams, etc., found in most of today’s organizations. To address this need, XML databases emerged to manage
unstructured data within a native XML format (see Chapter 14, Database Connectivity and Web Technologies, for more
D A T A
M O D E L S
information about XML). At the same time, O/R DBMSs added support for XML-based documents within their relational
data structure.
2.5.6 The Future of Data Models
Today the O/R DBMS is the dominant database for business applications. Its success could be attributed to the model’s
conceptual simplicity, easy-to-use query language, high transaction performance, high availability, security, scalability,
and expandability. In contrast, the OO DBMS is popular in niche markets such as computer-aided drawing/computeraided manufacturing (CAD/CAM), geographic information systems (GIS), telecommunications, and multimedia, which
require support for complex objects.
The OO and the relational data models have two totally different approaches. The OO data model was created to
address very specific engineering needs, not the wide-ranging needs of general data management tasks. The relational
model was created with a focus on better data management based on a sound mathematical foundation. Given these
differences, it is not surprising that the growth of the OO market has been slow compared to the rapid growth of the
relational data model.
One area in which OO concepts have been very influential is systems development and programming languages. Most
contemporary programming languages are object-oriented (Java, Ruby, Perl, C#, .NET, to name a few). Also, there
is an increasing need to manage an organization’s unstructured data.
It is difficult to speculate on the future development of database models. Will unstructured data management overcome
structured data management? We think that each approach complements and augments the other. O/R databases have
proven to efficiently support structured and unstructured data management. Furthermore, history has shown that O/R
DBMS are remarkably adaptable in supporting ever-evolving data management needs. Two examples of this evolution are:
쐌
Hybrid DBMSs are emerging that retain the advantages of the relational model and at the same time provide
programmers with an object-oriented view of the underlying data. These types of databases preserve the
performance characteristics of the relational model and the semantically rich programmatic support of the
object-oriented model.
쐌
SQL data services, such as Microsoft SQL Data Services (SDS) on its Azure Services Platform, are becoming
a critical component of relational database vendors’ Internet service strategies. These “cloud-based” (that is,
remotely processed and Internet-based) data services make it possible for companies of any size to store their
data in relational databases without incurring expensive hardware, software, and personnel costs, while having
access to high-end database features such as failover, backup, high transaction rates, and global data
distribution. Companies can use a “pay as you go” system based primarily on their storage and bandwidth
utilization and the features used.
2.5.7 Data Models: A Summary
The evolution of DBMSs has always been driven by the search for new ways of modeling increasingly complex
real-world data. A summary of the most commonly recognized data models is shown in Figure 2.5.
In the evolution of data models, there are some common characteristics that data models must have in order to be
widely accepted:
쐌
A data model must show some degree of conceptual simplicity without compromising the semantic
completeness of the database. It does not make sense to have a data model that is more difficult to
conceptualize than the real world.
쐌
A data model must represent the real world as closely as possible. This goal is more easily realized by adding
more semantics to the model’s data representation. (Semantics concern the dynamic data behavior, while data
representation constitutes the static aspect of the real-world scenario.)
쐌
Representation of the real-world transformations (behavior) must be in compliance with the consistency and
integrity characteristics of any data model.
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C H A P T E R
2
FIGURE
The evolution of data models
2.5
Semantics in
Data Model
Comments
least
Hierarchical
Network
Relational
Entity Relationship
• Difficult to represent M:N relationships
(hierarchical only)
• Structural level dependence
• No ad hoc queries (record-at-a-time access)
• Access path predefined (navigational access)
• Conceptual simplicity (structual independence)
• Provides ad hoc queries (SQL)
• Set-oriented access
• Easy to understand (more semantics)
• Limited to conceptual modeling
(no implementation component)
Semantic
most
Object-Oriented
Extended Relational
(O/R DBMS)
• More semantics in data model
• Support for complex objects
• Inheritance (class hierarchy)
• Behavior
• Unstructured data (XML)
• XML data exchanges
Each new data model addresses the shortcomings of previous models. The network model replaced the hierarchical
model because the former made it much easier to represent complex (many-to-many) relationships. In turn, the relational
model offers several advantages over the hierarchical and network models through its simpler data representation,
superior data independence, and easy-to-use query language; these features made it the preferred data model for business
applications. The OO data model introduced support for complex data within a rich semantic framework. The ERDM
added many of the OO features to the relational model and allowed it to maintain its strong market share within the
business environment. And in recent years, successful data models have facilitated the development of database products
that incorporate unstructured data as well as provide support for easy data exchanges via XML.
It is important to note that not all data models are created equal; some data models are better suited than others for
some tasks. For example, conceptual models are better suited for high-level data modeling, while implementation
models are better for managing stored data for implementation purposes. The entity relationship model is an example
of a conceptual model, while the hierarchical and network models are examples of implementation models. At the
same time, some models, such as the relational model and the OODM, could be used as both conceptual and
implementation models. Table 2.2 summarizes the advantages and disadvantages of the various database models.
Yes
Yes
Yes
Yes
Network
Relational
Entity
relationship
Objectoriented
1. Semantic content is added.
2. Visual representation includes semantic content.
3. Inheritance promotes data integrity.
1. Conceptual simplicity is at least equal to that of the
hierarchical model.
2. It handles more relationship types, such as M:N and multiparent.
3. Data access is more flexible than in hierarchical and file system
models.
4. Data Owner/Member relationship promotes data integrity.
5. There is conformance to standards.
6. It includes data definition language (DDL) and data manipulation
language (DML) in DBMS.
1. Structural independence is promoted by the use of independent
tables. Changes in a table’s structure do not affect data access or
application programs.
2. Tabular view substantially improves conceptual simplicity, thereby
promoting easier database design, implementation, management,
and use.
3. Ad hoc query capability is based on SQL.
4. Powerful RDBMS isolates the end user from physical-level details
and improves implementation and management simplicity.
1. Visual modeling yields exceptional conceptual simplicity.
2. Visual representation makes it an effective communication tool.
3. It is integrated with dominant relational model.
2.
3.
4.
1.
1.
2.
3.
4.
There is limited constraint representation.
There is limited relationship representation.
There is no data manipulation language.
Loss of information content occurs when attributes are removed
from entities to avoid crowded displays. (This limitation has been
addressed in subsequent graphical versions.)
Slow development of standards caused vendors to supply their
own enhancements, thus eliminating a widely accepted standard.
It is a complex navigational system.
There is a steep learning curve.
High system overhead slows transactions.
1. The RDBMS requires substantial hardware and system software
overhead.
2. Conceptual simplicity gives relatively untrained people the tools to
use a good system poorly, and if unchecked, it may produce the
same data anomalies found in file systems.
3. It may promote “islands of information” problems as individuals
and departments can easily develop their own applications.
Note: All databases assume the use of a common data pool within the database. Therefore, all database models promote data sharing, thus eliminating the potential
problem of islands of information.
Yes
Yes
Yes
No
No
1. Complex implementation requires knowledge of physical data
storage characteristics.
2. Navigational system yields complex application development,
management, and use; requires knowledge of hierarchical path.
3. Changes in structure require changes in all application programs.
4. There are implementation limitations (no multiparent or M:N
relationships).
5. There is no data definition or data manipulation language in
the DBMS.
6. There is a lack of standards.
1. System complexity limits efficiency—still a navigational system.
2. Navigational system yields complex implementation, application
development, and management.
3. Structural changes require changes in all application programs.
Yes
Hierarchical
DISADVANTAGES
1.
2.
3.
4.
5.
It promotes data sharing.
Parent/Child relationship promotes conceptual simplicity.
Database security is provided and enforced by DBMS.
Parent/Child relationship promotes data integrity.
It is efficient with 1:M relationships.
ADVANTAGES
DATA
INDEPENDENCE
DATA
MODEL
STRUCTURAL
INDEPENDENCE
Advantages and Disadvantages of Various Database Models
2.2
TABLE
D A T A
M O D E L S
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2
Thus far, you have been introduced to the basic constructs of the more prominent data models. Each model uses such
constructs to capture the meaning of the real-world data environment. Table 2.3 shows the basic terminology used by
the various data models.
TABLE
2.3
Data Model Basic Terminology Comparison
REAL
EXAMPLE
FILE
HIERARCHICAL
NETWORK
RELATIONAL ER MODEL
OO
WORLD
PROCESSING
MODEL
MODEL
MODEL
MODEL
Vendor
File
Segment
Record
Table
Entity
Class
A group of
vendors
file cabinet
type
type
Set
A single
Global
Record
Segment
Current
Row
Entity
Object
vendor
supplies
occurrence
record
(tuple)
occurrence
instance
The contact
Johnny
Field
Segment
Record
Table
Entity
Object
name
Ventura
field
field
attribute
Attribute
attribute
The vendor
G12987
Index
Sequence
Record
Key
Entity
Object
identifier
field
key
Identifier
identifier
Note: For additional information about the terms used in this table, please consult the corresponding chapters and online appendixes accompanying this book. For example, if you want to know more about the OO model, refer to Appendix G, Object-Oriented
Databases.
2.6 DEGREES OF DATA ABSTRACTION
If you ask 10 database designers what a data model is, you will end up with 10 different answers—depending on the
degree of data abstraction. To illustrate the meaning of data abstraction, consider the example of automotive design.
A car designer begins by drawing the concept of the car that is to be produced. Next, engineers design the details that
help transfer the basic concept into a structure that can be produced. Finally, the engineering drawings are translated
into production specifications to be used on the factory floor. As you can see, the process of producing the car begins
at a high level of abstraction and proceeds to an ever-increasing level of detail. The factory floor process cannot
proceed unless the engineering details are properly specified, and the engineering details cannot exist without the basic
conceptual framework created by the designer. Designing a usable database follows the same basic process. That is,
a database designer starts with an abstract view of the overall data environment and adds details as the design comes
closer to implementation. Using levels of abstraction can also be very helpful in integrating multiple (and sometimes
conflicting) views of data as seen at different levels of an organization.
In the early 1970s, the American National Standards Institute (ANSI) Standards Planning and Requirements
Committee (SPARC) defined a framework for data modeling based on degrees of data abstraction. The ANSI/SPARC
architecture (as it is often referred to) defines three levels of data abstraction: external, conceptual, and internal. You
can use this framework to better understand database models, as shown in Figure 2.6. In the figure, the ANSI/SPARC
framework has been expanded with the addition of a physical model to explicitly address physical-level implementation
details of the internal model.
2.6.1 The External Model
The external model is the end users’ view of the data environment. The term end users refers to people who use
the application programs to manipulate the data and generate information. End users usually operate in an
environment in which an application has a specific business unit focus. Companies are generally divided into several
business units, such as sales, finance, and marketing. Each business unit is subject to specific constraints and
requirements, and each one uses a data subset of the overall data in the organization. Therefore, end users working
within those business units view their data subsets as separate from or external to other units within the organization.
D A T A
FIGURE
M O D E L S
Data abstraction levels
2.6
End-User View
End-User View
External
Model
External
Model
Degree of
Abstraction
Conceptual
Model
Designer’s
View
High
ER
Object-Oriented
Characteristics
Hardware-independent
Software-independent
Logical independence
Internal
Model
Medium
Relational
Hardware-independent
Software-dependent
Low
Network
Hierarchical
Hardware-dependent
Software-dependent
DBMS
View
Physical independence
Physical
Model
Because data are being modeled, ER diagrams will be used to represent the external views. A specific representation
of an external view is known as an external schema. To illustrate the external model’s view, examine the data
environment of Tiny College. Figure 2.7 presents the external schemas for two Tiny College business units: student
registration and class scheduling. Each external schema includes the appropriate entities, relationships, processes, and
constraints imposed by the business unit. Also note that although the application views are isolated from each other,
each view shares a common entity with the other view. For example, the registration and scheduling external
schemas share the entities CLASS and COURSE.
Note the entity relationships represented in Figure 2.7. For example:
쐌
A PROFESSOR may teach many CLASSes, and each CLASS is taught by only one PROFESSOR; that is,
there is a 1:M relationship between PROFESSOR and CLASS.
쐌
A CLASS may ENROLL many students, and each student may ENROLL in many CLASSes, thus creating an
M:N relationship between STUDENT and CLASS. (You will learn about the precise nature of the ENROLL
entity in Chapter 4.)
쐌
Each COURSE may generate many CLASSes, but each CLASS references a single COURSE. For example,
there may be several classes (sections) of a database course having a course code of CIS-420. One of those classes
might be offered on MWF from 8:00 a.m. to 8:50 a.m., another might be offered on MWF from 1:00 p.m. to
1:50 p.m., while a third might be offered on Thursdays from 6:00 p.m. to 8:40 p.m. Yet all three classes have
the course code CIS-420.
쐌
Finally, a CLASS requires one ROOM, but a ROOM may be scheduled for many CLASSes. That is, each
classroom may be used for several classes: one at 9:00 a.m., one at 11:00 a.m., and one at 1 p.m., for
example. In other words, there is a 1:M relationship between ROOM and CLASS.
47
48
C H A P T E R
FIGURE
2
External models for Tiny College
2.7
The use of external views representing subsets of the database has some important advantages:
쐌
It makes it easy to identify specific data required to support each business unit’s operations.
쐌
It makes the designer’s job easy by providing feedback about the model’s adequacy. Specifically, the model can
be checked to ensure that it supports all processes as defined by their external models, as well as all operational
requirements and constraints.
쐌
It helps to ensure security constraints in the database design. Damaging an entire database is more difficult
when each business unit works with only a subset of data.
쐌
It makes application program development much simpler.
2.6.2 The Conceptual Model
Having identified the external views, a conceptual model is used, graphically represented by an ERD (as in Figure 2.8),
to integrate all external views into a single view. The conceptual model represents a global view of the entire
database as viewed by the entire organization. That is, the conceptual model integrates all external views (entities,
relationships, constraints, and processes) into a single global view of the data in the enterprise. Also known as a
conceptual schema, it is the basis for the identification and high-level description of the main data objects (avoiding
any database model–specific details).
The most widely used conceptual model is the ER model. Remember that the ER model is illustrated with the help of the
ERD, which is, in effect, the basic database blueprint. The ERD is used to graphically represent the conceptual schema.
The conceptual model yields some very important advantages. First, it provides a relatively easily understood bird’s-eye
(macro level) view of the data environment. For example, you can get a summary of Tiny College’s data environment
by examining the conceptual model presented in Figure 2.8.
Second, the conceptual model is independent of both software and hardware. Software independence means that
the model does not depend on the DBMS software used to implement the model. Hardware independence means
that the model does not depend on the hardware used in the implementation of the model. Therefore, changes in
D A T A
FIGURE
Conceptual model for Tiny College
2.8
M O D E L S
either the hardware or the DBMS software will
have no effect on the database design at the
conceptual level. Generally, the term logical
design is used to refer to the task of creating a
conceptual data model that could be implemented in any DBMS.
2.6.3 The Internal Model
Once a specific DBMS has been selected, the
internal model maps the conceptual model to
the DBMS. The internal model is the representation of the database as “seen” by the
DBMS. In other words, the internal model
requires the designer to match the conceptual
model’s characteristics and constraints to those
of the selected implementation model. An
internal schema depicts a specific representation of an internal model, using the database
constructs supported by the chosen database.
Because this book focuses on the relational model, a relational database was chosen to implement the internal model.
Therefore, the internal schema should map the conceptual model to the relational model constructs. In particular, the
entities in the conceptual model are mapped to tables in the relational model. Likewise, because a relational database
has been selected, the internal schema is expressed using SQL, the standard language for relational databases. In the
case of the conceptual model for Tiny College depicted in Figure 2.8, the internal model was implemented by creating
the tables PROFESSOR, COURSE, CLASS, STUDENT, ENROLL, and ROOM. A simplified version of the internal
model for Tiny College is shown in Figure 2.9.
The development of a detailed internal model is especially important to database designers who work with hierarchical
or network models because those models require very precise specification of data storage location and data access
paths. In contrast, the relational model requires less detail in its internal model because most RDBMSs handle data
access path definition transparently; that is, the designer need not be aware of the data access path details.
Nevertheless, even relational database software usually requires data storage location specification, especially in a
mainframe environment. For example, DB2 requires that you specify the data storage group, the location of the
database within the storage group, and the location of the tables within the database.
Because the internal model depends on specific database software, it is said to be software-dependent. Therefore, a
change in the DBMS software requires that the internal model be changed to fit the characteristics and requirements
of the implementation database model. When you can change the internal model without affecting the conceptual
model, you have logical independence. However, the internal model is still hardware-independent because it is
unaffected by the choice of the computer on which the software is installed. Therefore, a change in storage devices
or even a change in operating systems will not affect the internal model.
2.6.4 The Physical Model
The physical model operates at the lowest level of abstraction, describing the way data are saved on storage media
such as disks or tapes. The physical model requires the definition of both the physical storage devices and the (physical)
access methods required to reach the data within those storage devices, making it both software- and hardwaredependent. The storage structures used are dependent on the software (the DBMS and the operating system) and on
the type of storage devices that the computer can handle. The precision required in the physical model’s definition
demands that database designers who work at this level have a detailed knowledge of the hardware and software used
to implement the database design.
49
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C H A P T E R
2
FIGURE
Internal model for Tiny College
2.9
Early data models forced the database designer to take the details of the physical model’s data storage requirements
into account. However, the now dominant relational model is aimed largely at the logical rather than the physical level;
therefore, it does not require the physical-level details common to its predecessors.
Although the relational model does not require the designer to be concerned about the data’s physical storage
characteristics, the implementation of a relational model may require physical-level fine-tuning for increased
performance. Fine-tuning is especially important when very large databases are installed in a mainframe environment.
Yet even such performance fine-tuning at the physical level does not require knowledge of physical data storage
characteristics.
As noted earlier, the physical model is dependent on the DBMS, methods of accessing files, and types of hardware
storage devices supported by the operating system. When you can change the physical model without affecting the
internal model, you have physical independence. Therefore, a change in storage devices or methods and even a
change in operating system will not affect the internal model.
A summary of the levels of data abstraction is given in Table 2.4.
TABLE
2.4
MODEL
External
Conceptual
Levels of Data Abstraction
DEGREE OF
ABSTRACTION
High
Internal
Physical
Low
FOCUS
End-user views
Global view of data
(database model−independent)
INDEPENDENT OF
Hardware and software
Hardware and software
Specific database model
Hardware
Storage and access methods
Neither hardware nor software
D A T A
M O D E L S
S u m m a r y
◗
◗
◗
◗
◗
◗
A data model is an abstraction of a complex real-world data environment. Database designers use data models to
communicate with applications programmers and end users. The basic data-modeling components are entities,
attributes, relationships, and constraints. Business rules are used to identify and define the basic modeling
components within a specific real-world environment.
The hierarchical and network data models were early data models that are no longer used, but some of the concepts
are found in current data models. The hierarchical model depicts a set of one-to-many (1:M) relationships between a
parent and its children segments. The network model uses sets to represent 1:M relationships between record types.
The relational model is the current database implementation standard. In the relational model, the end user
perceives the data as being stored in tables. Tables are related to each other by means of common values in
common attributes. The entity relationship (ER) model is a popular graphical tool for data modeling that
complements the relational model. The ER model allows database designers to visually present different views of
the data—as seen by database designers, programmers, and end users—and to integrate the data into a common
framework.
The object-oriented data model (OODM) uses objects as the basic modeling structure. An object resembles an entity
in that it includes the facts that define it. But unlike an entity, the object also includes information about
relationships between the facts, as well as relationships with other objects, thus giving its data more meaning.
The relational model has adopted many object-oriented (OO) extensions to become the extended relational data
model (ERDM). Object/relational database management systems (O/R DBMS) were developed to implement the
ERDM. At this point, the OODM is largely used in specialized engineering and scientific applications, while the
ERDM is primarily geared to business applications. Although the most likely future scenario is an increasing merger
of OODM and ERDM technologies, both are overshadowed by the need to develop Internet access strategies for
databases. Usually OO data models are depicted using Unified Modeling Language (UML) class diagrams.
Data-modeling requirements are a function of different data views (global vs. local) and the level of data abstraction.
The American National Standards Institute Standards Planning and Requirements Committee (ANSI/SPARC)
describes three levels of data abstraction: external, conceptual, and internal. There is also a fourth level of data
abstraction, the physical level. This lowest level of data abstraction is concerned exclusively with physical storage
methods.
K e y
T e r m s
American National Standards
Institute (ANSI), 46
Crow’s Foot notation, 39
entity set, 39
data definition language (DDL), 36
attribute, 31
business rule, 32
data management language
(DML), 36
extended relational data model
(ERDM), 42
Chen notation, 39
data model, 30
external schema, 47
class, 41
entity, 31
hardware independence, 48
class diagram, 41
entity instance, 39
hierarchical model, 35
class hierarchy, 41
entity occurrence, 39
hybrid DBMS , 43
conceptual model, 48
entity relationship diagram
(ERD), 38
inheritance, 41
entity relationship (ER) model
(ERM), 38
internal schema, 49
conceptual schema, 48
connectivity, 39
constraint, 32
external model, 46
internal model, 49
logical design, 49
logical independence, 49
51
52
C H A P T E R
2
many-to-many (M:N or *..*)
relationship, 32
one-to-many (1:M or 1..*)
relationship, 32
relationship, 31
method, 41
segment, 35
network model, 35
one-to-one (1:1 or 1..1)
relationship, 32
object, 40
physical independence, 50
software independence, 48
object-oriented data model
(OODM), 40
physical model, 49
subschema, 36
relation, 36
SQL data services, 43
object-oriented database
management system
(OODBMS), 40
relational database management
system (RDBMS), 36
table, 36
object/relational database
management system
(O/R DBMS), 42
relational model, 36
relational diagram, 37
schema, 35
semantic data model, 41
tuple, 36
Unified Modeling Language
(UML), 41
Online Content
Answers to selected Review Questions and Problems for this chapter are contained in the Premium Website for
this book.
R e v i e w
Q u e s t i o n s
1. Discuss the importance of data modeling.
2. What is a business rule, and what is its purpose in data modeling?
3. How do you translate business rules into data model components?
4. What languages emerged to standardize the basic network data model, and why was such standardization
important to users and designers?
5. Describe the basic features of the relational data model and discuss their importance to the end user and the designer.
6. Explain how the entity relationship (ER) model helped produce a more structured relational database design
environment.
7. Use the scenario described by “A customer can make many payments, but each payment is made by only one
customer” as the basis for an entity relationship diagram (ERD) representation.
8. Why is an object said to have greater semantic content than an entity?
9. What is the difference between an object and a class in the object-oriented data model (OODM)?
10. How would you model Question 7 with an OODM? (Use Figure 2.4 as your guide.)
11. What is an ERDM, and what role does it play in the modern (production) database environment?
12. In terms of data and structural independence, compare file system data management with the five data models
discussed in this chapter.
13. What is a relationship, and what three types of relationships exist?
14. Give an example of each of the three types of relationships.
15. What is a table, and what role does it play in the relational model?
16. What is a relational diagram? Give an example.
17. What is logical independence?
18. What is physical independence?
19. What is connectivity? (Use a Crow’s Foot ERD to illustrate connectivity.)
D A T A
M O D E L S
P r o b l e m s
Use the contents of Figure 2.1 to work Problems 1−3.
1. Write the business rule(s) that govern the relationship between AGENT and CUSTOMER.
2. Given the business rule(s) you wrote in Problem 1, create the basic Crow’s Foot ERD.
3. Using the ERD you drew in Problem 2, create the equivalent object representation and UML class diagram. (Use
Figure 2.4 as your guide.)
Using Figure P2.4 as your guide, work Problems 4–5. The DealCo relational diagram shows the initial entities and
attributes for the DealCo stores, located in two regions of the country.
FIGURE
The DealCo relational diagram
P2.4
4. Identify each relationship type and write all of the business rules.
5. Create the basic Crow’s Foot ERD for DealCo.
Using Figure P2.6 as your guide, work Problems 6−8. The Tiny College relational diagram shows the initial entities
and attributes for Tiny College.
FIGURE
The Tiny College relational diagram
P2.6
6. Identify each relationship type and write all of the business rules.
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C H A P T E R
2
7. Create the basic Crow’s Foot ERD for Tiny College.
8. Create the UML class diagram that reflects the entities and relationships you identified in the relational diagram.
9. Typically, a patient staying in a hospital receives medications that have been ordered by a particular doctor.
Because the patient often receives several medications per day, there is a 1:M relationship between PATIENT and
ORDER. Similarly, each order can include several medications, creating a 1:M relationship between ORDER and
MEDICATION.
a. Identify the business rules for PATIENT, ORDER, and MEDICATION.
b. Create a Crow’s Foot ERD that depicts a relational database model to capture these business rules.
10. United Broke Artists (UBA) is a broker for not-so-famous artists. UBA maintains a small database to track
painters, paintings, and galleries. A painting is painted by a particular artist, and that painting is exhibited in a
particular gallery. A gallery can exhibit many paintings, but each painting can be exhibited in only one gallery.
Similarly, a painting is painted by a single painter, but each painter can paint many paintings. Using PAINTER,
PAINTING, and GALLERY, in terms of a relational database:
a. What tables would you create, and what would the table components be?
b. How might the (independent) tables be related to one another?
11. Using the ERD from Problem 10, create the relational schema. (Create an appropriate collection of attributes for
each of the entities. Make sure you use the appropriate naming conventions to name the attributes.)
12. Convert the ERD from Problem 10 into the corresponding UML class diagram.
13. Describe the relationships (identify the business rules) depicted in the Crow’s Foot ERD shown in Figure P2.13.
FIGURE
P2.13
The Crow’s Foot ERD
for Problem 13
14. Create a Crow’s Foot ERD to include the following business rules for the ProdCo company:
a. Each sales representative writes many invoices.
b. Each invoice is written by one sales representative.
c. Each sales representative is assigned to one department.
d. Each department has many sales representatives.
e. Each customer can generate many invoices.
f.
Each invoice is generated by one customer.
D A T A
M O D E L S
15. Write the business rules that are reflected in the ERD shown in Figure P2.15. (Note that the ERD reflects some
simplifying assumptions. For example, each book is written by only one author. Also, remember that the ERD
is always read from the “1” to the “M” side, regardless of the orientation of the ERD components.)
FIGURE
P2.15
The Crow’s Foot ERD
for Problem 15
16. Create a Crow’s Foot ERD for each of the following descriptions. (Note: The word many merely means “more
than one” in the database modeling environment.)
a. Each of the MegaCo Corporation’s divisions is composed of many departments. Each department has many
employees assigned to it, but each employee works for only one department. Each department is managed
by one employee, and each of those managers can manage only one department at a time.
b. During some period of time, a customer can rent many videotapes from the BigVid store. Each of BigVid’s
videotapes can be rented to many customers during that period of time.
c. An airliner can be assigned to fly many flights, but each flight is flown by only one airliner.
d. The KwikTite Corporation operates many factories. Each factory is located in a region. Each region can be
“home” to many of KwikTite’s factories. Each factory employs many employees, but each of those
employees is employed by only one factory.
e. An employee may have earned many degrees, and each degree may have been earned by many employees.
55
PART
II
Design Concepts
The Relational Database Model
3
Entity Relationship (ER) Modeling
4
Advanced Data Modeling
5
Normalization of Database Tables
6
BP’s Data Modeling Initiative
British Petroleum is one of the largest energy companies in the world, engaged in fuel
exploration and production in 29 countries and actively developing alternative energy
sources such as solar and wind energy and biofuels. In this large, diverse corporation,
management is decentralized and IT expenditure and infrastructure development has
historically been project-driven. As a result, BP’s Information Technology and Services
(IT&S) division was unable to implement uniform IT standards and platforms throughout
the company. The company had adopted well over 5,000 software applications.
The decentralized company structure strongly impacted database development. Each
project created its own data models.The extent and approach to data modeling differed
with each project. Project managers used a large variety of data modeling tools, including
System Architecture, ERWin, ARIS, Enterprise Architecture, Visio, and even PowerPoint.
Moreover, there was no central repository where models and data definitions could be
stored. Once a project was finished, these models were frequently lost. So, BP suffered
from inconsistent data definitions, data duplication, and quality problems.
In 2003, BP decided to change all that. The company set a goal to manage data and
information “as a shared corporate asset that is easily accessible.” It created an Enterprise
Architecture team to identify common IT standards. By the end of 2005, the team had
conducted a cross-company data modeling study and created a list of agreed upon
requirements.The idea was to establish “data modeling as a service” to all business units.
The function of the Enterprise Architecture team would not be to enforce standards and
procedures, but to train, support, and provide resources.
Since potential users were located all over the globe, the team decided to build a data
modeling portal that would house all data modeling related resources: standards and
guidelines, discussion boards, registration for trainings, and a large data model repository
where data models are automatically uploaded and shared. To support this effort, BP
adopted a single data modeling tool, ER/Studio. Users could work in ER/Studio and the
data models would automatically be published to Microsoft SharePoint. By 2009, the
repository contained 235 models for over 50,000 entities.
The response from users has been very positive. A recent survey found that nearly all
users agree that they are benefiting from the use of a common modeling tool, a common
repository, and common standards and guidelines. In addition, the number of employees
using the portal has increased. These two indicators strongly suggest that BP’s “data
modeling as a service” strategy is overcoming the disadvantages created by its policies of
decentralized management and voluntary adoption.
B
V
usiness
ignette
T H R E E
3
The Relational Database Model
In this chapter, you will learn:
쐍 That the relational database model offers a logical view of data
쐍 About the relational model’s basic component: relations
쐍 That relations are logical constructs composed of rows (tuples) and columns (attributes)
쐍 That relations are implemented as tables in a relational DBMS
쐍 About relational database operators, the data dictionary, and the system catalog
쐍 How data redundancy is handled in the relational database model
쐍 Why indexing is important
In Chapter 2, Data Models, you learned that the relational data model’s structural and data
independence allow you to examine the model’s logical structure without considering the
physical aspects of data storage and retrieval. You also learned that entity relationship
diagrams (ERDs) may be used to depict entities and their relationships graphically. In this
chapter, you will learn some important details about the relational model’s logical structure
and more about how the ERD can be used to design a relational database.
You will also learn how the relational database’s basic data components fit into a logical
construct known as a table.You will discover that one important reason for the relational
database model’s simplicity is that its tables can be treated as logical rather than physical
units.You will also learn how the independent tables within the database can be related to
one another.
After learning about tables, their components, and their relationships, you will be introduced
to the basic concepts that shape the design of tables. Because the table is such an integral
part of relational database design, you will also learn the characteristics of well-designed and
poorly designed tables.
Finally, you will be introduced to some basic concepts that will become your gateway to the
next few chapters. For example, you will examine different kinds of relationships and the
way those relationships might be handled in the relational database environment.
P
review
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
Note
The relational model, introduced by E. F. Codd in 1970, is based on predicate logic and set theory. Predicate logic,
used extensively in mathematics, provides a framework in which an assertion (statement of fact) can be verified as
either true or false. For example, suppose that a student with a student ID of 12345678 is named Melissa Sanduski.
This assertion can easily be demonstrated to be true or false. Set theory is a mathematical science that deals with
sets, or groups of things, and is used as the basis for data manipulation in the relational model. For example,
assume that set A contains three numbers: 16, 24, and 77. This set is represented as A(16, 24, 77). Furthermore,
set B contains four numbers: 44, 77, 90, and 11, and so is represented as B(44, 77, 90, 11). Given this information,
you can conclude that the intersection of A and B yields a result set with a single number, 77. This result can be
expressed as A 艚 B = 77. In other words, A and B share a common value, 77.
Based on these concepts, the relational model has three well-defined components:
1. A logical data structure represented by relations (Sections 3.1, 3.2, and 3.5).
2. A set of integrity rules to enforce that the data are and remain consistent over time (Sections 3.3, 3.6, 3.7,
and 3.8).
3. A set of operations that defines how data are manipulated (Section 3.4).
3.1 A LOGICAL VIEW OF DATA
In Chapter 1, Database Systems, you learned that a database stores and manages both data and metadata. You also
learned that the DBMS manages and controls access to the data and the database structure. Such an arrangement—
placing the DBMS between the application and the database—eliminates most of the file system’s inherent limitations.
The result of such flexibility, however, is a far more complex physical structure. In fact, the database structures required
by both the hierarchical and network database models often become complicated enough to diminish efficient database
design. The relational data model changed all of that by allowing the designer to focus on the logical representation
of the data and its relationships, rather than on the physical storage details. To use an automotive analogy, the
relational database uses an automatic transmission to relieve you of the need to manipulate clutch pedals and
gearshifts. In short, the relational model enables you to view data logically rather than physically.
The practical significance of taking the logical view is that it serves as a reminder of the simple file concept of data
storage. Although the use of a table, quite unlike that of a file, has the advantages of structural and data independence,
a table does resemble a file from a conceptual point of view. Because you can think of related records as being stored
in independent tables, the relational database model is much easier to understand than the hierarchical and network
models. Logical simplicity tends to yield simple and effective database design methodologies.
Because the table plays such a prominent role in the relational model, it deserves a closer look. Therefore, our
discussion begins with an exploration of the details of table structure and contents.
3.1.1 Tables and Their Characteristics
The logical view of the relational database is facilitated by the creation of data relationships based on a logical construct
known as a relation. Because a relation is a mathematical construct, end users find it much easier to think of a relation
as a table. A table is perceived as a two-dimensional structure composed of rows and columns. A table is also called
a relation because the relational model’s creator, E. F. Codd, used the term relation as a synonym for table. You can
think of a table as a persistent representation of a logical relation, that is, a relation whose contents can be
permanently saved for future use. As far as the table’s user is concerned, a table contains a group of related entity
occurrences, that is, an entity set. For example, a STUDENT table contains a collection of entity occurrences, each
representing a student. For that reason, the terms entity set and table are often used interchangeably.
59
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C H A P T E R
3
Note
The word relation, also known as a dataset in Microsoft Access, is based on the mathematical set theory from
which Codd derived his model. Because the relational model uses attribute values to establish relationships
among tables, many database users incorrectly assume that the term relation refers to such relationships. Many
then incorrectly conclude that only the relational model permits the use of relationships.
You will discover that the table view of data makes it easy to spot and define entity relationships, thereby greatly
simplifying the task of database design. The characteristics of a relational table are summarized in Table 3.1.
TABLE
3.1
1
2
3
4
5
6
7
8
Characteristics of a Relational Table
A table is perceived as a two-dimensional structure composed of rows and columns.
Each table row (tuple) represents a single entity occurrence within the entity set.
Each table column represents an attribute, and each column has a distinct name.
Each row/column intersection represents a single data value.
All values in a column must conform to the same data format.
Each column has a specific range of values known as the attribute domain.
The order of the rows and columns is immaterial to the DBMS.
Each table must have an attribute or a combination of attributes that uniquely identifies each row.
The table shown in Figure 3.1 illustrates the characteristics listed in Table 3.1.
FIGURE
STUDENT table attribute values
3.1
Table name: STUDENT
STU_NUM
STU_LNAME
STU_FNAME
STU_INIT
STU_DOB
STU_HRS
STU_CLASS
STU_GPA
STU_TRANSFER
DEPT_CODE
STU_PHONE
PROF_NUM
= Student number
= Student last name
= Student first name
= Student middle initial
= Student date of birth
= Credit hours earned
= Student classification
= Grade point average
= Student transferred from another institution
= Department code
= 4-digit campus phone extension
= Number of the professor who is the student’s advisor
Database name: Ch03_TinyCollege
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
Note
Relational database terminology is very precise. Unfortunately, file system terminology sometimes creeps into
the database environment. Thus, rows are sometimes referred to as records and columns are sometimes labeled
as fields. Occasionally, tables are labeled files. Technically speaking, this substitution of terms is not always
appropriate; the database table is a logical rather than a physical concept, and the terms file, record, and field
describe physical concepts. Nevertheless, as long as you recognize that the table is actually a logical rather than
a physical construct, you may (at the conceptual level) think of table rows as records and of table columns as
fields. In fact, many database software vendors still use this familiar file system terminology.
Online Content
All of the databases used to illustrate the material in this chapter are found in the Premium Website for this book.
The database names used in the folder match the database names used in the figures. For example, the source
of the tables shown in Figure 3.1 is the Ch03_TinyCollege database.
Using the STUDENT table shown in Figure 3.1, you can draw the following conclusions corresponding to the points
in Table 3.1:
1.
The STUDENT table is perceived to be a two-dimensional structure composed of eight rows (tuples) and twelve
columns (attributes).
2.
Each row in the STUDENT table describes a single entity occurrence within the entity set. (The entity set is
represented by the STUDENT table.) For example, row 4 in Figure 3.1 describes a student named Walter H.
Oblonski. Given the table contents, the STUDENT entity set includes eight distinct entities (rows), or students.
3.
Each column represents an attribute, and each column has a distinct name.
4.
All of the values in a column match the attribute’s characteristics. For example, the grade point average
(STU_GPA) column contains only STU_GPA entries for each of the table rows. Data must be classified
according to their format and function. Although various DBMSs can support different data types, most
support at least the following:
a. Numeric. Numeric data are data on which you can perform meaningful arithmetic procedures. For
example, in Figure 3.1, STU_HRS and STU_GPA are numeric attributes.
b. Character. Character data, also known as text data or string data, can contain any character or symbol not
intended for mathematical manipulation. In Figure 3.1, STU_CLASS and STU_PHONE are examples of
character attributes.
c. Date. Date attributes contain calendar dates stored in a special format known as the Julian date format.
For example, STU_DOB in Figure 3.1 is a date attribute.
d. Logical. Logical data can only have true or false (yes or no) values. In Figure 3.1, the STU_TRANSFER
attribute uses a logical data format.
5.
The column’s range of permissible values is known as its domain. Because the STU_GPA values are limited
to the range 0–4, inclusive, the domain is [0,4].
6.
The order of rows and columns is immaterial to the user.
7.
Each table must have a primary key. In general terms, the primary key (PK) is an attribute (or a combination
of attributes) that uniquely identifies any given row. In this case, STU_NUM (the student number) is the primary
key. Using the data presented in Figure 3.1, observe that a student’s last name (STU_LNAME) would not be
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a good primary key because it is possible to find several students whose last name is Smith. Even the
combination of the last name and first name (STU_FNAME) would not be an appropriate primary key because,
as Figure 3.1 shows, it is quite possible to find more than one student named John Smith.
3.2 KEYS
In the relational model, keys are important because they are used to ensure that each row in a table is uniquely
identifiable. They are also used to establish relationships among tables and to ensure the integrity of the data.
Therefore, a proper understanding of the concept and use of keys in the relational model is very important. A key
consists of one or more attributes that determine other attributes. For example, an invoice number identifies all of the
invoice attributes, such as the invoice date and the customer name.
One type of key, the primary key, has already been introduced. Given the structure of the STUDENT table shown in
Figure 3.1, defining and describing the primary key seem simple enough. However, because the primary key plays such
an important role in the relational environment, you will examine the primary key’s properties more carefully. In this
section, you also will become acquainted with superkeys, candidate keys, and secondary keys.
The key’s role is based on a concept known as determination. In the context of a database table, the statement “A
determines B” indicates that if you know the value of attribute A, you can look up (determine) the value of attribute
B. For example, knowing the STU_NUM in the STUDENT table (see Figure 3.1) means that you are able to look up
(determine) that student’s last name, grade point average, phone number, and so on. The shorthand notation for “A
determines B” is A → B. If A determines B, C, and D, you write A → B, C, D. Therefore, using the attributes of the
STUDENT table in Figure 3.1, you can represent the statement “STU_NUM determines STU_LNAME” by writing:
STU_NUM → STU_LNAME
In fact, the STU_NUM value in the STUDENT table determines all of the student’s attribute values. For example, you
can write:
STU_NUM → STU_LNAME, STU_FNAME, STU_INIT
and
STU_NUM → STU_LNAME, STU_FNAME, STU_INIT, STU_DOB, STU_TRANSFER
In contrast, STU_NUM is not determined by STU_LNAME because it is quite possible for several students to have the
last name Smith.
The principle of determination is very important because it is used in the definition of a central relational database
concept known as functional dependence. The term functional dependence can be defined most easily this way: the
attribute B is functionally dependent on A if A determines B. More precisely:
The attribute B is functionally dependent on the attribute A if each value in column A determines
one and only one value in column B.
Using the contents of the STUDENT table in Figure 3.1, it is appropriate to say that STU_PHONE is functionally
dependent on STU_NUM. For example, the STU_NUM value 321452 determines the STU_PHONE value 2134. On
the other hand, STU_NUM is not functionally dependent on STU_PHONE because the STU_PHONE value 2267 is
associated with two STU_NUM values: 324274 and 324291. (This could happen when roommates share a single land
line phone number.) Similarly, the STU_NUM value 324273 determines the STU_LNAME value Smith. But the
STU_NUM value is not functionally dependent on STU_LNAME because more than one student may have the last
name Smith.
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
The functional dependence definition can be generalized to cover the case in which the determining attribute values
occur more than once in a table. Functional dependence can then be defined this way:1
Attribute A determines attribute B (that is, B is functionally dependent on A) if all of the rows in the
table that agree in value for attribute A also agree in value for attribute B.
Be careful when defining the dependency’s direction. For example, Gigantic State University determines its student
classification based on hours completed; these are shown in Table 3.2.
TABLE
3.2
Therefore, you can write:
Student Classification
HOURS COMPLETED
Less than 30
30−59
60−89
90 or more
CLASSIFICATION
Fr
So
Jr
Sr
STU_HRS → STU_CLASS
But the specific number of hours is not dependent on the
classification. It is quite possible to find a junior with 62
completed hours or one with 84 completed hours. In other
words, the classification (STU_CLASS) does not determine
one and only one value for completed hours (STU_HRS).
Keep in mind that it might take more than a single attribute
to define functional dependence; that is, a key may be composed of more than one attribute. Such a multiattribute key
is known as a composite key.
Any attribute that is part of a key is known as a key attribute. For instance, in the STUDENT table, the student’s
last name would not be sufficient to serve as a key. On the other hand, the combination of last name, first name, initial,
and phone is very likely to produce unique matches for the remaining attributes. For example, you can write:
STU_LNAME, STU_FNAME, STU_INIT, STU_PHONE → STU_HRS, STU_CLASS
or
STU_LNAME, STU_FNAME, STU_INIT, STU_PHONE → STU_HRS, STU_CLASS, STU_GPA
or
STU_LNAME, STU_FNAME, STU_INIT, STU_PHONE → STU_HRS, STU_CLASS, STU_GPA, STU_DOB
Given the possible existence of a composite key, the notion of functional dependence can be further refined by
specifying full functional dependence:
If the attribute (B) is functionally dependent on a composite key (A) but not on any subset of that
composite key, the attribute (B) is fully functionally dependent on (A).
Within the broad key classification, several specialized keys can be defined. For example, a superkey is any key that
uniquely identifies each row. In short, the superkey functionally determines all of a row’s attributes. In the STUDENT
table, the superkey could be any of the following:
STU_NUM
STU_NUM, STU_LNAME
STU_NUM, STU_LNAME, STU_INIT
In fact, STU_NUM, with or without additional attributes, can be a superkey even when the additional attributes are
redundant.
1 SQL:2003 ANSI standard specification. ISO/IEC 9075-2:2003 – SQL/Foundation.
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A candidate key can be described as a superkey without unnecessary attributes, that is, a minimal superkey. Using
this distinction, note that the composite key
STU_NUM, STU_LNAME
is a superkey, but it is not a candidate key because STU_NUM by itself is a candidate key! The combination
STU_LNAME, STU_FNAME, STU_INIT, STU_PHONE
might also be a candidate key, as long as you discount the possibility that two students share the same last name, first
name, initial, and phone number.
If the student’s Social Security number had been included as one of the attributes in the STUDENT table in
Figure 3.1—perhaps named STU_SSN—both it and STU_NUM would have been candidate keys because either one
would uniquely identify each student. In that case, the selection of STU_NUM as the primary key would be driven by
the designer’s choice or by end-user requirements. In short, the primary key is the candidate key chosen to be the
unique row identifier. Note, incidentally, that a primary key is a superkey as well as a candidate key.
Within a table, each primary key value must be unique to ensure that each row is uniquely identified by the primary
key. In that case, the table is said to exhibit entity integrity. To maintain entity integrity, a null (that is, no data entry
at all) is not permitted in the primary key.
Note
A null is no value at all. It does not mean a zero or a space. A null is created when you press the Enter key or
the Tab key to move to the next entry without making a prior entry of any kind. Pressing the Spacebar creates
a blank (or a space).
Nulls can never be part of a primary key, and they should be avoided—to the greatest extent possible—in other
attributes, too. There are rare cases in which nulls cannot be reasonably avoided when you are working with nonkey
attributes. For example, one of an EMPLOYEE table’s attributes is likely to be the EMP_INITIAL. However, some
employees do not have a middle initial. Therefore, some of the EMP_INITIAL values may be null. You will also discover
later in this section that there may be situations in which a null exists because of the nature of the relationship between
two entities. In any case, even if nulls cannot always be avoided, they must be used sparingly. In fact, the existence of
nulls in a table is often an indication of poor database design.
Nulls, if used improperly, can create problems because they have many different meanings. For example, a null can
represent:
쐌
An unknown attribute value.
쐌
A known, but missing, attribute value.
쐌
A “not applicable” condition.
Depending on the sophistication of the application development software, nulls can create problems when functions
such as COUNT, AVERAGE, and SUM are used. In addition, nulls can create logical problems when relational tables
are linked.
Controlled redundancy makes the relational database work. Tables within the database share common attributes that
enable the tables to be linked together. For example, note that the PRODUCT and VENDOR tables in Figure 3.2 share
a common attribute named VEND_CODE. And note that the PRODUCT table’s VEND_CODE value 232 occurs more
than once, as does the VEND_CODE value 235. Because the PRODUCT table is related to the VENDOR table
through these VEND_CODE values, the multiple occurrence of the values is required to make the 1:M relationship
between VENDOR and PRODUCT work. Each VEND_CODE value in the VENDOR table is unique—the VENDOR
is the “1” side in the VENDOR-PRODUCT relationship. But any given VEND_CODE value from the VENDOR table
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
may occur more than once in the PRODUCT table, thus providing evidence that PRODUCT is the “M” side of the
VENDOR-PRODUCT relationship. In database terms, the multiple occurrences of the VEND_CODE values in the
PRODUCT table are not redundant because they are required to make the relationship work. You should recall from
Chapter 2 that data redundancy exists only when there is unnecessary duplication of attribute values.
FIGURE
An example of a simple relational database
3.2
Table name: PRODUCT
Primary key: PROD_CODE
Foreign key: VEND_CODE
Database name: Ch03_SaleCo
link
Table name: VENDOR
Primary key: VEND_CODE
Foreign key: none
As you examine Figure 3.2, note that the VEND_CODE value in one table can be used to point to the corresponding
value in the other table. For example, the VEND_CODE value 235 in the PRODUCT table points to vendor Henry
Ortozo in the VENDOR table. Consequently, you discover that the product “Houselite chain saw, 16-in. bar” is
delivered by Henry Ortozo and that he can be contacted by calling 615-899-3425. The same connection can be made
for the product “Steel tape, 12-ft. length” in the PRODUCT table.
Remember the naming convention—the prefix PROD was used in Figure 3.2 to indicate that the attributes “belong”
to the PRODUCT table. Therefore, the prefix VEND in the PRODUCT table’s VEND_CODE indicates that
VEND_CODE points to some other table in the database. In this case, the VEND prefix is used to point to the
VENDOR table in the database.
A relational database can also be represented by a relational schema. A relational schema is a textual representation
of the database tables where each table is listed by its name followed by the list of its attributes in parentheses. The
primary key attribute(s) is (are) underlined. You will see such schemas in Chapter 6, Normalization of Database Tables.
For example, the relational schema for Figure 3.2 would be shown as:
VENDOR (VEND_CODE, VEND_CONTACT, VEND_AREACODE, VEND_PHONE)
PRODUCT (PROD_CODE, PROD_DESCRIPT, PROD_PRICE, PROD_ON_HAND, VEND_CODE)
The link between the PRODUCT and VENDOR tables in Figure 3.2 can also be represented by the relational diagram
shown in Figure 3.3. In this case, the link is indicated by the line that connects the VENDOR and PRODUCT tables.
Note that the link in Figure 3.3 is the equivalent of the relationship line in an ERD. This link is created when two tables
share an attribute with common values. More specifically, the primary key of one table (VENDOR) appears as the
foreign key in a related table (PRODUCT). A foreign key (FK) is an attribute whose values match the primary key
values in the related table. For example, in Figure 3.2, the VEND_CODE is the primary key in the VENDOR table,
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FIGURE
3.3
The relational diagram for
the Ch03_SaleCo database
and it occurs as a foreign key in the PRODUCT table.
Because the VENDOR table is not linked to a third table, the
VENDOR table shown in Figure 3.2 does not contain a
foreign key.
If the foreign key contains either matching values or nulls,
the table that makes use of that foreign key is said to exhibit
referential integrity. In other words, referential integrity
means that if the foreign key contains a value, that value
refers to an existing valid tuple (row) in another relation.
Note that referential integrity is maintained between the
PRODUCT and VENDOR tables shown in Figure 3.2.
Finally, a secondary key is defined as a key that is used strictly for data retrieval purposes. Suppose customer data
are stored in a CUSTOMER table in which the customer number is the primary key. Do you suppose that most
customers will remember their numbers? Data retrieval for a customer can be facilitated when the customer’s last name
and phone number are used. In that case, the primary key is the customer number; the secondary key is the
combination of the customer’s last name and phone number. Keep in mind that a secondary key does not necessarily
yield a unique outcome. For example, a customer’s last name and home telephone number could easily yield several
matches where one family lives together and shares a phone line. A less efficient secondary key would be the
combination of the last name and zip code; this could yield dozens of matches, which could then be combed for a
specific match.
A secondary key’s effectiveness in narrowing down a search depends on how restrictive that secondary key is. For
instance, although the secondary key CUS_CITY is legitimate from a database point of view, the attribute values “New
York” or “Sydney” are not likely to produce a usable return unless you want to examine millions of possible matches.
(Of course, CUS_CITY is a better secondary key than CUS_COUNTRY.)
Table 3.3 summarizes the various relational database table keys.
TABLE
3.3
Relational Database Keys
KEY TYPE
Superkey
Candidate key
Primary key
Secondary key
Foreign key
DEFINITION
An attribute (or combination of attributes) that uniquely identifies each row in a table.
A minimal (irreducible) superkey. A superkey that does not contain a subset of attributes
that is itself a superkey.
A candidate key selected to uniquely identify all other attribute values in any given row.
Cannot contain null entries.
An attribute (or combination of attributes) used strictly for data retrieval purposes.
An attribute (or combination of attributes) in one table whose values must either match the
primary key in another table or be null.
3.3 INTEGRITY RULES
Relational database integrity rules are very important to good database design. Many (but by no means all) RDBMSs
enforce integrity rules automatically. However, it is much safer to make sure that your application design conforms to
the entity and referential integrity rules mentioned in this chapter. Those rules are summarized in Table 3.4.
T H E
TABLE
R E L A T I O N A L
D A T A B A S E
M O D E L
Integrity Rules
3.4
ENTITY INTEGRITY
Requirement
Purpose
Example
REFERENCE INTEGRITY
Requirement
Purpose
Example
DESCRIPTION
All primary key entries are unique, and no part of a primary key may
be null.
Each row will have a unique identity, and foreign key values can properly
reference primary key values.
No invoice can have a duplicate number, nor can it be null. In short, all
invoices are uniquely identified by their invoice number.
DESCRIPTION
A foreign key may have either a null entry, as long as it is not a part of its
table’s primary key, or an entry that matches the primary key value in a
table to which it is related. (Every non-null foreign key value must reference an existing primary key value.)
It is possible for an attribute NOT to have a corresponding value, but it will
be impossible to have an invalid entry. The enforcement of the referential
integrity rule makes it impossible to delete a row in one table whose primary key has mandatory matching foreign key values in another table.
A customer might not yet have an assigned sales representative (number),
but it will be impossible to have an invalid sales representative (number).
The integrity rules summarized in Table 3.4 are illustrated in Figure 3.4.
FIGURE
An illustration of integrity rules
3.4
Table name: CUSTOMER
Primary key: CUS_CODE
Foreign key: AGENT_CODE
Database name: Ch03_InsureCo
Table name: AGENT
Primary key: AGENT_CODE
Foreign key: none
Note the following features of Figure 3.4.
1.
Entity integrity. The CUSTOMER table’s primary key is CUS_CODE. The CUSTOMER primary key column
has no null entries, and all entries are unique. Similarly, the AGENT table’s primary key is AGENT_CODE, and
this primary key column is also free of null entries.
2.
Referential integrity. The CUSTOMER table contains a foreign key, AGENT_CODE, which links entries in
the CUSTOMER table to the AGENT table. The CUS_CODE row that is identified by the (primary key) number
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10013 contains a null entry in its AGENT_CODE foreign key because Mr. Paul F. Olowski does not yet have
a sales representative assigned to him. The remaining AGENT_CODE entries in the CUSTOMER table all
match the AGENT_CODE entries in the AGENT table.
To avoid nulls, some designers use special codes, known as flags, to indicate the absence of some value. Using
Figure 3.4 as an example, the code -99 could be used as the AGENT_CODE entry of the fourth row of the
CUSTOMER table to indicate that customer Paul Olowski does not yet have an agent assigned to him. If such a flag
is used, the AGENT table must contain a dummy row with an AGENT_CODE value of -99. Thus, the AGENT table’s
first record might contain the values shown in Table 3.5.
TABLE
A Dummy Variable Value Used as a Flag
3.5
AGENT_CODE
-99
AGENT_AREACODE
000
AGENT_PHONE
000-0000
AGENT_LNAME
None
AGENT_YTD_SALES
$0.00
Chapter 4, Entity Relationship (ER) Modeling, discusses several ways in which nulls may be handled.
Other integrity rules that can be enforced in the relational model are the NOT NULL and UNIQUE constraints. The
NOT NULL constraint can be placed on a column to ensure that every row in the table has a value for that column.
The UNIQUE constraint is a restriction placed on a column to ensure that no duplicate values exist for that column.
3.4 RELATIONAL SET OPERATORS
The data in relational tables are of limited value unless the data can be manipulated to generate useful information. This
section describes the basic data manipulation capabilities of the relational model. Relational algebra defines the
theoretical way of manipulating table contents using the eight relational operators: SELECT, PROJECT, JOIN,
INTERSECT, UNION, DIFFERENCE, PRODUCT, and DIVIDE. In Chapter 7, Introduction to Structured Query
Language (SQL), and Chapter 8, Advanced SQL, you will learn how SQL commands can be used to accomplish
relational algebra operations.
Note
The degree of relational completeness can be defined by the extent to which relational algebra is supported. To
be considered minimally relational, the DBMS must support the key relational operators SELECT, PROJECT, and
JOIN. Very few DBMSs are capable of supporting all eight relational operators.
The relational operators have the property of closure; that is, the use of relational algebra operators on existing
relations (tables) produces new relations. There is no need to examine the mathematical definitions, properties, and
characteristics of those relational algebra operators. However, their use can easily be illustrated as follows:
1.
SELECT, also known as RESTRICT, yields values for all rows found in a table that satisfy a given condition.
SELECT can be used to list all of the row values, or it can yield only those row values that match a specified
criterion. In other words, SELECT yields a horizontal subset of a table. The effect of a SELECT is shown in
Figure 3.5.
2.
PROJECT yields all values for selected attributes. In other words, PROJECT yields a vertical subset of a table.
The effect of a PROJECT is shown in Figure 3.6.
T H E
FIGURE
R E L A T I O N A L
D A T A B A S E
M O D E L
SELECT
3.5
Original table
New table
SELECT ALL yields
SELECT only PRICE less than $2.00 yields
SELECT only P_CODE = 311452 yields
FIGURE
PROJECT
3.6
Original table
New table
PROJECT PRICE yields
PROJECT P_DESCRIPT and PRICE yields
PROJECT P_CODE and PRICE yields
3.
UNION combines all rows from two tables, excluding duplicate rows. The tables must have the same attribute
characteristics (the columns and domains must be compatible) to be used in the UNION. When two or more
tables share the same number of columns, and when their corresponding columns share the same (or
compatible) domains, they are said to be union-compatible. The effect of a UNION is shown in Figure 3.7.
4.
INTERSECT yields only the rows that appear in both tables. As was true in the case of UNION, the tables must
be union-compatible to yield valid results. For example, you cannot use INTERSECT if one of the attributes is
numeric and one is character-based. The effect of an INTERSECT is shown in Figure 3.8.
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FIGURE
3
UNION
3.7
UNION
FIGURE
yields
INTERSECT
3.8
INTERSECT
5.
yields
DIFFERENCE yields all rows in one table that are not found in the other table; that is, it subtracts one table
from the other. As was true in the case of UNION, the tables must be union-compatible to yield valid results.
The effect of a DIFFERENCE is shown in Figure 3.9. However, note that subtracting the first table from the
second table is not the same as subtracting the second table from the first table.
FIGURE
DIFFERENCE
3.9
DIFFERENCE
yields
6.
PRODUCT yields all possible pairs of rows from two tables—also known as the Cartesian product. Therefore,
if one table has six rows and the other table has three rows, the PRODUCT yields a list composed of 6 × 3
= 18 rows. The effect of a PRODUCT is shown in Figure 3.10.
7.
JOIN allows information to be combined from two or more tables. JOIN is the real power behind the relational
database, allowing the use of independent tables linked by common attributes. The CUSTOMER and AGENT
tables shown in Figure 3.11 will be used to illustrate several types of joins.
T H E
FIGURE
R E L A T I O N A L
D A T A B A S E
M O D E L
PRODUCT
3.10
PRODUCT
FIGURE
yields
Two tables that will be used in join illustrations
3.11
Table name: CUSTOMER
Table name: AGENT
A natural join links tables by selecting only the rows with common values in their common attribute(s). A
natural join is the result of a three-stage process:
a. First, a PRODUCT of the tables is created, yielding the results shown in Figure 3.12.
b. Second, a SELECT is performed on the output of Step a to yield only the rows for which the
AGENT_CODE values are equal. The common columns are referred to as the join columns. Step b yields
the results shown in Figure 3.13.
c. A PROJECT is performed on the results of Step b to yield a single copy of each attribute, thereby
eliminating duplicate columns. Step c yields the output shown in Figure 3.14.
The final outcome of a natural join yields a table that does not include unmatched pairs and provides only the copies
of the matches.
Note a few crucial features of the natural join operation:
쐌
If no match is made between the table rows, the new table does not include the unmatched row. In that case,
neither AGENT_CODE 421 nor the customer whose last name is Smithson is included. Smithson’s
AGENT_CODE 421 does not match any entry in the AGENT table.
쐌
The column on which the join was made—that is, AGENT_CODE—occurs only once in the new table.
쐌
If the same AGENT_CODE were to occur several times in the AGENT table, a customer would be listed for
each match. For example, if the AGENT_CODE 167 were to occur three times in the AGENT table, the
customer named Rakowski, who is associated with AGENT_CODE 167, would occur three times in the
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FIGURE
3
Natural join, Step 1: PRODUCT
3.12
FIGURE
Natural join, Step 2: SELECT
3.13
FIGURE
3.14
Natural join, Step 3: PROJECT
resulting table. (A good AGENT table
cannot, of course, yield such a result
because it would contain unique primary key values.)
Another form of join, known as equijoin, links
tables on the basis of an equality condition that
compares specified columns of each table. The
outcome of the equijoin does not eliminate
duplicate columns, and the condition or criterion used to join the tables must be explicitly
defined. The equijoin takes its name from the equality comparison operator (=) used in the condition. If any other
comparison operator is used, the join is called a theta join.
Each of the preceding joins is often classified as an inner join. An inner join is a join that only returns matched records
from the tables that are being joined. In an outer join, the matched pairs would be retained, and any unmatched
values in the other table would be left null. It is an easy mistake to think that an outer join is the opposite of an
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
inner join. However, it is more accurate to think of an outer join as an “inner join plus.” The outer join still returns
all of the matched records that the inner join returns, plus it returns the unmatched records from one of the tables.
More specifically, if an outer join is produced for tables CUSTOMER and AGENT, two scenarios are possible:
A left outer join yields all of the rows in the CUSTOMER table, including those that do not have a matching value
in the AGENT table. An example of such a join is shown in Figure 3.15.
FIGURE
A right outer join yields all of the rows
in the AGENT table, including those
that do not have matching values in the
CUSTOMER table. An example of such
a join is shown in Figure 3.16.
Left outer join
3.15
FIGURE
Generally speaking, outer joins operate
like equijoins. The outer join does not
drop one copy of the common attribute,
and it requires the specification of the
join condition. Figures 3.15 and 3.16
illustrate the result of outer joins after a
relational PROJECT operation is applied
to them to manually remove the duplicate column.
Right outer join
3.16
Outer joins are especially useful when
you are trying to determine what value(s) in related tables cause(s) referential
integrity problems. Such problems are
created when foreign key values do not
match the primary key values in the
related table(s). In fact, if you are asked
to convert large spreadsheets or other
nondatabase data into relational database tables, you will discover that the outer joins save you vast amounts
of time and uncounted headaches when you encounter referential integrity errors after the conversions.
You may wonder why the outer joins are labeled left and right. The labels refer to the order in which the tables
are listed in the SQL command. Chapter 8 explores such joins in more detail.
8.
The DIVIDE operation uses one single-column table (e.g., column “a”) as the divisor and one 2-column table
(i.e., columns “a” and “b”) as the dividend. The tables must have a common column (e.g., column “a”). The
output of the DIVIDE operation is a single column with the values of column “a” from the dividend table rows
where the value of the common column (i.e., column “a”) in both tables matches. Figure 3.17 shows a DIVIDE.
FIGURE
DIVIDE
3.17
DIVIDE
yields
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Using the example shown in Figure 3.17, note that:
a. Table 1 is “divided” by Table 2 to produce Table 3. Tables 1 and 2 both contain the column CODE but do
not share LOC.
b. To be included in the resulting Table 3, a value in the unshared column (LOC) must be associated (in the
dividing Table 2) with every value in Table 1.
c. The only value associated with both A and B is 5.
3.5 THE DATA DICTIONARY AND THE SYSTEM CATALOG
The data dictionary provides a detailed description of all tables found within the user/designer-created database.
Thus, the data dictionary contains at least all of the attribute names and characteristics for each table in the system.
In short, the data dictionary contains metadata—data about data. Using the small database presented in Figure 3.4,
you might picture its data dictionary as shown in Table 3.6.
Note
The data dictionary in Table 3.6 is an example of the human view of the entities, attributes, and relationships.
The purpose of this data dictionary is to ensure that all members of database design and implementation teams
use the same table and attribute names and characteristics. The DBMS’s internally stored data dictionary
contains additional information about relationship types, entity and referential integrity checks and enforcement, and index types and components. This additional information is generated during the database implementation stage.
The data dictionary is sometimes described as “the database designer’s database” because it records the design
decisions about tables and their structures.
Like the data dictionary, the system catalog contains metadata. The system catalog can be described as a detailed
system data dictionary that describes all objects within the database, including data about table names, the table’s
creator and creation date, the number of columns in each table, the data type corresponding to each column, index
filenames, index creators, authorized users, and access privileges. Because the system catalog contains all required data
dictionary information, the terms system catalog and data dictionary are often used interchangeably. In fact, current
relational database software generally provides only a system catalog, from which the designer’s data dictionary
information may be derived. The system catalog is actually a system-created database whose tables store the
user/designer-created database characteristics and contents. Therefore, the system catalog tables can be queried just
like any user/designer-created table.
In effect, the system catalog automatically produces database documentation. As new tables are added to the database,
that documentation also allows the RDBMS to check for and eliminate homonyms and synonyms. In general terms,
homonyms are similar-sounding words with different meanings, such as boar and bore, or identically spelled words
with different meanings, such as fair (meaning “just”) and fair (meaning “festival”). In a database context, the word
homonym indicates the use of the same attribute name to label different attributes. For example, you might use
C_NAME to label a customer name attribute in a CUSTOMER table and also use C_NAME to label a consultant name
attribute in a CONSULTANT table. To lessen confusion, you should avoid database homonyms; the data dictionary is
very useful in this regard.
In a database context, a synonym is the opposite of a homonym and indicates the use of different names to describe
the same attribute. For example, car and auto refer to the same object. Synonyms must be avoided. You will discover
why using synonyms is a bad idea when you work through Problem 27 at the end of this chapter.
=
=
=
=
=
Y
Y
Y
10000−99999
99999
Xxxxxxxx
Xxxxxxxx
X
dd-mmm-yyyy
999
999
999
999-9999
Xxxxxxxx
9,999,999.99
CHAR(5)
VARCHAR(20)
VARCHAR(20)
CHAR(1)
DATE
CHAR(3)
CHAR(3)
CHAR(3)
CHAR(8)
VARCHAR(20)
NUMBER(9,2)
Customer account code
Customer last name
Customer first name
Customer initial
Customer insurance
renewal date
Agent code
Agent code
Agent area code
Agent telephone number
Agent last name
Agent year-to-date sales
FK
PK
PK
OR
FK
PK
AGENT_CODE
FK
REFERENCED
TABLE
Foreign key
Primary key
Fixed character length data (1−255 characters)
Variable character length data (1−2,000 characters)
Numeric data (NUMBER(9,2)) are used to specify numbers with two decimal places and up to nine digits, including the decimal places.
Some RDBMSs permit the use of a MONEY or CURRENCY data type.
Y
Y
Y
Y
Y
REQUIRED
RANGE
FORMAT
TYPE
CONTENTS
Note: Telephone area codes are always composed of digits 0−9. Because area codes are not used arithmetically, they are most efficiently stored as character data. Also,
the area codes are always composed of three digits. Therefore, the area code data type is defined as CHAR(3). On the other hand, names do not conform to some standard
length. Therefore, the customer first names are defined as VARCHAR(20), thus indicating that up to 20 characters may be used to store the names. Character data are
shown as left-justified.
FK
PK
CHAR
VARCHAR
NUMBER
AGENT_CODE
AGENT_CODE
AGENT_AREACODE
AGENT_PHONE
AGENT_LNAME
AGENT_YTD_SLS
CUS_CODE
CUS_LNAME
CUS_FNAME
CUS_INITIAL
CUS_RENEW_DATE
CUSTOMER
AGENT
ATTRIBUTE NAME
A Sample Data Dictionary
TABLE NAME
3.6
TABLE
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
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3.6 RELATIONSHIPS WITHIN THE RELATIONAL DATABASE
You already know that relationships are classified as one-to-one (1:1), one-to-many (1:M), and many-to-many (M:N or
M:M). This section explores those relationships further to help you apply them properly when you start developing
database designs, focusing on the following points:
쐌
The 1:M relationship is the relational modeling ideal. Therefore, this relationship type should be the norm in
any relational database design.
쐌
The 1:1 relationship should be rare in any relational database design.
쐌
M:N relationships cannot be implemented as such in the relational model. Later in this section, you will see
how any M:N relationship can be changed into two 1:M relationships.
3.6.1 The 1:M Relationship
The 1:M relationship is the relational database norm. To see how such a relationship is modeled and implemented,
consider the PAINTER paints PAINTING example shown in Figure 3.18.
FIGURE
3.18
The 1:M relationship between
PAINTER and PAINTING
Compare the data model in Figure 3.18 with its implementation in Figure 3.19.
As you examine the PAINTER and PAINTING table contents
in Figure 3.19, note the following features:
쐌
FIGURE
Each painting is painted by one and only one
painter, but each painter could have painted many
paintings. Note that painter 123 (Georgette P. Ross)
has three paintings stored in the PAINTING table.
The implemented 1:M relationship between PAINTER and PAINTING
3.19
Table name: PAINTER
Primary key: PAINTER_NUM
Foreign key: none
Database name: Ch03_Museum
Table name: PAINTING
Primary key: PAINTING_NUM
Foreign key: PAINTER_NUM
쐌
There is only one row in the PAINTER table for any given row in the PAINTING table, but there may be many
rows in the PAINTING table for any given row in the PAINTER table.
The 1:M relationship is found in any database environment. Students in a typical college or university will discover that
each COURSE can generate many CLASSes but that each CLASS refers to only one COURSE. For example, an
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
Note
The one-to-many (1:M) relationship is easily implemented in the relational model by putting the primary key of
the 1 side in the table of the many side as a foreign key.
Accounting II course might yield two classes: one offered on Monday, Wednesday, and Friday (MWF) from 10:00 a.m.
to 10:50 a.m. and one offered on Thursday (Th) from 6:00 p.m. to 8:40 p.m. Therefore, the 1:M relationship
between COURSE and CLASS might be described this way:
쐌
Each COURSE can have many CLASSes, but each CLASS references only one COURSE.
쐌
There will be only one row in the COURSE table for any given row in the CLASS table, but there can be many
rows in the CLASS table for any given row in the COURSE table.
Figure 3.20 maps the ERM (entity relationship model) for the 1:M relationship between COURSE and CLASS.
FIGURE
The 1:M relationship between COURSE and CLASS is
further illustrated in Figure 3.21.
3.20
The 1:M relationship between
COURSE and CLASS
FIGURE
The implemented 1:M relationship between COURSE and CLASS
3.21
Table name: COURSE
Primary key: CRS_CODE
Foreign key: none
Table name: CLASS
Primary key: CLASS_CODE
Foreign key: CRS_CODE
Database name: Ch03_TinyCollege
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Using Figure 3.21, take a minute to review some important terminology. Note that CLASS_CODE in the CLASS table
uniquely identifies each row. Therefore, CLASS_CODE has been chosen to be the primary key. However, the
combination CRS_CODE and CLASS_SECTION will also uniquely identify each row in the class table. In other words,
the composite key composed of CRS_CODE and CLASS_SECTION is a candidate key. Any candidate key must have
the not null and unique constraints enforced. (You will see how this is done when you learn SQL in Chapter 7.)
For example, note in Figure 3.19 that the PAINTER table’s primary key, PAINTER_NUM, is included in the PAINTING
table as a foreign key. Similarly, in Figure 3.21, the COURSE table’s primary key, CRS_CODE, is included in the
CLASS table as a foreign key.
3.6.2 The 1:1 Relationship
As the 1:1 label implies, in this relationship, one entity can be related to only one other entity, and vice versa. For
example, one department chair—a professor—can chair only one department, and one department can have only one
department chair. The entities PROFESSOR and DEPARTMENT thus exhibit a 1:1 relationship. (You might argue that
not all professors chair a department and professors cannot be required to chair a department. That is, the relationship
between the two entities is optional. However, at this stage of the discussion, you should focus your attention on the
basic 1:1 relationship. Optional relationships will be addressed in Chapter 4.) The basic 1:1 relationship is modeled
in Figure 3.22, and its implementation is shown in Figure 3.23.
FIGURE
3.22
The 1:1 relationship between
PROFESSOR and DEPARTMENT
As you examine the tables in Figure 3.23, note that there
are several important features:
쐌
Each professor is a Tiny College employee. Therefore,
the professor identification is through the EMP_NUM.
(However, note that not all employees are
professors—there’s another optional relationship.)
쐌
The 1:1 PROFESSOR chairs DEPARTMENT relationship is implemented by having the EMP_NUM
foreign key in the DEPARTMENT table. Note that
the 1:1 relationship is treated as a special case of the
1:M relationship in which the “many” side is restricted to a single occurrence. In this case, DEPARTMENT
contains the EMP_NUM as a foreign key to indicate that it is the department that has a chair.
쐌
Also note that the PROFESSOR table contains the DEPT_CODE foreign key to implement the 1:M
DEPARTMENT employs PROFESSOR relationship. This is a good example of how two entities can participate
in two (or even more) relationships simultaneously.
The preceding “PROFESSOR chairs DEPARTMENT” example illustrates a proper 1:1 relationship. In fact, the use
of a 1:1 relationship ensures that two entity sets are not placed in the same table when they should not be.
However, the existence of a 1:1 relationship sometimes means that the entity components were not defined properly.
It could indicate that the two entities actually belong in the same table!
As rare as 1:1 relationships should be, certain conditions absolutely require their use. In Chapter 5, Advanced Data
Modeling, we will explore a concept called a generalization hierarchy, which is a powerful tool for improving our
database designs under specific conditions to avoid a proliferation of nulls. One of the characteristics of generalization
hierarchies is that they are implemented as 1:1 relationships.
3.6.3 The M:N Relationship
A many-to-many (M:N) relationship is not supported directly in the relational environment. However, M:N relationships
can be implemented by creating a new entity in 1:M relationships with the original entities.
T H E
FIGURE
R E L A T I O N A L
D A T A B A S E
M O D E L
The implemented 1:1 relationship between PROFESSOR and DEPARTMENT
3.23
Table name: PROFESSOR
Primary key: EMP_NUM
Foreign key: DEPT_CODE
Database name: Ch03_TinyCollege
The 1:M DEPARTMENT employs PROFESSOR relationship is implemented through
the placement of the DEPT_CODE foreign key in the PROFESSOR table.
Table name: DEPARTMENT
Primary key: DEPT_CODE
Foreign key: EMP_NUM
The 1:1 PROFESSOR chairs DEPARTMENT relationship
is implemented through the placement of the
EMP_NUM foreign key in the DEPARTMENT table.
Online Content
If you open the Ch03_TinyCollege database in the Premium Website, you’ll see that the STUDENT and CLASS
entities still use PROF_NUM as their foreign key. PROF_NUM and EMP_NUM are labels for the same attribute,
which is an example of the use of synonyms; that is, different names for the same attribute. These synonyms will
be eliminated in future chapters as the Tiny College database continues to be improved.
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Online Content
If you look at the Ch03_AviaCo database in the Premium Website, you will see the implementation of the 1:1
PILOT to EMPLOYEE relationship. This relationship is based on a concept known as “generalization hierarchy,”
which you will learn about in Chapter 5.
To explore the many-to-many (M:N) relationship, consider a rather typical college environment in which each
STUDENT can take many CLASSes, and each CLASS can contain many STUDENTs. The ER model in Figure 3.24
shows this M:N relationship.
FIGURE
3.24
The ERM’s M:N relationship
between STUDENT and CLASS
Note the features of the ERM in Figure 3.24.
쐌
Each CLASS can have many STUDENTs, and each
STUDENT can take many CLASSes.
쐌
There can be many rows in the CLASS table for any
given row in the STUDENT table, and there can be
many rows in the STUDENT table for any given row
in the CLASS table.
To examine the M:N relationship more closely, imagine a
small college with two students, each of whom takes three
classes. Table 3.7 shows the enrollment data for the two
students.
TABLE
Sample Student Enrollment Data
3.7
STUDENT'S LAST NAME
Bowser
Smithson
SELECTED CLASSES
Accounting 1, ACCT-211, code 10014
Intro to Microcomputing, CIS-220, code 10018
Intro to Statistics, QM-261, code 10021
Accounting 1, ACCT-211, code 10014
Intro to Microcomputing, CIS-220, code 10018
Intro to Statistics, QM-261, code 10021
Given such a data relationship and the sample data in Table 3.7, you could wrongly assume that you could implement
this M:N relationship by simply adding a foreign key in the many side of the relationship that points to the primary
key of the related table, as shown in Figure 3.25.
However, the M:N relationship should not be implemented as shown in Figure 3.25 for two good reasons:
쐌
The tables create many redundancies. For example, note that the STU_NUM values occur many times in the
STUDENT table. In a real-world situation, additional student attributes such as address, classification, major,
and home phone would also be contained in the STUDENT table, and each of those attribute values would be
repeated in each of the records shown here. Similarly, the CLASS table contains many duplications: each
student taking the class generates a CLASS record. The problem would be even worse if the CLASS table
included such attributes as credit hours and course description. Those redundancies lead to the anomalies
discussed in Chapter 1.
쐌
Given the structure and contents of the two tables, the relational operations become very complex and are
likely to lead to system efficiency errors and output errors.
T H E
FIGURE
R E L A T I O N A L
D A T A B A S E
M O D E L
The wrong implementation of the M:N relationship between STUDENT and CLASS
3.25
Table name: STUDENT
Primary key: STU_NUM
Foreign key: none
Database name: Ch03_CollegeTry
Table name: CLASS
Primary key: CLASS_CODE
Foreign key: STU_NUM
Fortunately, the problems inherent in the many-to-many (M:N) relationship can easily be avoided by creating a
composite entity (also referred to as a bridge entity or an associative entity). Because such a table is used to link
the tables that were originally related in an M:N relationship, the composite entity structure includes—as foreign
keys—at least the primary keys of the tables that are to be linked. The database designer has two main options when
defining a composite table’s primary key: use the combination of those foreign keys or create a new primary key.
Remember that each entity in the ERM is represented by a table. Therefore, you can create the composite ENROLL
table shown in Figure 3.26 to link the tables CLASS and STUDENT. In this example, the ENROLL table’s primary key
is the combination of its foreign keys CLASS_CODE and STU_NUM. But the designer could have decided to create
a single-attribute new primary key such as ENROLL_LINE, using a different line value to identify each ENROLL table
row uniquely. (Microsoft Access users might use the Autonumber data type to generate such line values automatically.)
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FIGURE
3
Converting the M:N relationship into two 1:M relationships
3.26
Table name: STUDENT
Primary key: STU_NUM
Foreign key: none
Database name: Ch03_CollegeTry2
Table name: ENROLL
Primary key: CLASS_CODE + STU_NUM
Foreign key: CLASS_CODE, STU_NUM
Table name: CLASS
Primary key: CLASS_CODE
Foreign key: CRS_CODE
Because the ENROLL table in Figure 3.26 links two tables, STUDENT and CLASS, it is also called a linking table.
In other words, a linking table is the implementation of a composite entity.
Note
In addition to the linking attributes, the composite ENROLL table can also contain such relevant attributes as the
grade earned in the course. In fact, a composite table can contain any number of attributes that the designer
wants to track. Keep in mind that the composite entity, although it is implemented as an actual table, is
conceptually a logical entity that was created as a means to an end: to eliminate the potential for multiple
redundancies in the original M:N relationship.
The ENROLL table shown in Figure 3.26 yields the required M:N to 1:M conversion. Observe that the composite
entity represented by the ENROLL table must contain at least the primary keys of the CLASS and STUDENT tables
(CLASS_CODE and STU_NUM, respectively) for which it serves as a connector. Also note that the STUDENT and
CLASS tables now contain only one row per entity. The ENROLL table contains multiple occurrences of the foreign
key values, but those controlled redundancies are incapable of producing anomalies as long as referential integrity is
enforced. Additional attributes may be assigned as needed. In this case, ENROLL_GRADE is selected to satisfy a
reporting requirement. Also note that the ENROLL table’s primary key consists of the two attributes CLASS_CODE
and STU_NUM because both the class code and the student number are needed to define a particular student’s grade.
Naturally, the conversion is reflected in the ERM, too. The revised relationship is shown in Figure 3.27.
As you examine Figure 3.27, note that the composite entity named ENROLL represents the linking table between
STUDENT and CLASS.
The 1:M relationship between COURSE and CLASS was first illustrated in Figure 3.20 and Figure 3.21. You can
increase the amount of available information even as you control the database’s redundancies. Thus, Figure 3.28
T H E
FIGURE
3.27
Changing the M:N relationship
to two 1:M relationships
R E L A T I O N A L
D A T A B A S E
M O D E L
shows the expanded ERM, including the 1:M relationship
between COURSE and CLASS shown in Figure 3.20. Note
that the model is able to handle multiple sections of a
CLASS while controlling redundancies by making sure that
all of the COURSE data common to each CLASS are kept
in the COURSE table.
The relational diagram that corresponds to the ERM in
Figure 3.28 is shown in Figure 3.29.
FIGURE
The ERM will be examined in greater detail in Chapter 4 to
show you how it is used to design more complex databases.
The ERM will also be used as the basis for the development
and implementation of a realistic database design in Appendixes B and C (see the Premium Website) for a university
computer lab.
3.28
The expanded entity
relationship model
FIGURE
The relational diagram for the Ch03_TinyCollege database
3.29
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3.7 DATA REDUNDANCY REVISITED
In Chapter 1 you learned that data redundancy leads to data anomalies. Those anomalies can destroy the effectiveness
of the database. You also learned that the relational database makes it possible to control data redundancies by using
common attributes that are shared by tables, called foreign keys.
The proper use of foreign keys is crucial to controlling data redundancy. Although the use of foreign keys does not
totally eliminate data redundancies, because the foreign key values can be repeated many times, the proper use of
foreign keys minimizes data redundancies, thus minimizing the chance that destructive data anomalies will develop.
Note
The real test of redundancy is not how many copies of a given attribute are stored, but whether the elimination
of an attribute will eliminate information. Therefore, if you delete an attribute and the original information can
still be generated through relational algebra, the inclusion of that attribute would be redundant. Given that view
of redundancy, proper foreign keys are clearly not redundant in spite of their multiple occurrences in a table.
However, even when you use this less restrictive view of redundancy, keep in mind that controlled redundancies
are often designed as part of the system to ensure transaction speed and/or information requirements. Exclusive
reliance on relational algebra to produce required information may lead to elegant designs that fail the test of
practicality.
You will learn in Chapter 4 that database designers must reconcile three often contradictory requirements: design
elegance, processing speed, and information requirements. And you will learn in Chapter 13, Business Intelligence and
Data Warehouses, that proper data warehousing design requires carefully defined and controlled data redundancies to
function properly. Regardless of how you describe data redundancies, the potential for damage is limited by proper
implementation and careful control.
As important as data redundancy control is, there are times when the level of data redundancy must actually be
increased to make the database serve crucial information purposes. You will learn about such redundancies in Chapter
13. There are also times when data redundancies seem to exist to preserve the historical accuracy of the data. For
example, consider a small invoicing system. The system includes the CUSTOMER, who may buy one or more
PRODUCTs, thus generating an INVOICE. Because a customer may buy more than one product at a time, an invoice
T H E
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D A T A B A S E
M O D E L
may contain several invoice LINEs, each providing details about the purchased product. The PRODUCT table should
contain the product price to provide a consistent pricing input for each product that appears on the invoice. The tables
that are part of such a system are shown in Figure 3.30. The system’s relational diagram is shown in Figure 3.31.
FIGURE
A small invoicing system
3.30
Table name: CUSTOMER
Primary key: CUS_CODE
Foreign key: none
Table name: INVOICE
Primary key: INV_NUMBER
Foreign key: CUS_CODE
Database name: Ch03_SaleCo
Table name: LINE
Primary key: INV_NUMBER + LINE_NUMBER
Foreign keys: INV_NUMBER, PROD_CODE
Table name: PRODUCT
Primary key: PROD_CODE
Foreign key: none
FIGURE
3.31
The relational diagram for the invoicing system
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As you examine the tables in the invoicing system in Figure 3.30 and the relationships depicted in Figure 3.31, note
that you can keep track of typical sales information. For example, by tracing the relationships among the four tables,
you discover that customer 10014 (Myron Orlando) bought two items on March 8, 2010, that were written to invoice
number 1001: one Houselite chain saw with a 16-inch bar and three rat-tail files. (Note: Trace the CUS_CODE
number 10014 in the CUSTOMER table to the matching CUS_CODE value in the INVOICE table. Next, take the
INV_NUMBER 1001 and trace it to the first two rows in the LINE table. Finally, match the two PROD_CODE values
in LINE with the PROD_CODE values in PRODUCT.) Application software will be used to write the correct bill by
multiplying each invoice line item’s LINE_UNITS by its LINE_PRICE, adding the results, applying appropriate taxes,
etc. Later, other application software might use the same technique to write sales reports that track and compare sales
by week, month, or year.
As you examine the sales transactions in Figure 3.30, you might reasonably suppose that the product price billed to
the customer is derived from the PRODUCT table because that’s where the product data are stored. But why does that
same product price occur again in the LINE table? Isn’t that a data redundancy? It certainly appears to be. But this
time, the apparent redundancy is crucial to the system’s success. Copying the product price from the PRODUCT table
to the LINE table maintains the historical accuracy of the transactions. Suppose, for instance, that you fail to write
the LINE_PRICE in the LINE table and that you use the PROD_PRICE from the PRODUCT table to calculate the sales
revenue. Now suppose that the PRODUCT table’s PROD_PRICE changes, as prices frequently do. This price change
will be properly reflected in all subsequent sales revenue calculations. However, the calculations of past sales revenues
will also reflect the new product price, which was not in effect when the transaction took place! As a result, the revenue
calculations for all past transactions will be incorrect, thus eliminating the possibility of making proper sales
comparisons over time. On the other hand, if the price data are copied from the PRODUCT table and stored with the
transaction in the LINE table, that price will always accurately reflect the transaction that took place at that time. You
will discover that such planned “redundancies” are common in good database design.
Finally, you might wonder why the LINE_NUMBER attribute was used in the LINE table in Figure 3.30. Wouldn’t the
combination of INV_NUMBER and PROD_CODE be a sufficient composite primary key—and, therefore, isn’t the
LINE_NUMBER redundant? Yes, the LINE_NUMBER is redundant, but this redundancy is quite common practice on
invoicing software that typically generates such line numbers automatically. In this case, the redundancy is not
necessary. But given its automatic generation, the redundancy is not a source of anomalies. The inclusion of
LINE_NUMBER also adds another benefit: the order of the retrieved invoicing data will always match the order in
which the data were entered. If product codes are used as part of the primary key, indexing will arrange those product
codes as soon as the invoice is completed and the data are stored. You can imagine the potential confusion when a
customer calls and says, “The second item on my invoice has an incorrect price” and you are looking at an invoice
whose lines show a different order from those on the customer’s copy!
3.8 INDEXES
Suppose you want to locate a particular book in a library. Does it make sense to look through every book in the library
until you find the one you want? Of course not; you use the library’s catalog, which is indexed by title, topic, and
author. The index (in either a manual or a computer system) points you to the book’s location, thereby making retrieval
of the book a quick and simple matter. An index is an orderly arrangement used to logically access rows in a table.
Or suppose you want to find a topic, such as “ER model,” in this book. Does it make sense to read through every page
until you stumble across the topic? Of course not; it is much simpler to go to the book’s index, look up the phrase ER
model, and read the page references that point you to the appropriate page(s). In each case, an index is used to locate
a needed item quickly.
Indexes in the relational database environment work like the indexes described in the preceding paragraphs. From a
conceptual point of view, an index is composed of an index key and a set of pointers. The index key is, in effect, the
T H E
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D A T A B A S E
M O D E L
index’s reference point. More formally, an index is an ordered arrangement of keys and pointers. Each key points to
the location of the data identified by the key.
For example, suppose you want to look up all of the paintings created by a given painter in the Ch03_Museum
database in Figure 3.19. Without an index, you must read each row in the PAINTING table and see if the
PAINTER_NUM matches the requested painter. However, if you index the PAINTER table and use the index key
PAINTER_NUM, you merely need to look up the appropriate PAINTER_NUM in the index and find the matching
pointers. Conceptually speaking, the index would resemble the presentation depicted in Figure 3.32.
FIGURE
Components of an index
3.32
Painting Table Index
Painting Table
123
1, 2, 4
126
3, 5
PAINTER_NUM
(index key)
Pointers to the
PAINTING
table rows
As you examine Figure 3.32, note that the first PAINTER_NUM index key value (123) is found in records 1, 2,
and 4 of the PAINTING table. The second PAINTER_NUM index key value (126) is found in records 3 and 5 of the
PAINTING table.
DBMSs use indexes for many different purposes. You just learned that an index can be used to retrieve data more
efficiently. But indexes can also be used by a DBMS to retrieve data ordered by a specific attribute or attributes. For
example, creating an index on a customer’s last name will allow you to retrieve the customer data alphabetically by the
customer’s last name. Also, an index key can be composed of one or more attributes. For example, in Figure 3.30,
you can create an index on VEND_CODE and PROD_CODE to retrieve all rows in the PRODUCT table ordered by
vendor, and within vendor, ordered by product.
Indexes play an important role in DBMSs for the implementation of primary keys. When you define a table’s primary
key, the DBMS automatically creates a unique index on the primary key column(s) you declared. For example, in Figure
3.30, when you declare CUS_CODE to be the primary key of the CUSTOMER table, the DBMS automatically creates
a unique index on that attribute. A unique index, as its name implies, is an index in which the index key can have
only one pointer value (row) associated with it. (The index in Figure 3.32 is not a unique index because the
PAINTER_NUM has multiple pointer values associated with it. For example, painter number 123 points to three
rows—1, 2, and 4—in the PAINTING table.)
A table can have many indexes, but each index is associated with only one table. The index key can have multiple
attributes (composite index). Creating an index is easy. You will learn in Chapter 7 that a simple SQL command
produces any required index.
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3.9 CODD’S RELATIONAL DATABASE RULES
In 1985, Dr. E. F. Codd published a list of 12 rules to define a relational database system.2 The reason Dr. Codd
published the list was his concern that many vendors were marketing products as “relational” even though those
products did not meet minimum relational standards. Dr. Codd’s list, shown in Table 3.8, serves as a frame of reference
for what a truly relational database should be. Bear in mind that even the dominant database vendors do not fully
support all 12 rules.
TABLE
3.8
RULE
1
Dr. Codd’s 12 Relational Database Rules
RULE NAME
Information
2
Guaranteed Access
3
Systematic Treatment of Nulls
4
Dynamic Online Catalog Based on
the Relational Model
5
Comprehensive Data Sublanguage
6
View Updating
7
High-Level Insert, Update, and Delete
8
Physical Data Independence
9
Logical Data Independence
10
Integrity Independence
11
Distribution Independence
12
Nonsubversion
Rule Zero
DESCRIPTION
All information in a relational database must be logically represented as column values in rows within tables.
Every value in a table is guaranteed to be accessible through a
combination of table name, primary key value, and column
name.
Nulls must be represented and treated in a systematic way,
independent of data type.
The metadata must be stored and managed as ordinary data,
that is, in tables within the database. Such data must be available to authorized users using the standard database relational language.
The relational database may support many languages. However, it must support one well-defined, declarative language
with support for data definition, view definition, data manipulation (interactive and by program), integrity constraints,
authorization, and transaction management (begin, commit,
and rollback).
Any view that is theoretically updatable must be updatable
through the system.
The database must support set-level inserts, updates, and
deletes.
Application programs and ad hoc facilities are logically unaffected when physical access methods or storage structures are
changed.
Application programs and ad hoc facilities are logically unaffected when changes are made to the table structures that
preserve the original table values (changing order of columns
or inserting columns).
All relational integrity constraints must be definable in the
relational language and stored in the system catalog, not at
the application level.
The end users and application programs are unaware and
unaffected by the data location (distributed vs. local
databases).
If the system supports low-level access to the data, there must
not be a way to bypass the integrity rules of the database.
All preceding rules are based on the notion that in order for a
database to be considered relational, it must use its relational
facilities exclusively to manage the database.
2Codd, E., “Is Your DBMS Really Relational?” and “Does Your DBMS Run by the Rules?” Computerworld, October 14 and October 21, 1985.
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
S u m m a r y
◗
◗
◗
◗
◗
◗
Tables are the basic building blocks of a relational database. A grouping of related entities, known as an entity set,
is stored in a table. Conceptually speaking, the relational table is composed of intersecting rows (tuples) and
columns. Each row represents a single entity, and each column represents the characteristics (attributes) of the
entities.
Keys are central to the use of relational tables. Keys define functional dependencies; that is, other attributes are
dependent on the key and can, therefore, be found if the key value is known. A key can be classified as a superkey,
a candidate key, a primary key, a secondary key, or a foreign key.
Each table row must have a primary key. The primary key is an attribute or a combination of attributes that uniquely
identifies all remaining attributes found in any given row. Because a primary key must be unique, no null values are
allowed if entity integrity is to be maintained.
Although the tables are independent, they can be linked by common attributes. Thus, the primary key of one table
can appear as the foreign key in another table to which it is linked. Referential integrity dictates that the foreign
key must contain values that match the primary key in the related table or must contain nulls.
The relational model supports relational algebra functions: SELECT, PROJECT, JOIN, INTERSECT, UNION,
DIFFERENCE, PRODUCT, and DIVIDE. A relational database performs much of the data manipulation work
behind the scenes. For example, when you create a database, the RDBMS automatically produces a structure to
house a data dictionary for your database. Each time you create a new table within the database, the RDBMS
updates the data dictionary, thereby providing the database documentation.
Once you know the relational database basics, you can concentrate on design. Good design begins by identifying
appropriate entities and their attributes and then the relationships among the entities. Those relationships (1:1,
1:M, and M:N) can be represented using ERDs. The use of ERDs allows you to create and evaluate simple logical
design. The 1:M relationship is most easily incorporated in a good design; you just have to make sure that the
primary key of the “1” is included in the table of the “many.”
K e y
T e r m s
associative entity, 81
functional dependence, 62
referential integrity, 66
attribute domain, 60
homonym, 74
relational algebra, 68
bridge entity, 81
index, 86
relational schema, 65
candidate key, 64
index key, 86
right outer join, 73
closure, 68
inner join, 72
secondary key, 66
composite entity, 81
join column(s), 71
set theory, 59
composite key, 63
key, 62
superkey, 63
data dictionary, 74
key attribute, 63
synonym, 74
determination, 62
left outer join, 73
system catalog, 74
domain, 61
linking table, 82
theta join, 72
entity integrity, 64
natural join, 71
tuple, 60
equijoin, 72
null, 64
union-compatible, 69
flags, 68
outer join, 72
unique index, 87
foreign key (FK), 65
predicate logic, 59
full functional dependence, 63
primary key (PK), 61
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Online Content
Answers to selected Review Questions and Problems for this chapter are contained in the Premium Website for
this book.
R e v i e w
Q u e s t i o n s
1. What is the difference between a database and a table?
2. What does it mean to say that a database displays both entity integrity and referential integrity?
3. Why are entity integrity and referential integrity important in a database?
4. What are the requirements that two relations must satisfy in order to be considered union-compatible?
5. Which relational algebra operators can be applied to a pair of tables that are not union-compatible?
6. Explain why the data dictionary is sometimes called “the database designer’s database.”
7. A database user manually notes that “The file contains two hundred records, each record containing nine fields.”
Use appropriate relational database terminology to “translate” that statement.
Online Content
All of the databases used in the questions and problems are found in the Premium Website for this book. The
database names used in the folder match the database names used in the figures. For example, the source of the
tables shown in Figure Q3.5 is the Ch03_CollegeQue database.
Use Figure Q3.8 to answer Questions 8–10.
FIGURE
Q3.8
The Ch03_CollegeQue
database tables
8. Using the STUDENT and PROFESSOR tables, illustrate the difference between a natural join, an equijoin,
and an outer join.
Database name: Ch03_CollegeQue
9. Create the basic ERD for the database shown in
Figure Q3.8.
Table name: STUDENT
10. Create the relational diagram for the database shown
in Figure Q3.8.
Use Figure Q3.11 to answer Questions 11–13.
11. Create the table that results from applying a
UNION relational operator to the tables shown in
Figure Q3.11.
Table name: PROFESSOR
12. Create the table that results from applying an
INTERSECT relational operator to the tables shown in
Figure Q3.11.
13. Using the tables in Figure Q3.11, create the table that
results from MACHINE DIFFERENCE BOOTH.
14. Suppose you have the ERM shown in Figure Q3.14.
How would you convert this model into an ERM that
displays only 1:M relationships? (Make sure you create
the revised ERM.)
T H E
FIGURE
R E L A T I O N A L
D A T A B A S E
M O D E L
The Ch03_VendingCo database tables
Q3.11
Database name: Ch03_VendingCo
Table name: BOOTH
FIGURE
Table name: MACHINE
The Crow’s Foot ERM for Question 14
Q3.14
15. What are homonyms and synonyms, and why should they be avoided in database design?
16. How would you implement a l:M relationship in a database composed of two tables? Give an example.
17. Identify and describe the components of the table shown in Figure Q3.17, using correct terminology. Use your
knowledge of naming conventions to identify the table’s probable foreign key(s).
FIGURE
The Ch03_NoComp database EMPLOYEE table
Q3.17
Table name: EMPLOYEE
Use the database shown in Figure Q3.18 to answer Questions 18–23.
18. Identify the primary keys.
19. Identify the foreign keys.
20. Create the ERM.
Database name: Ch03_NoComp
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FIGURE
Q3.18
3
The Ch03_Theater database
tables
Database name: Ch03_Theater
Table name: DIRECTOR
21. Create the relational diagram to show the relationship
between DIRECTOR and PLAY.
22. Suppose you wanted quick lookup capability to get a
listing of all plays directed by a given director. Which
table would be the basis for the INDEX table, and what
would be the index key?
23. What would be the conceptual view of the INDEX table
that is described in Question 22? Depict the contents
of the conceptual INDEX table.
Table name: PLAY
P r o b l e m s
Use the database shown in Figure P3.1 to answer Problems 1−9.
1. For each table, identify the primary key and the foreign key(s). If a table does not have a foreign key, write None
in the space provided.
TABLE
EMPLOYEE
STORE
REGION
PRIMARY KEY
FOREIGN KEY(S)
2. Do the tables exhibit entity integrity? Answer yes or no, and then explain your answer.
TABLE
EMPLOYEE
STORE
REGION
ENTITY INTEGRITY
EXPLANATION
3. Do the tables exhibit referential integrity? Answer yes or no, and then explain your answer. Write NA (Not
Applicable) if the table does not have a foreign key.
TABLE
EMPLOYEE
STORE
REGION
REFERENTIAL INTEGRITY
FOREIGN KEY(S)
4. Describe the type(s) of relationship(s) between STORE and REGION.
5. Create the ERD to show the relationship between STORE and REGION.
6. Create the relational diagram to show the relationship between STORE and REGION.
T H E
FIGURE
R E L A T I O N A L
D A T A B A S E
M O D E L
The Ch03_StoreCo database tables
P3.1
Table name: EMPLOYEE
Database name: Ch03_StoreCo
Table name: STORE
Table name: REGION
7. Describe the type(s) of relationship(s) between EMPLOYEE and STORE. (Hint: Each store employs many
employees, one of whom manages the store.)
8. Create the ERD to show the relationships among EMPLOYEE, STORE, and REGION.
9. Create the relational diagram to show the relationships among EMPLOYEE, STORE, and REGION.
Use the database shown in Figure P3.10 to work Problems 10−16. Note that the database is composed of four tables
that reflect these relationships:
쐌
An EMPLOYEE has only one JOB_CODE, but a JOB_CODE can be held by many EMPLOYEEs.
쐌
An EMPLOYEE can participate in many PLANs, and any PLAN can be assigned to many EMPLOYEEs.
Note also that the M:N relationship has been broken down into two 1:M relationships for which the BENEFIT table
serves as the composite or bridge entity.
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FIGURE
3
The Ch03_BeneCo database tables
P3.10
Database name: Ch03_BeneCo
Table name: EMPLOYEE
Table name: BENEFIT
Table name: JOB
Table name: PLAN
10. For each table in the database, identify the primary key and the foreign key(s). If a table does not have a foreign
key, write None in the space provided.
TABLE
EMPLOYEE
BENEFIT
JOB
PLAN
PRIMARY KEY
FOREIGN KEY(S)
11. Create the ERD to show the relationship between EMPLOYEE and JOB.
12. Create the relational diagram to show the relationship between EMPLOYEE and JOB.
13. Do the tables exhibit entity integrity? Answer yes or no, and then explain your answer.
TABLE
EMPLOYEE
BENEFIT
JOB
PLAN
ENTITY INTEGRITY
EXPLANATION
14. Do the tables exhibit referential integrity? Answer yes or no, and then explain your answer. Write NA (Not
Applicable) if the table does not have a foreign key.
TABLE
EMPLOYEE
BENEFIT
JOB
PLAN
REFERENTIAL INTEGRITY
EXPLANATION
15. Create the ERD to show the relationships among EMPLOYEE, BENEFIT, JOB, and PLAN.
T H E
R E L A T I O N A L
D A T A B A S E
M O D E L
16. Create the relational diagram to show the relationships among EMPLOYEE, BENEFIT, JOB, and PLAN.
Use the database shown in Figure P3.17 to answer Problems 17−23.
FIGURE
The Ch03_TransCo database tables
P3.17
Table name: TRUCK
Primary key: TRUCK_NUM
Foreign key: BASE_CODE, TYPE_CODE
Database name: Ch03_TransCo
Table name: BASE
Primary key: BASE_CODE
Foreign key: none
Table name: TYPE
Primary key: TYPE_CODE
Foreign key: none
17. For each table, identify the primary key and the foreign key(s). If a table does not have a foreign key, write None
in the space provided.
TABLE
TRUCK
BASE
TYPE
PRIMARY KEY
FOREIGN KEY(S)
18. Do the tables exhibit entity integrity? Answer yes or no, and then explain your answer.
TABLE
TRUCK
BASE
TYPE
ENTITY INTEGRITY
EXPLANATION
19. Do the tables exhibit referential integrity? Answer yes or no, and then explain your answer. Write NA (Not
Applicable) if the table does not have a foreign key.
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TABLE
TRUCK
BASE
TYPE
REFERENTIAL INTEGRITY
EXPLANATION
20. Identify the TRUCK table’s candidate key(s).
21. For each table, identify a superkey and a secondary key.
TABLE
TRUCK
BASE
TYPE
SUPERKEY
SECONDARY KEY
22. Create the ERD for this database.
23. Create the relational diagram for this database.
Use the database shown in Figure P3.24 to answer Problems 24−31. ROBCOR is an aircraft charter company that
supplies on-demand charter flight services using a fleet of four aircraft. Aircraft are identified by a unique registration
number. Therefore, the aircraft registration number is an appropriate primary key for the AIRCRAFT table.
FIGURE
The Ch03_AviaCo database tables
P3.24
Table name: CHARTER
Database name: Ch03_AviaCo
The destinations are indicated by standard three-letter airport codes. For example,
STL = St. Louis, MO
ATL = Atlanta, GA
BNA = Nashville, TN
Table name: AIRCRAFT
AC-TTAF
AC-TTEL
AC_TTER
= Aircraft total time, airframe (hours)
= Total time, left engine (hours)
= Total time, right engine (hours)
In a fully developed system, such attribute values
would be updated by application software when the
CHARTER table entries were posted.
Table name: MODEL
Customers are charged per round-trip mile, using the MOD_CHG_MILE rate. The MOD_SEATS gives the total number of
seats in the airplane, including the pilot and copilot seats. Therefore, a PA31-350 trip that is flown by a pilot and a copilot
has six passenger seats available.
T H E
FIGURE
R E L A T I O N A L
D A T A B A S E
M O D E L
The Ch03_AviaCo database tables (continued)
P3.24
Table name: PILOT
Database name: Ch03_AviaCo
The pilot licenses shown in the PILOT table include the ATP = Airline Transport Pilot and COMM = Commercial Pilot.
Businesses that operate on-demand air services are governed by Part 135 of the Federal Air Regulations (FARs) that are
enforced by the Federal Aviation Administration (FAA). Such businesses are known as “Part 135 operators.” Part 125
operations require that pilots successfully complete flight proficiency checks every six months. The “Part 135” flight
proficiency check data are recorded in PIL_PT135_DATE. To fly commercially, pilots must have at least a commercial
license and a second-class medical certificate (PIL_MED_TYPE = 2).
The PIL_RATINGS include
SEL
= Single Engine, Land
SES
= Single Engine, Sea
CFI
= Certified Flight Instructor
MEL = Multiengine, Land
Instr. = Instrument
CFII = Certified Flight Instructor, Instrument
Table name: EMPLOYEE
Table name: CUSTOMER
The nulls in the CHARTER table’s CHAR_COPILOT column indicate that a copilot is not required for some charter
trips or for some aircraft. Federal Aviation Administration (FAA) rules require a copilot on jet aircraft and on aircraft
having a gross take-off weight over 12,500 pounds. None of the aircraft in the AIRCRAFT table is governed by this
requirement; however, some customers may require the presence of a copilot for insurance reasons. All charter trips
are recorded in the CHARTER table.
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Note
Earlier in the chapter, it was stated that it is best to avoid homonyms and synonyms. In this problem, both the
pilot and the copilot are pilots in the PILOT table, but EMP_NUM cannot be used for both in the CHARTER
table. Therefore, the synonyms CHAR_PILOT and CHAR_COPILOT were used in the CHARTER table.
Although the solution works in this case, it is very restrictive and it generates nulls when a copilot is not
required. Worse, such nulls proliferate as crew requirements change. For example, if the AviaCo charter
company grows and starts using larger aircraft, crew requirements may increase to include flight engineers and
load masters. The CHARTER table would then have to be modified to include the additional crew assignments;
such attributes as CHAR_FLT_ENGINEER and CHAR_LOADMASTER would have to be added to the CHARTER
table. Given this change, each time a smaller aircraft flew a charter trip without the number of crew members
required in larger aircraft, the missing crew members would yield additional nulls in the CHARTER table.
You will have a chance to correct those design shortcomings in Problem 27. The problem illustrates two
important points:
1. Don’t use synonyms. If your design requires the use of synonyms, revise the design!
2. To the greatest possible extent, design the database to accommodate growth without requiring structural
changes in the database tables. Plan ahead and try to anticipate the effects of change on the database.
24. For each table, where possible, identify:
a. The primary key.
b. A superkey.
c. A candidate key.
d. The foreign key(s).
e. A secondary key.
25. Create the ERD. (Hint: Look at the table contents. You will discover that an AIRCRAFT can fly many CHARTER
trips but that each CHARTER trip is flown by one AIRCRAFT, that a MODEL references many AIRCRAFT but
that each AIRCRAFT references a single MODEL, etc.)
26. Create the relational diagram.
27. Modify the ERD you created in Problem 25 to eliminate the problems created by the use of synonyms. (Hint:
Modify the CHARTER table structure by eliminating the CHAR_PILOT and CHAR_COPILOT attributes; then
create a composite table named CREW to link the CHARTER and EMPLOYEE tables. Some crew members,
such as flight attendants, may not be pilots. That’s why the EMPLOYEE table enters into this relationship.)
28. Create the relational diagram for the design you revised in Problem 27. (After you have had a chance to revise
the design, your instructor will show you the results of the design change, using a copy of the revised database
named Ch03_AviaCo_2.)
You are interested in seeing data on charters flown by either Mr. Robert Williams (employee number 105) or Ms. Elizabeth
Travis (employee number 109) as pilot or copilot, but not charters flown by both of them. Complete problems 29−31 to
find these data.
29. Create the table that would result from applying the SELECT and PROJECT relational operators to the
CHARTER table to return only the CHAR_TRIP, CHAR_PILOT, and CHAR_COPILOT attributes for charters
flown by either employee 104 or employee 109.
30. Create the table that would result from applying the SELECT and PROJECT relational operators to the
CHARTER table to return only the CHAR_TRIP, CHAR_PILOT, and CHAR_COPILOT attributes for charters
flown by both employee 104 and employee 109.
31. Create the table that would result from applying a DIFFERENCE relational operator of your result from problem
29 to your result from problem 30.
In this chapter, you will learn:
쐍 The main characteristics of entity relationship components
쐍 How relationships between entities are defined, refined, and incorporated into the database
design process
쐍 How ERD components affect database design and implementation
쐍 That real-world database design often requires the reconciliation of conflicting goals
This chapter expands coverage of the data-modeling aspect of database design. Data
modeling is the first step in the database design journey, serving as a bridge between
real-world objects and the database model that is implemented in the computer. Therefore,
P
review
the importance of data-modeling details, expressed graphically through entity relationship
diagrams (ERDs), cannot be overstated.
Most of the basic concepts and definitions used in the entity relationship model (ERM) were
introduced in Chapter 2, Data Models. For example, the basic components of entities and
relationships and their representation should now be familiar to you. This chapter goes
much deeper and further, analyzing the graphic depiction of relationships among the entities
and showing how those depictions help you summarize the wealth of data required to
implement a successful design.
Finally, the chapter illustrates how conflicting goals can be a challenge in database design,
possibly requiring you to make design compromises.
4
F O U R
Entity Relationship (ER) Modeling
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Note
Because this book generally focuses on the relational model, you might be tempted to conclude that the ERM
is exclusively a relational tool. Actually, conceptual models such as the ERM can be used to understand and
design the data requirements of an organization. Therefore, the ERM is independent of the database type.
Conceptual models are used in the conceptual design of databases, while relational models are used in the
logical design of databases. However, because you are now familiar with the relational model from the previous
chapter, the relational model is used extensively in this chapter to explain ER constructs and the way they are
used to develop database designs.
4.1 THE ENTITY RELATIONSHIP MODEL (ERM)
You should remember from Chapter 2, Data Models, and Chapter 3, The Relational Database Model, that the ERM
forms the basis of an ERD. The ERD represents the conceptual database as viewed by the end user. ERDs depict the
database’s main components: entities, attributes, and relationships. Because an entity represents a real-world object,
the words entity and object are often used interchangeably. Thus, the entities (objects) of the Tiny College database
design developed in this chapter include students, classes, teachers, and classrooms. The order in which the ERD
components are covered in the chapter is dictated by the way the modeling tools are used to develop ERDs that can
form the basis for successful database design and implementation.
In Chapter 2, you also learned about the various notations used with ERDs—the original Chen notation and the newer
Crow’s Foot and UML notations. The first two notations are used at the beginning of this chapter to introduce some
basic ER modeling concepts. Some conceptual database modeling concepts can be expressed only using the Chen
notation. However, because the emphasis is on design and implementation of databases, the Crow’s Foot and UML
class diagram notations are used for the final Tiny College ER diagram example. Because of its implementation
emphasis, the Crow’s Foot notation can represent only what could be implemented. In other words:
쐌
The Chen notation favors conceptual modeling.
쐌
The Crow’s Foot notation favors a more implementation-oriented approach.
쐌
The UML notation can be used for both conceptual and implementation modeling.
Online Content
To learn how to create ER diagrams with the help of Microsoft Visio, see the Premium Website for this book:
• Appendix A, Designing Databases with Visio Professional: A Tutorial shows you how to create Crow’s
Foot ERDs.
• Appendix H, Unified Modeling Language (UML), shows you how to create UML class diagrams.
4.1.1 Entities
Recall that an entity is an object of interest to the end user. In Chapter 2, you learned that at the ER modeling level,
an entity actually refers to the entity set and not to a single entity occurrence. In other words, the word entity in the
ERM corresponds to a table—not to a row—in the relational environment. The ERM refers to a table row as an entity
instance or entity occurrence. In both the Chen and Crow’s Foot notations, an entity is represented by a rectangle
containing the entity’s name. The entity name, a noun, is usually written in all capital letters.
E N T I T Y
R E L A T I O N S H I P
( E R )
M O D E L I N G
4.1.2 Attributes
Attributes are characteristics of entities. For example, the STUDENT entity includes, among many others, the
attributes STU_LNAME, STU_FNAME, and STU_INITIAL. In the original Chen notation, attributes are represented
by ovals and are connected to the entity rectangle with a line. Each oval contains the name of the attribute it represents.
In the Crow’s Foot notation, the attributes are written in the attribute box below the entity rectangle. (See Figure 4.1.)
Because the Chen representation is rather space-consuming, software vendors have adopted the Crow’s Foot attribute
display.
FIGURE
The attributes of the STUDENT entity: Chen and Crow’s Foot
4.1
Chen Model
Crow’s Foot Model
STU_INITIAL
STU_EMAIL
STU_FNAME
STU_LNAME
STUDENT
STU_PHONE
Required and Optional Attributes
A required attribute is an attribute that must have a value; in other words, it cannot be left empty. As shown in
Figure 4.1, there are two boldfaced attributes in the Crow’s Foot notation. This indicates that a data entry will be
required. In this example, STU_LNAME and STU_FNAME require data entries because of the assumption that all
students have a last name and a first name. But students might not have a middle name, and perhaps they do not (yet)
have a phone number and an e-mail address. Therefore, those attributes are not presented in boldface in the entity
box. An optional attribute is an attribute that does not require a value; therefore, it can be left empty.
Domains
Attributes have a domain. As you learned in Chapter 3, a domain is the set of possible values for a given attribute.
For example, the domain for the grade point average (GPA) attribute is written (0,4) because the lowest possible GPA
value is 0 and the highest possible value is 4. The domain for the gender attribute consists of only two possibilities: M
or F (or some other equivalent code). The domain for a company’s date of hire attribute consists of all dates that fit
in a range (for example, company startup date to current date).
Attributes may share a domain. For instance, a student address and a professor address share the same domain of all
possible addresses. In fact, the data dictionary may let a newly declared attribute inherit the characteristics of an existing
attribute if the same attribute name is used. For example, the PROFESSOR and STUDENT entities may each have an
attribute named ADDRESS and could therefore share a domain.
Identifiers (Primary Keys)
The ERM uses identifiers, that is, one or more attributes that uniquely identify each entity instance. In the relational
model, such identifiers are mapped to primary keys (PKs) in tables. Identifiers are underlined in the ERD. Key attributes
are also underlined in a frequently used table structure shorthand notation using the format:
TABLE NAME (KEY_ATTRIBUTE 1, ATTRIBUTE 2, ATTRIBUTE 3, . . . ATTRIBUTE K)
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For example, a CAR entity may be represented by:
CAR (CAR_VIN, MOD_CODE, CAR_YEAR, CAR_COLOR)
(Each car is identified by a unique vehicle identification number, or CAR_VIN.)
Composite Identifiers
Ideally, an entity identifier is composed of only a single attribute. For example, the table in Figure 4.2 uses a
single-attribute primary key named CLASS_CODE. However, it is possible to use a composite identifier, that is, a
primary key composed of more than one attribute. For instance, the Tiny College database administrator may decide
to identify each CLASS entity instance (occurrence) by using a composite primary key composed of the combination
of CRS_CODE and CLASS_SECTION instead of using CLASS_CODE. Either approach uniquely identifies each entity
instance. Given the current structure of the CLASS table shown in Figure 4.2, CLASS_CODE is the primary key, and
the combination of CRS_CODE and CLASS_SECTION is a proper candidate key. If the CLASS_CODE attribute is
deleted from the CLASS entity, the candidate key (CRS_CODE and CLASS_SECTION) becomes an acceptable
composite primary key.
FIGURE
The CLASS table (entity) components and contents
4.2
Note
Remember that Chapter 3 made a commonly accepted distinction between COURSE and CLASS. A CLASS
constitutes a specific time and place of a COURSE offering. A class is defined by the course description and its
time and place, or section. Consider a professor who teaches Database I, Section 2; Database I, Section 5;
Database I, Section 8; and Spreadsheet II, Section 6. That instructor teaches two courses (Database I and
Spreadsheet II), but four classes. Typically, the COURSE offerings are printed in a course catalog, while the
CLASS offerings are printed in a class schedule for each semester, trimester, or quarter.
If the CLASS_CODE in Figure 4.2 is used as the primary key, the CLASS entity may be represented in shorthand
form by:
CLASS (CLASS_CODE, CRS_CODE, CLASS_SECTION, CLASS_TIME, ROOM_CODE, PROF_NUM)
E N T I T Y
R E L A T I O N S H I P
( E R )
M O D E L I N G
On the other hand, if CLASS_CODE is deleted, and the composite primary key is the combination of CRS_CODE and
CLASS_SECTION, the CLASS entity may be represented by:
CLASS (CRS_CODE, CLASS_SECTION, CLASS_TIME, ROOM_CODE, PROF_NUM)
Note that both key attributes are underlined in the entity notation.
Composite and Simple Attributes
Attributes are classified as simple or composite. A composite attribute, not to be confused with a composite key,
is an attribute that can be further subdivided to yield additional attributes. For example, the attribute ADDRESS can
be subdivided into street, city, state, and zip code. Similarly, the attribute PHONE_NUMBER can be subdivided into
area code and exchange number. A simple attribute is an attribute that cannot be subdivided. For example, age, sex,
and marital status would be classified as simple attributes. To facilitate detailed queries, it is wise to change composite
attributes into a series of simple attributes.
Single-Valued Attributes
A single-valued attribute is an attribute that can have only a single value. For example, a person can have only one
Social Security number, and a manufactured part can have only one serial number. Keep in mind that a single-valued
attribute is not necessarily a simple attribute. For instance, a part’s serial number, such as SE-08-02-189935, is
single-valued, but it is a composite attribute because it can be subdivided into the region in which the part was produced
(SE), the plant within that region (08), the shift within the plant (02), and the part number (189935).
Multivalued Attributes
Multivalued attributes are attributes that can have many values. For instance, a person may have several college
degrees, and a household may have several different phones, each with its own number. Similarly, a car’s color may
be subdivided into many colors (that is, colors for the roof, body, and trim). In the Chen ERM, the multivalued attributes
are shown by a double line connecting the attribute to the entity. The Crow’s Foot notation does not identify
multivalued attributes. The ERD in Figure 4.3 contains all of the components introduced thus far. In Figure 4.3, note
that CAR_VIN is the primary key, and CAR_COLOR is a multivalued attribute of the CAR entity.
FIGURE
A multivalued attribute in an entity
4.3
Chen Model
CAR_YEAR
MOD_CODE
CAR_VIN
Crow’s Foot Model
CAR
CAR_COLOR
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Note
In the ERD models in Figure 4.3, the CAR entity’s foreign key (FK) has been typed as MOD_CODE. This attribute
was manually added to the entity. Actually, proper use of database modeling software will automatically
produce the FK when the relationship is defined. In addition, the software will label the FK appropriately and
write the FK’s implementation details in a data dictionary. Therefore, when you use database modeling software
like Visio Professional, never type the FK attribute yourself; let the software handle that task when the
relationship between the entities is defined. (You can see how that's done in Appendix A, Designing Databases
with Visio Professional: A Tutorial, in the Premium Website.)
Implementing Multivalued Attributes
Although the conceptual model can handle M:N relationships and multivalued attributes, you should not implement
them in the RDBMS. Remember from Chapter 3 that in the relational table, each column/row intersection represents
a single data value. So if multivalued attributes exist, the designer must decide on one of two possible courses of action:
1.
Within the original entity, create several new attributes, one for each of the original multivalued attribute’s
components. For example, the CAR entity’s attribute CAR_COLOR can be split to create the new attributes
CAR_TOPCOLOR, CAR_BODYCOLOR, and CAR_TRIMCOLOR, which are then assigned to the CAR
entity. (See Figure 4.4.)
FIGURE
Splitting the multivalued attribute into new attributes
4.4
Chen Model
Crow’s Foot Model
CAR_YEAR
MOD_CODE
CAR_VIN
CAR_TOPCOLOR
CAR
CAR_TRIMCOLOR
CAR_BODYCOLOR
Although this solution seems to work, its adoption can lead to major structural problems in the table. For
example, if additional color components—such as a logo color—are added for some cars, the table structure
must be modified to accommodate the new color section. In that case, cars that do not have such color sections
generate nulls for the nonexisting components, or their color entries for those sections are entered as N/A to
indicate “not applicable.” (Imagine how the solution in Figure 4.4—splitting a multivalued attribute into new
attributes—would cause problems if it were applied to an employee entity containing employee degrees and
certifications. If some employees have 10 degrees and certifications while most have fewer or none, the
number of degree/certification attributes would number 10, and most of those attribute values would be null
for most of the employees.) In short, although you have seen solution 1 applied, it is not an acceptable solution.
2.
Create a new entity composed of the original multivalued attribute’s components. This new entity allows the
designer to define color for different sections of the car. (See Table 4.1.) Then, this new CAR_COLOR entity
is related to the original CAR entity in a 1:M relationship.
E N T I T Y
TABLE
4.1
SECTION
Top
Body
Trim
Interior
FIGURE
Components of the
Multivalued Attribute
COLOR
White
Blue
Gold
Blue
R E L A T I O N S H I P
( E R )
M O D E L I N G
Using the approach illustrated in Table 4.1, you even get a
fringe benefit: you are now able to assign as many colors as
necessary without having to change the table structure. Note
that the ERM shown in Figure 4.5 reflects the components
listed in Table 4.1. This is the preferred way to deal with
multivalued attributes. Creating a new entity in a 1:M relationship with the original entity yields several benefits: it’s a
more flexible, expandable solution, and it is compatible with
the relational model!
A new entity set composed of a multivalued attribute’s components
4.5
Derived Attributes
Finally, an attribute may be classified as a derived attribute. A derived attribute is an attribute whose value is
calculated (derived) from other attributes. The derived attribute need not be physically stored within the database;
instead, it can be derived by using an algorithm. For example, an employee’s age, EMP_AGE, may be found by
computing the integer value of the difference between the current date and the EMP_DOB. If you use Microsoft
Access, you would use the formula INT((DATE() – EMP_DOB)/365). In Microsoft SQL Server, you would use SELECT
DATEDIFF(“YEAR”, EMP_DOB, GETDATE()), where DATEDIFF is a function that computes the difference between
dates. The first parameter indicates the measurement, in this case, years.
If you use Oracle, you would use SYSDATE instead of DATE(). (You are assuming, of course, that the EMP_DOB was
stored in the Julian date format.) Similarly, the total cost of an order can be derived by multiplying the quantity ordered
by the unit price. Or the estimated average speed can be derived by dividing trip distance by the time spent en route.
A derived attribute is indicated in the Chen notation by a dashed line connecting the attribute and the entity. (See
Figure 4.6.) The Crow’s Foot notation does not have a method for distinguishing the derived attribute from other
attributes.
Derived attributes are sometimes referred to as computed attributes. A derived attribute computation can be as simple
as adding two attribute values located on the same row, or it can be the result of aggregating the sum of values located
on many table rows (from the same table or from a different table). The decision to store derived attributes in database
tables depends on the processing requirements and the constraints placed on a particular application. The designer
should be able to balance the design in accordance with such constraints. Table 4.2 shows the advantages and
disadvantages of storing (or not storing) derived attributes in the database.
4.1.3 Relationships
Recall from Chapter 2 that a relationship is an association between entities. The entities that participate in a
relationship are also known as participants, and each relationship is identified by a name that describes the
relationship. The relationship name is an active or passive verb; for example, a STUDENT takes a CLASS, a
PROFESSOR teaches a CLASS, a DEPARTMENT employs a PROFESSOR, a DIVISION is managed by an
EMPLOYEE, and an AIRCRAFT is flown by a CREW.
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FIGURE
Depiction of a derived attribute
4.6
Chen Model
EMP_INITIAL
EMP_FNAME
EMP_DOB
EMP_LNAME
EMP_NUM
TABLE
Crow’s Foot Model
EMPLOYEE
EMP_AGE
Advantages and Disadvantages of Storing Derived Attributes
4.2
Advantage
Disadvantage
DERIVED ATTRIBUTE
STORED
NOT STORED
Saves storage space
Saves CPU processing cycles
Computation always yields current value
Saves data access time
Data value is readily available
Can be used to keep track of historical data
Uses CPU processing cycles
Requires constant maintenance to ensure
derived value is current, especially if any values Increases data access time
Adds coding complexity to queries
used in the calculation change
Relationships between entities always operate in both directions. That is, to define the relationship between the entities
named CUSTOMER and INVOICE, you would specify that:
쐌
A CUSTOMER may generate many INVOICEs.
쐌
Each INVOICE is generated by one CUSTOMER.
Because you know both directions of the relationship between CUSTOMER and INVOICE, it is easy to see that this
relationship can be classified as 1:M.
The relationship classification is difficult to establish if you know only one side of the relationship. For example, if you
specify that:
A DIVISION is managed by one EMPLOYEE.
You don’t know if the relationship is 1:1 or 1:M. Therefore, you should ask the question “Can an employee manage
more than one division?” If the answer is yes, the relationship is 1:M, and the second part of the relationship is then
written as:
An EMPLOYEE may manage many DIVISIONs.
If an employee cannot manage more than one division, the relationship is 1:1, and the second part of the relationship
is then written as:
An EMPLOYEE may manage only one DIVISION.
E N T I T Y
R E L A T I O N S H I P
( E R )
M O D E L I N G
4.1.4 Connectivity and Cardinality
You learned in Chapter 2 that entity relationships may be classified as one-to-one, one-to-many, or many-to-many. You
also learned how such relationships were depicted in the Chen and Crow’s Foot notations. The term connectivity is
used to describe the relationship classification.
Cardinality expresses the minimum and maximum number of entity occurrences associated with one occurrence of
the related entity. In the ERD, cardinality is indicated by placing the appropriate numbers beside the entities, using the
format (x,y). The first value represents the minimum number of associated entities, while the second value represents
the maximum number of associated entities. Many database designers who use Crow’s Foot modeling notation do not
depict the specific cardinalities on the ER diagram itself because the specific limits described by the cardinalities cannot
be implemented directly through the database design. Correspondingly, some Crow’s Foot ER modeling tools do not
print the numeric cardinality range in the diagram; instead, you can add it as text if you want to have it shown. When
the specific cardinalities are not included on the diagram in Crow’s Foot notation, cardinality is implied by the use of
the symbols shown in Figure 4.7, which describe the connectivity and participation (discussed below). The numeric
cardinality range has been added using the Visio text drawing tool.
FIGURE
4.7
Connectivity and cardinality in
an ERD
Knowing the minimum and maximum number of entity
occurrences is very useful at the application software level.
For example, Tiny College might want to ensure that a class
is not taught unless it has at least 10 students enrolled.
Similarly, if the classroom can hold only 30 students, the
application software should use that cardinality to limit
enrollment in the class. However, keep in mind that the
DBMS cannot handle the implementation of the cardinalities
at the table level—that capability is provided by the application software or by triggers. You will learn how to create and
execute triggers in Chapter 8, Advanced SQL.
As you examine the Crow’s Foot diagram in Figure 4.7,
keep in mind that the cardinalities represent the number of occurrences in the related entity. For example, the
cardinality (1,4) written next to the CLASS entity in the “PROFESSOR teaches CLASS” relationship indicates that
each professor teaches up to four classes, which means that the PROFESSOR table’s primary key value occurs at least
once and no more than four times as foreign key values in the CLASS table. If the cardinality had been written as (1,N),
there would be no upper limit to the number of classes a professor might teach. Similarly, the cardinality (1,1) written
next to the PROFESSOR entity indicates that each class is taught by one and only one professor. That is, each CLASS
entity occurrence is associated with one and only one entity occurrence in PROFESSOR.
Connectivities and cardinalities are established by very concise statements known as business rules, which were
introduced in Chapter 2. Such rules, derived from a precise and detailed description of an organization’s data
environment, also establish the ERM’s entities, attributes, relationships, connectivities, cardinalities, and constraints.
Because business rules define the ERM’s components, making sure that all appropriate business rules are identified is
a very important part of a database designer’s job.
Note
The placement of the cardinalities in the ER diagram is a matter of convention. The Chen notation places the
cardinalities on the side of the related entity. The Crow’s Foot and UML diagrams place the cardinalities next to
the entity to which the cardinalities apply.
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Online Content
Because the careful definition of complete and accurate business rules is crucial to good database design, their
derivation is examined in detail in Appendix B, The University Lab: Conceptual Design. The modeling skills
you are learning in this chapter are applied in the development of a real database design in Appendix B. The
initial design shown in Appendix B is then modified in Appendix C, The University Lab: Conceptual Design
Verification, Logical Design, and Implementation. (Both appendixes are found in the Premium Website.)
4.1.5 Existence Dependence
An entity is said to be existence-dependent if it can exist in the database only when it is associated with another
related entity occurrence. In implementation terms, an entity is existence-dependent if it has a mandatory foreign
key—that is, a foreign key attribute that cannot be null. For example, if an employee wants to claim one or more
dependents for tax-withholding purposes, the relationship “EMPLOYEE claims DEPENDENT” would be appropriate.
In that case, the DEPENDENT entity is clearly existence-dependent on the EMPLOYEE entity because it is impossible
for the dependent to exist apart from the EMPLOYEE in the database.
If an entity can exist apart from all of its related entities (it is existence-independent), then that entity is referred to
as a strong entity or regular entity. For example, suppose that the XYZ Corporation uses parts to produce its
products. Furthermore, suppose that some of those parts are produced in-house and other parts are bought from
vendors. In that scenario, it is quite possible for a PART to exist independently from a VENDOR in the relationship
“PART is supplied by VENDOR,” because at least some of the parts are not supplied by a vendor. Therefore, PART
is existence-independent from VENDOR.
Note
The relationship strength concept is not part of the original ERM. Instead, this concept applies directly to Crow’s
Foot diagrams. Because Crow’s Foot diagrams are used extensively to design relational databases, it is important
to understand relationship strength as it affects database implementation. The Chen ERD notation is oriented
toward conceptual modeling and therefore does not distinguish between weak and strong relationships.
4.1.6 Relationship Strength
The concept of relationship strength is based on how the primary key of a related entity is defined. To implement a
relationship, the primary key of one entity appears as a foreign key in the related entity. For example, the 1:M
relationship between VENDOR and PRODUCT in Chapter 3, Figure 3.3, is implemented by using the VEND_CODE
primary key in VENDOR as a foreign key in PRODUCT. There are times when the foreign key also is a primary key
component in the related entity. For example, in Figure 4.5, the CAR entity primary key (CAR_VIN) appears as both
a primary key component and a foreign key in the CAR_COLOR entity. In this section, you will learn how various
relationship strength decisions affect primary key arrangement in database design.
E N T I T Y
R E L A T I O N S H I P
( E R )
M O D E L I N G
Weak (Non-identifying) Relationships
A weak relationship, also known as a non-identifying relationship, exists if the PK of the related entity does not
contain a PK component of the parent entity. By default, relationships are established by having the PK of the parent
entity appear as an FK on the related entity. For example, suppose that the COURSE and CLASS entities are
defined as:
COURSE(CRS_CODE, DEPT_CODE, CRS_DESCRIPTION, CRS_CREDIT)
CLASS(CLASS_CODE, CRS_CODE, CLASS_SECTION, CLASS_TIME, ROOM_CODE, PROF_NUM)
In this case, a weak relationship exists between COURSE and CLASS because the CLASS_CODE is the CLASS
entity’s PK, while the CRS_CODE in CLASS is only an FK. In this example, the CLASS PK did not inherit the PK
component from the COURSE entity.
Figure 4.8 shows how the Crow’s Foot notation depicts a weak relationship by placing a dashed relationship line
between the entities. The tables shown below the ERD illustrate how such a relationship is implemented.
FIGURE
A weak (non-identifying) relationship between COURSE and CLASS
4.8
Table name: COURSE
Table name: CLASS
Database name: Ch04_TinyCollege
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Online Content
All of the databases used to illustrate the material in this chapter are found in the Premium Website.
Note
If you are used to looking at relational diagrams such as the ones produced by Microsoft Access, you expect to
see the relationship line in the relational diagram drawn from the PK to the FK. However, the relational diagram
convention is not necessarily reflected in the ERD. In an ERD, the focus is on the entities and the relationships
between them, rather than on the way those relationships are anchored graphically. You will discover that the
placement of the relationship lines in a complex ERD that includes both horizontally and vertically placed
entities is largely dictated by the designer’s decision to improve the readability of the design. (Remember that
the ERD is used for communication between the designer(s) and end users.)
Strong (Identifying) Relationships
A strong relationship, also known as an identifying relationship, exists when the PK of the related entity contains
a PK component of the parent entity. For example, the definitions of the COURSE and CLASS entities
COURSE(CRS_CODE, DEPT_CODE, CRS_DESCRIPTION, CRS_CREDIT)
CLASS(CRS_CODE, CLASS_SECTION , CLASS_TIME, ROOM_CODE, PROF_NUM)
indicate that a strong relationship exists between COURSE and CLASS, because the CLASS entity’s composite PK
is composed of CRS_CODE + CLASS_SECTION. (Note that the CRS_CODE in CLASS is also the FK to the
COURSE entity.)
The Crow’s Foot notation depicts the strong (identifying) relationship with a solid line between the entities, shown in
Figure 4.9. Whether the relationship between COURSE and CLASS is strong or weak depends on how the CLASS
entity’s primary key is defined.
Keep in mind that the order in which the tables are created and loaded is very important. For example, in the
“COURSE generates CLASS” relationship, the COURSE table must be created before the CLASS table. After all, it
would not be acceptable to have the CLASS table’s foreign key reference a COURSE table that did not yet exist. In
fact, you must load the data of the “1” side first in a 1:M relationship to avoid the possibility of referential
integrity errors, regardless of whether the relationships are weak or strong.
As you examine Figure 4.9 you might wonder what the O symbol next to the CLASS entity signifies. You will discover
the meaning of this cardinality in Section 4.1.8, Relationship Participation.
Remember that the nature of the relationship is often determined by the database designer, who must use professional
judgment to determine which relationship type and strength best suit the database transaction, efficiency, and
information requirements. That point will often be emphasized in detail!
4.1.7 Weak Entities
In contrast to the strong or regular entity mentioned in Section 4.1.5, a weak entity is one that meets two conditions:
1.
The entity is existence-dependent; that is, it cannot exist without the entity with which it has a relationship.
2.
The entity has a primary key that is partially or totally derived from the parent entity in the relationship.
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
A strong (identifying) relationship between COURSE and CLASS
4.9
Table name: COURSE
Database name: Ch04_TinyCollege_Alt
Table name: CLASS
For example, a company insurance policy insures an employee and his/her dependents. For the purpose of describing
an insurance policy, an EMPLOYEE might or might not have a DEPENDENT, but the DEPENDENT must be
associated with an EMPLOYEE. Moreover, the DEPENDENT cannot exist without the EMPLOYEE; that is, a person
cannot get insurance coverage as a dependent unless s(he) happens to be a dependent of an employee. DEPENDENT
is the weak entity in the relationship “EMPLOYEE has DEPENDENT.” This relationship is shown in Figure 4.10.
Note that the Chen notation in Figure 4.10 identifies the weak entity by using a double-walled entity rectangle. The Crow’s
Foot notation generated by Visio Professional uses the relationship line and the PK/FK designation to indicate whether the
related entity is weak. A strong (identifying) relationship indicates that the related entity is weak. Such a relationship means
that both conditions for the weak entity definition have been met—the related entity is existence-dependent, and the PK
of the related entity contains a PK component of the parent entity. (Some versions of the Crow’s Foot ERD depict the
weak entity by drawing a short line segment in each of the four corners of the weak entity box.)
Remember that the weak entity inherits part of its primary key from its strong counterpart. For example, at least part
of the DEPENDENT entity’s key shown in Figure 4.10 was inherited from the EMPLOYEE entity:
EMPLOYEE (EMP_NUM, EMP_LNAME, EMP_FNAME, EMP_INITIAL, EMP_DOB, EMP_HIREDATE)
DEPENDENT (EMP_NUM, DEP_NUM, DEP_FNAME, DEP_DOB)
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4
A weak entity in an ERD
4.10
Chen Model
M
1
has
EMPLOYEE
(0,N)
EMP_NUM
EMP_LNAME
EMP_FNAME
EMP_INITIAL
EMP_DOB
EMP_HIREDATE
DEPENDENT
(1,1)
EMP_NUM
DEP_NUM
DEP_FNAME
DEP_DOB
Crow’s Foot Model
Figure 4.11 illustrates the implementation of the relationship between the weak entity (DEPENDENT) and its parent
or strong counterpart (EMPLOYEE). Note that DEPENDENT’s primary key is composed of two attributes, EMP_NUM
and DEP_NUM, and that EMP_NUM was inherited from EMPLOYEE.
FIGURE
A weak entity in a strong relationship
4.11
Table name: EMPLOYEE
Table name: DEPENDENT
Database name: Ch04_ShortCo
E N T I T Y
R E L A T I O N S H I P
( E R )
M O D E L I N G
Given this scenario, and with the help of this relationship, you can determine that:
Jeanine J. Callifante claims two dependents, Annelise and Jorge.
Keep in mind that the database designer usually determines whether an entity can be described as weak based on the
business rules. An examination of the relationship between COURSE and CLASS in Figure 4.8 might cause you to
conclude that CLASS is a weak entity to COURSE. After all, in Figure 4.8, it seems clear that a CLASS cannot exist
without a COURSE; so there is existence dependence. For example, a student cannot enroll in the Accounting I class
ACCT-211, Section 3 (CLASS_CODE 10014) unless there is an ACCT-211 course. However, note that the CLASS
table’s primary key is CLASS_CODE, which is not derived from the COURSE parent entity. That is, CLASS may be
represented by:
CLASS (CLASS_CODE, CRS_CODE, CLASS_SECTION, CLASS_TIME, ROOM_CODE, PROF_NUM)
The second weak entity requirement has not been met; therefore, by definition, the CLASS entity in Figure 4.8 may
not be classified as weak. On the other hand, if the CLASS entity’s primary key had been defined as a composite key,
composed of the combination CRS_CODE and CLASS_SECTION, CLASS could be represented by:
CLASS (CRS_CODE, CLASS_SECTION, CLASS_TIME, ROOM_CODE, PROF_NUM)
In that case, illustrated in Figure 4.9, the CLASS primary key is partially derived from COURSE because CRS_CODE
is the COURSE table’s primary key. Given this decision, CLASS is a weak entity by definition. (In Visio Professional
Crow’s Foot terms, the relationship between COURSE and CLASS is classified as strong, or identifying.) In any case,
CLASS is always existence-dependent on COURSE, whether or not it is defined as weak.
4.1.8 Relationship Participation
Participation in an entity relationship is either optional or mandatory. Recall that relationships are bidirectional; that
is, they operate in both directions. If COURSE is related to CLASS, then by definition, CLASS is related to COURSE.
Because of the bidirectional nature of relationships, it is necessary to determine the connectivity of the relationship
from COURSE to CLASS and the connectivity of the relationship from CLASS to COURSE. Similarly, the specific
maximum and minimum cardinalities must be determined in each direction for the relationship. Once again, you must
consider the bidirectional nature of the relationship when determining participation.
Optional participation means that one entity occurrence does not require a corresponding entity occurrence in a
particular relationship. For example, in the “COURSE generates CLASS” relationship, you noted that at least some
courses do not generate a class. In other words, an entity occurrence (row) in the COURSE table does not necessarily
require the existence of a corresponding entity occurrence in the CLASS table. (Remember that each entity is
implemented as a table.) Therefore, the CLASS entity is considered to be optional to the COURSE entity. In Crow’s
Foot notation, an optional relationship between entities is shown by drawing a small circle (O) on the side of the
optional entity, as illustrated in Figure 4.9. The existence of an optional entity indicates that the minimum cardinality
is 0 for the optional entity. (The term optionality is used to label any condition in which one or more optional
relationships exist.)
Note
Remember that the burden of establishing the relationship is always placed on the entity that contains the
foreign key. In most cases, that will be the entity on the “many” side of the relationship.
Mandatory participation means that one entity occurrence requires a corresponding entity occurrence in a
particular relationship. If no optionality symbol is depicted with the entity, the entity is assumed to exist in a mandatory
relationship with the related entity. If the mandatory participation is depicted graphically, it is typically shown as a small
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hash mark across the relationship line, similar to the Crow’s Foot depiction of a connectivity of 1. The existence of
a mandatory relationship indicates that the minimum cardinality is at least 1 for the mandatory entity.
Note
You might be tempted to conclude that relationships are weak when they occur between entities in an optional
relationship and that relationships are strong when they occur between entities in a mandatory relationship.
However, this conclusion is not warranted. Keep in mind that relationship participation and relationship
strength do not describe the same thing. You are likely to encounter a strong relationship when one entity is
optional to another. For example, the relationship between EMPLOYEE and DEPENDENT is clearly a strong one,
but DEPENDENT is clearly optional to EMPLOYEE. After all, you cannot require employees to have dependents.
And it is just as possible for a weak relationship to be established when one entity is mandatory to another. The
relationship strength depends on how the PK of the related entity is formulated, while the relationship
participation depends on how the business rule is written. For example, the business rules “Each part must be
supplied by a vendor” and “A part may or may not be supplied by a vendor” create different optionalities for
the same entities! Failure to understand this distinction may lead to poor design decisions that cause major
problems when table rows are inserted or deleted.
When you create a relationship in MS Visio, the default relationship will be mandatory on the “1” side and optional
on the “many” side. Table 4.3 shows the various connectivity and participation combinations that are supported by the
Crow’s Foot notation. Recall that these combinations are often referred to as cardinality in Crow’s Foot notation when
specific cardinalities are not used.
TABLE
4.3
Crow’s Foot Symbols
CROW’S FOOT SYMBOL
CARDINALITY
(0,N)
COMMENT
Zero or many. Many side is optional.
(1,N)
One or many. Many side is mandatory.
(1,1)
One and only one. 1 side is mandatory.
(0,1)
Zero or one. 1 side is optional.
Because relationship participation turns out to be a very important component of the database design process, let’s
examine a few more scenarios. Suppose that Tiny College employs some professors who conduct research without
teaching classes. If you examine the “PROFESSOR teaches CLASS” relationship, it is quite possible for a
PROFESSOR not to teach a CLASS. Therefore, CLASS is optional to PROFESSOR. On the other hand, a CLASS
must be taught by a PROFESSOR. Therefore, PROFESSOR is mandatory to CLASS. Note that the ERD model in
Figure 4.12 shows the cardinality next to CLASS to be (0,3), thus indicating that a professor may teach no classes at
all or as many as three classes. And each CLASS table row will reference one and only one PROFESSOR
row—assuming each class is taught by one and only one professor—represented by the (1,1) cardinality next to the
PROFESSOR table.
Failure to understand the distinction between mandatory and optional participation in relationships might yield
designs in which awkward (and unnecessary) temporary rows (entity instances) must be created just to accommodate
the creation of required entities. Therefore, it is important that you clearly understand the concepts of mandatory and
optional participation.
It is also important to understand that the semantics of a problem might determine the type of participation in a
relationship. For example, suppose that Tiny College offers several courses; each course has several classes. Note
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
An optional CLASS entity in the relationship “PROFESSOR teaches CLASS”
4.12
again the distinction between class and course in this discussion: a CLASS constitutes a specific offering (or section)
of a COURSE. (Typically, courses are listed in the university’s course catalog, while classes are listed in the class
schedules that students use to register for their classes.)
Analyzing the CLASS entity’s contribution to the “COURSE generates CLASS” relationship, it is easy to see that a
CLASS cannot exist without a COURSE. Therefore, you can conclude that the COURSE entity is mandatory in the
relationship. But two scenarios for the CLASS entity may be written, shown in Figures 4.13 and 4.14.
FIGURE
CLASS is optional to COURSE
4.13
FIGURE
COURSE and CLASS in a mandatory relationship
4.14
The different scenarios are a function of the semantics of the problem; that is, they depend on how the relationship
is defined.
1.
CLASS is optional. It is possible for the department to create the entity COURSE first and then create the
CLASS entity after making the teaching assignments. In the real world, such a scenario is very likely; there may
be courses for which sections (classes) have not yet been defined. In fact, some courses are taught only once
a year and do not generate classes each semester.
2.
CLASS is mandatory. This condition is created by the constraint that is imposed by the semantics of the
statement “Each COURSE generates one or more CLASSes.” In ER terms, each COURSE in the “generates”
relationship must have at least one CLASS. Therefore, a CLASS must be created as the COURSE is created,
in order to comply with the semantics of the problem.
Keep in mind the practical aspects of the scenario presented in Figure 4.14. Given the semantics of this relationship,
the system should not accept a course that is not associated with at least one class section. Is such a rigid environment
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desirable from an operational point of view? For example, when a new COURSE is created, the database first updates
the COURSE table, thereby inserting a COURSE entity that does not yet have a CLASS associated with it. Naturally,
the apparent problem seems to be solved when CLASS entities are inserted into the corresponding CLASS table.
However, because of the mandatory relationship, the system will be in temporary violation of the business rule
constraint. For practical purposes, it would be desirable to classify the CLASS as optional in order to produce a more
flexible design.
Finally, as you examine the scenarios presented in Figures 4.13 and 4.14, keep in mind the role of the DBMS. To
maintain data integrity, the DBMS must ensure that the “many” side (CLASS) is associated with a COURSE through
the foreign key rules.
4.1.9 Relationship Degree
A relationship degree indicates the number of entities or participants associated with a relationship. A unary
relationship exists when an association is maintained within a single entity. A binary relationship exists when two
entities are associated. A ternary relationship exists when three entities are associated. Although higher degrees
exist, they are rare and are not specifically named. (For example, an association of four entities is described simply as
a four-degree relationship.) Figure 4.15 shows these types of relationship degrees.
Unary Relationships
In the case of the unary relationship shown in Figure 4.15, an employee within the EMPLOYEE entity is the manager
for one or more employees within that entity. In this case, the existence of the “manages” relationship means that
EMPLOYEE requires another EMPLOYEE to be the manager—that is, EMPLOYEE has a relationship with itself. Such
a relationship is known as a recursive relationship. The various cases of recursive relationships will be explored in
Section 4.1.10.
Binary Relationships
A binary relationship exists when two entities are associated in a relationship. Binary relationships are most common.
In fact, to simplify the conceptual design, whenever possible, most higher-order (ternary and higher) relationships are
decomposed into appropriate equivalent binary relationships. In Figure 4.15, the relationship “a PROFESSOR teaches
one or more CLASSes” represents a binary relationship.
Ternary and Higher-Degree Relationships
Although most relationships are binary, the use of ternary and higher-order relationships does allow the designer some
latitude regarding the semantics of a problem. A ternary relationship implies an association among three different
entities. For example, note the relationships (and their consequences) in Figure 4.16, which are represented by the
following business rules:
쐌
A DOCTOR writes one or more PRESCRIPTIONs.
쐌
A PATIENT may receive one or more PRESCRIPTIONs.
쐌
A DRUG may appear in one or more PRESCRIPTIONs. (To simplify this example, assume that the business
rule states that each prescription contains only one drug. In short, if a doctor prescribes more than one drug,
a separate prescription must be written for each drug.)
As you examine the table contents in Figure 4.16, note that it is possible to track all transactions. For instance, you
can tell that the first prescription was written by doctor 32445 for patient 102, using the drug DRZ.
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
Three types of relationship degree
4.15
4.1.10 Recursive Relationships
As was previously mentioned, a recursive relationship is one in which a relationship can exist between occurrences
of the same entity set. (Naturally, such a condition is found within a unary relationship.) For example, a 1:M unary
relationship can be expressed by “an EMPLOYEE may manage many EMPLOYEEs, and each EMPLOYEE is
managed by one EMPLOYEE.” And as long as polygamy is not legal, a 1:1 unary relationship may be expressed by
“an EMPLOYEE may be married to one and only one other EMPLOYEE.” Finally, the M:N unary relationship may
be expressed by “a COURSE may be a prerequisite to many other COURSEs, and each COURSE may have many
other COURSEs as prerequisites.” Those relationships are shown in Figure 4.17.
The 1:1 relationship shown in Figure 4.17 can be implemented in the single table shown in Figure 4.18. Note that
you can determine that James Ramirez is married to Louise Ramirez, who is married to James Ramirez. And Anne
Jones is married to Anton Shapiro, who is married to Anne Jones.
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FIGURE
4
The implementation of a ternary relationship
4.16
Database name: Ch04_Clinic
Table name: DRUG
Table name: PATIENT
Table name: DOCTOR
FIGURE
Table name: PRESCRIPTION
An ER representation of recursive relationships
4.17
FIGURE
4.18
The 1:1 recursive relationship
“EMPLOYEE is married to
EMPLOYEE”
Database name: CH04_PartCo
Table name: EMPLOYEE_V1
Unary relationships are common in manufacturing
industries. For example, Figure 4.19 illustrates that a rotor
assembly (C-130) is composed of many parts, but each part
is used to create only one rotor assembly. Figure 4.19
indicates that a rotor assembly is composed of four 2.5-cm
washers, two cotter pins, one 2.5-cm steel shank, four
10.25-cm rotor blades, and two 2.5-cm hex nuts. The
relationship implemented in Figure 4.19 thus enables you to
track each part within each rotor assembly.
If a part can be used to assemble several different kinds of
other parts and is itself composed of many parts, two tables
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
Another unary relationship: “PART contains PART”
4.19
Table name: PART_V1
Database name; CH04_PartCo
are required to implement the “PART contains PART” relationship. Figure 4.20 illustrates such an environment. Parts
tracking is increasingly important as managers become more aware of the legal ramifications of producing more
complex output. In fact, in many industries, especially those involving aviation, full parts tracking is required by law.
FIGURE
Implementation of the M:N recursive relationship “PART contains PART”
4.20
Table name: COMPONENT
Database name: Ch04_PartCo
Table name: PART
The M:N recursive relationship might be more familiar in a school environment. For instance, note how the M:N
“COURSE requires COURSE” relationship illustrated in Figure 4.17 is implemented in Figure 4.21. In this example,
MATH-243 is a prerequisite to QM-261 and QM-362, while both MATH-243 and QM-261 are prerequisites to
QM-362.
Finally, the 1:M recursive relationship “EMPLOYEE manages EMPLOYEE,” shown in Figure 4.17, is implemented in
Figure 4.22.
One common pitfall when working with unary relationships is to confuse participation with referential integrity. In
theory, participation and referential integrity are very different concepts and are normally easy to distinguish in binary
relationships. In practical terms, conversely, participation and referential integrity are very similar because they are
both implemented through constraints on the same set of attributes. This similarity often leads to confusion when the
concepts are applied within the limited structure of a unary relationship. Consider the unary 1:1 relationship described
in Figure 4.18 of a spousal relationship between employees. Participation, as described above, is bidirectional,
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FIGURE
4
Implementation of the M:N recursive relationship “COURSE requires COURSE”
4.21
Table name: COURSE
Database name: Ch04_TinyCollege
Table name: PREREQ
FIGURE
Implementation of the 1:M recursive relationship “EMPLOYEE manages EMPLOYEE”
4.22
Database name: Ch04_PartCo
Table name: EMPLOYEE_V2
meaning that it must be addressed in both directions along the relationship. Participation in Figure 4.18 addresses the
questions:
쐌
Must every employee have a spouse who is an employee?
쐌
Must every employee be a spouse to another employee?
For the data shown in Figure 4.18, the correct answer to both of those questions is “No.” It is possible to be an
employee and not have another employee as a spouse. Also, it is possible to be an employee and not be the spouse
of another employee.
Referential integrity deals with the correspondence of values in the foreign key with values in the related primary key.
Referential integrity is not bidirectional, and therefore has only one question that it answers.
쐌
Must every employee spouse be a valid employee?
For the data shown in Figure 4.18, the correct answer is “Yes.” Another way to frame this question is to consider
whether or not every value provided for the EMP_SPOUSE attribute must match some value in the EMP_NUM
attribute.
In practical terms, both participation and referential integrity involve the values used as primary key/foreign key to
implement the relationship. Referential integrity requires that the values in the foreign key correspond to values in the
primary key. In one direction, participation considers whether or not the foreign key can contain a null. In Figure 4.18,
E N T I T Y
R E L A T I O N S H I P
( E R )
M O D E L I N G
for example, employee Robert Delaney is not required to have a value in EMP_SPOUSE. In the other direction,
participation considers whether or not every value in the primary key must appear as a value in the foreign key. In
Figure 4.18, for example, employee Robert Delaney’s value for EMP_NUM (348) is not required to appear as a value
in EMP_SPOUSE for any other employee.
4.1.11 Associative (Composite) Entities
In the original ERM described by Chen, relationships do not contain attributes. You should recall from Chapter 3 that
the relational model generally requires the use of 1:M relationships. (Also, recall that the 1:1 relationship has its place,
but it should be used with caution and proper justification.) If M:N relationships are encountered, you must create a
bridge between the entities that display such relationships. The associative entity is used to implement a M:N
relationship between two or more entities. This associative entity (also known as a composite or bridge entity) is
composed of the primary keys of each of the entities to be connected. An example of such a bridge is shown in Figure
4.23. The Crow’s Foot notation does not identify the composite entity as such. Instead, the composite entity is
identified by the solid relationship line between the parent and child entities, thereby indicating the presence of a strong
(identifying) relationship.
FIGURE
Converting the M:N relationship into two 1:M relationships
4.23
Table name: STUDENT
Database name: Ch04_CollegeTry
Table name: ENROLL
Table name: CLASS
Note that the composite ENROLL entity in Figure 4.23 is existence-dependent on the other two entities; the
composition of the ENROLL entity is based on the primary keys of the entities that are connected by the composite
entity. The composite entity may also contain additional attributes that play no role in the connective process. For
example, although the entity must be composed of at least the STUDENT and CLASS primary keys, it may also
include such additional attributes as grades, absences, and other data uniquely identified by the student’s performance
in a specific class.
Finally, keep in mind that the ENROLL table’s key (CLASS_CODE and STU_NUM) is composed entirely of the
primary keys of the CLASS and STUDENT tables. Therefore, no null entries are possible in the ENROLL table’s key
attributes.
Implementing the small database shown in Figure 4.23 requires that you define the relationships clearly. Specifically,
you must know the “1” and the “M” sides of each relationship, and you must know whether the relationships are
mandatory or optional. For example, note the following points:
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쐌
4
A class may exist (at least at the start of registration) even though it contains no students. Therefore, if you
examine Figure 4.24, an optional symbol should appear on the STUDENT side of the M:N relationship
between STUDENT and CLASS.
You might argue that to be classified as a STUDENT, a person must be enrolled in at least one CLASS.
Therefore, CLASS is mandatory to STUDENT from a purely conceptual point of view. However, when a
student is admitted to college, that student has not (yet) signed up for any classes. Therefore, at least initially,
CLASS is optional to STUDENT. Note that the practical considerations in the data environment help dictate
the use of optionalities. If CLASS is not optional to STUDENT—from a database point of view—a class
assignment must be made when the student is admitted. But that’s not how the process actually works, and
the database design must reflect this. In short, the optionality reflects practice.
FIGURE
The M:N relationship between STUDENT and CLASS
4.24
Because the M:N relationship between STUDENT and CLASS is decomposed into two 1:M relationships
through ENROLL, the optionalities must be transferred to ENROLL. (See Figure 4.25.) In other words, it now
becomes possible for a class not to occur in ENROLL if no student has signed up for that class. Because a class
need not occur in ENROLL, the ENROLL entity becomes optional to CLASS. And because the ENROLL
entity is created before any students have signed up for a class, the ENROLL entity is also optional to
STUDENT, at least initially.
FIGURE
A composite entity in an ERD
4.25
쐌
As students begin to sign up for their classes, they will be entered into the ENROLL entity. Naturally, if a
student takes more than one class, that student will occur more than once in ENROLL. For example, note that
in the ENROLL table in Figure 4.23, STU_NUM = 321452 occurs three times. On the other hand, each
student occurs only once in the STUDENT entity. (Note that the STUDENT table in Figure 4.23 has only one
STU_NUM = 321452 entry.) Therefore, in Figure 4.25, the relationship between STUDENT and ENROLL
is shown to be 1:M, with the M on the ENROLL side.
E N T I T Y
쐌
R E L A T I O N S H I P
( E R )
M O D E L I N G
As you can see in Figure 4.23, a class can occur more than once in the ENROLL table. For example,
CLASS_CODE = 10014 occurs twice. However, CLASS_CODE = 10014 occurs only once in the CLASS
table to reflect that the relationship between CLASS and ENROLL is 1:M. Note that in Figure 4.25, the M is
located on the ENROLL side, while the 1 is located on the CLASS side.
4.2 DEVELOPING AN ER DIAGRAM
The process of database design is an iterative rather than a linear or sequential process. The verb iterate means “to
do again or repeatedly.” An iterative process is, thus, one based on repetition of processes and procedures. Building
an ERD usually involves the following activities:
쐌
Create a detailed narrative of the organization’s description of operations.
쐌
Identify the business rules based on the description of operations.
쐌
Identify the main entities and relationships from the business rules.
쐌
Develop the initial ERD.
쐌
Identify the attributes and primary keys that adequately describe the entities.
쐌
Revise and review the ERD.
During the review process, it is likely that additional objects, attributes, and relationships will be uncovered. Therefore,
the basic ERM will be modified to incorporate the newly discovered ER components. Subsequently, another round of
reviews might yield additional components or clarification of the existing diagram. The process is repeated until the end
users and designers agree that the ERD is a fair representation of the organization’s activities and functions.
During the design process, the database designer does not depend simply on interviews to help define entities,
attributes, and relationships. A surprising amount of information can be gathered by examining the business forms and
reports that an organization uses in its daily operations.
To illustrate the use of the iterative process that ultimately yields a workable ERD, let’s start with an initial interview
with the Tiny College administrators. The interview process yields the following business rules:
1.
Tiny College (TC) is divided into several schools: a school of business, a school of arts and sciences, a school
of education, and a school of applied sciences. Each school is administered by a dean who is a professor. Each
professor can be the dean of only one school, and a professor is not required to be the dean of any school.
Therefore, a 1:1 relationship exists between PROFESSOR and SCHOOL. Note that the cardinality can be
expressed by writing (1,1) next to the entity PROFESSOR and (0,1) next to the entity SCHOOL.
2.
Each school comprises several departments. For example, the school of business has an accounting
department, a management/marketing department, an economics/finance department, and a computer
information systems department. Note again the cardinality rules: The smallest number of departments
operated by a school is one, and the largest number of departments is indeterminate (N). On the other hand,
each department belongs to only a single school; thus, the cardinality is expressed by (1,1). That is, the
minimum number of schools that a department belongs to is one, as is the maximum number. Figure 4.26
illustrates these first two business rules.
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FIGURE
4
The first Tiny College ERD segment
4.26
Note
It is again appropriate to evaluate the reason for maintaining the 1:1 relationship between PROFESSOR and
SCHOOL in the PROFESSOR is dean of SCHOOL relationship. It is worth repeating that the existence of 1:1
relationships often indicates a misidentification of attributes as entities. In this case, the 1:1 relationship could
easily be eliminated by storing the dean’s attributes in the SCHOOL entity. This solution would also make it
easier to answer the queries, “Who is the dean?” and “What are that dean’s credentials?” The downside of this
solution is that it requires the duplication of data that are already stored in the PROFESSOR table, thus setting
the stage for anomalies. However, because each school is run by a single dean, the problem of data duplication
is rather minor. The selection of one approach over another often depends on information requirements,
transaction speed, and the database designer’s professional judgment. In short, do not use 1:1 relationships
lightly, and make sure that each 1:1 relationship within the database design is defensible.
3.
Each department may offer courses. For example, the management/marketing department offers courses such
as Introduction to Management, Principles of Marketing, and Production Management. The ERD segment for
this condition is shown in Figure 4.27. Note that this relationship is based on the way Tiny College operates.
If, for example, Tiny College had some departments that were classified as “research only,” those departments
would not offer courses; therefore, the COURSE entity would be optional to the DEPARTMENT entity.
4.
The relationship between COURSE and CLASS was illustrated in Figure 4.9. Nevertheless, it is worth
repeating that a CLASS is a section of a COURSE. That is, a department may offer several sections (classes)
of the same database course. Each of those classes is taught by a professor at a given time in a given place.
In short, a 1:M relationship exists between COURSE and CLASS. However, because a course may exist in
Tiny College’s course catalog even when it is not offered as a class in a current class schedule, CLASS is
optional to COURSE. Therefore, the relationship between COURSE and CLASS looks like Figure 4.28.
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
The second Tiny College ERD segment
4.27
FIGURE
The third Tiny College ERD segment
4.28
5.
Each department should have one or more professors assigned to it. One and only one of those professors
chairs the department, and no professor is required to accept the chair position. Therefore, DEPARTMENT
is optional to PROFESSOR in the “chairs” relationship. Those relationships are summarized in the ER segment
shown in Figure 4.29.
FIGURE
The fourth Tiny College ERD segment
4.29
6.
Each professor may teach up to four classes; each class is a section of a course. A professor may also be on
a research contract and teach no classes at all. The ERD segment in Figure 4.30 depicts those conditions.
7.
A student may enroll in several classes but takes each class only once during any given enrollment period. For
example, during the current enrollment period, a student may decide to take five classes—Statistics,
Accounting, English, Database, and History—but that student would not be enrolled in the same Statistics class
five times during the enrollment period! Each student may enroll in up to six classes, and each class may have
up to 35 students, thus creating an M:N relationship between STUDENT and CLASS. Because a CLASS can
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FIGURE
4
The fifth Tiny College ERD segment
4.30
initially exist (at the start of the enrollment period) even though no students have enrolled in it, STUDENT is
optional to CLASS in the M:N relationship. This M:N relationship must be divided into two 1:M relationships
through the use of the ENROLL entity, shown in the ERD segment in Figure 4.31. But note that the optional
symbol is shown next to ENROLL. If a class exists but has no students enrolled in it, that class doesn’t occur
in the ENROLL table. Note also that the ENROLL entity is weak: it is existence-dependent, and its (composite)
PK is composed of the PKs of the STUDENT and CLASS entities. You can add the cardinalities (0,6) and
(0,35) next to the ENROLL entity to reflect the business rule constraints, as shown in Figure 4.31. (Visio
Professional does not automatically generate such cardinalities, but you can use a text box to accomplish
that task.)
FIGURE
The sixth Tiny College ERD segment
4.31
8.
Each department has several (or many) students whose major is offered by that department. However, each
student has only a single major and is, therefore, associated with a single department. (See Figure 4.32.)
However, in the Tiny College environment, it is possible—at least for a while—for a student not to declare a
major field of study. Such a student would not be associated with a department; therefore, DEPARTMENT is
optional to STUDENT. It is worth repeating that the relationships between entities and the entities themselves
reflect the organization’s operating environment. That is, the business rules define the ERD components.
9.
Each student has an advisor in his or her department; each advisor counsels several students. An advisor is also
a professor, but not all professors advise students. Therefore, STUDENT is optional to PROFESSOR in the
“PROFESSOR advises STUDENT” relationship. (See Figure 4.33.)
10.
As you can see in Figure 4.34, the CLASS entity contains a ROOM_CODE attribute. Given the naming
conventions, it is clear that ROOM_CODE is an FK to another entity. Clearly, because a class is taught in a
room, it is reasonable to assume that the ROOM_CODE in CLASS is the FK to an entity named ROOM. In
turn, each room is located in a building. So the last Tiny College ERD is created by observing that a BUILDING
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
The seventh Tiny College ERD segment
4.32
FIGURE
The eighth Tiny College ERD segment
4.33
can contain many ROOMs, but each ROOM is found in a single BUILDING. In this ERD segment, it is clear
that some buildings do not contain (class) rooms. For example, a storage building might not contain any named
rooms at all.
FIGURE
The ninth Tiny College ERD segment
4.34
Using the preceding summary, you can identify the following entities:
SCHOOL
COURSE
DEPARTMENT
CLASS
PROFESSOR
STUDENT
BUILDING
ROOM
ENROLL (the associative entity between STUDENT and CLASS)
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Once you have discovered the relevant entities, you can define the initial set of relationships among them. Next, you
describe the entity attributes. Identifying the attributes of the entities helps you to better understand the relationships
among entities. Table 4.4 summarizes the ERM’s components, and names the entities and their relations.
TABLE
4.4
Components of the ERM
ENTITY
RELATIONSHIP
CONNECTIVITY
SCHOOL
operates
1:M
DEPARTMENT
has
1:M
DEPARTMENT
employs
1:M
DEPARTMENT
offers
1:M
COURSE
generates
1:M
PROFESSOR
is dean of
1:1
PROFESSOR
chairs
1:1
PROFESSOR
teaches
1:M
PROFESSOR
advises
1:M
STUDENT
enrolls in
M:N
BUILDING
contains
1:M
ROOM
is used for
1:M
Note: ENROLL is the composite entity that implements the M:N relationship “STUDENT
ENTITY
DEPARTMENT
STUDENT
PROFESSOR
COURSE
CLASS
SCHOOL
DEPARTMENT
CLASS
STUDENT
CLASS
ROOM
CLASS
enrolls in CLASS.”
You must also define the connectivity and cardinality for the just-discovered relations based on the business rules.
However, to avoid crowding the diagram, the cardinalities are not shown. Figure 4.35 shows the Crow’s Foot ERD for
Tiny College. Note that this is an implementation-ready model. Therefore it shows the ENROLL composite entity.
Figure 4.36 shows the conceptual UML class diagram for Tiny College. Note that this class diagram depicts the M:N
relationship between STUDENT and CLASS. Figure 4.37 shows the implementation-ready UML class diagram for
Tiny College (note that the ENROLL composite entity is shown in this class diagram.
4.3 DATABASE DESIGN CHALLENGES: CONFLICTING GOALS
Database designers often must make design compromises that are triggered by conflicting goals, such as adherence to
design standards (design elegance), processing speed, and information requirements.
쐌
Design standards. The database design must conform to design standards. Such standards have guided you
in developing logical structures that minimize data redundancies, thereby minimizing the likelihood that
destructive data anomalies will occur. You have also learned how standards prescribe avoiding nulls to the
greatest extent possible. In fact, you have learned that design standards govern the presentation of all
components within the database design. In short, design standards allow you to work with well-defined
components and to evaluate the interaction of those components with some precision. Without design
standards, it is nearly impossible to formulate a proper design process, to evaluate an existing design, or to
trace the likely logical impact of changes in design.
쐌
Processing speed. In many organizations, particularly those generating large numbers of transactions, high
processing speeds are often a top priority in database design. High processing speed means minimal access
time, which may be achieved by minimizing the number and complexity of logically desirable relationships. For
example, a “perfect” design might use a 1:1 relationship to avoid nulls, while a higher transaction-speed design
might combine the two tables to avoid the use of an additional relationship, using dummy entries to avoid the
nulls. If the focus is on data-retrieval speed, you might also be forced to include derived attributes in the design.
쐌
Information requirements. The quest for timely information might be the focus of database design. Complex
information requirements may dictate data transformations, and they may expand the number of entities and
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
The completed Tiny College ERD
4.35
attributes within the design. Therefore, the database may have to sacrifice some of its “clean” design structures
and/or some of its high transaction speed to ensure maximum information generation. For example, suppose
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FIGURE
4
The conceptual UML class diagram for Tiny College
4.36
that a detailed sales report must be generated periodically. The sales report includes all invoice subtotals, taxes,
and totals; even the invoice lines include subtotals. If the sales report includes hundreds of thousands (or even
millions) of invoices, computing the totals, taxes, and subtotals is likely to take some time. If those computations
had been made and the results had been stored as derived attributes in the INVOICE and LINE tables at the
time of the transaction, the real-time transaction speed might have declined. But that loss of speed would only
be noticeable if there were many simultaneous transactions. The cost of a slight loss of transaction speed at the
front end and the addition of multiple derived attributes is likely to pay off when the sales reports are generated
(not to mention the fact that it will be simpler to generate the queries).
A design that meets all logical requirements and design conventions is an important goal. However, if this perfect
design fails to meet the customer’s transaction speed and/or information requirements, the designer will not have done
a proper job from the end user’s point of view. Compromises are a fact of life in the real world of database design.
Even while focusing on the entities, attributes, relationships, and constraints, the designer should begin thinking about
end-user requirements such as performance, security, shared access, and data integrity. The designer must consider
processing requirements and verify that all update, retrieval, and deletion options are available. Finally, a design is of
little value unless the end product is capable of delivering all specified query and reporting requirements.
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
The implementation-ready UML class diagram for Tiny College
4.37
You are quite likely to discover that even the best design process produces an ERD that requires further changes
mandated by operational requirements. Such changes should not discourage you from using the process. ER modeling
is essential in the development of a sound design that is capable of meeting the demands of adjustment and growth.
Using ERDs yields perhaps the richest bonus of all: a thorough understanding of how an organization really functions.
There are occasional design and implementation problems that do not yield “clean” implementation solutions. To get
a sense of the design and implementation choices a database designer faces, let’s revisit the 1:1 recursive relationship
“EMPLOYEE is married to EMPLOYEE” first examined in Figure 4.18. Figure 4.38 shows three different ways of
implementing such a relationship.
Note that the EMPLOYEE_V1 table in Figure 4.38 is likely to yield data anomalies. For example, if Anne Jones
divorces Anton Shapiro, two records must be updated—by setting the respective EMP_SPOUSE values to null—to
properly reflect that change. If only one record is updated, inconsistent data occur. The problem becomes even worse
if several of the divorced employees then marry each other. In addition, that implementation also produces undesirable
nulls for employees who are not married to other employees in the company.
Another approach would be to create a new entity shown as MARRIED_V1 in a 1:M relationship with EMPLOYEE.
(See Figure 4.38.) This second implementation does eliminate the nulls for employees who are not married to
somebody working for the same company. (Such employees would not be entered in the MARRIED_V1 table.)
However, this approach still yields possible duplicate values. For example, the marriage between employees 345 and
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FIGURE
4
Various implementations of the 1:1 recursive relationship
4.38
Table name: EMPLOYEE_V1
Database name: Ch04_PartCo
First implementation
Table name: EMPLOYEE
Table name: MARRIED_V1
Second implementation
Table name: MARRIAGE
Table name: MARPART
Table name: EMPLOYEE
The relational diagram for the third implementation
Third implementation
347 may still appear twice, once as 345,347 and once as 347,345. (Since each of those permutations is unique the
first time it appears, the creation of a unique index will not solve the problem.)
As you can see, the first two implementations yield several problems:
쐌
Both solutions use synonyms. The EMPLOYEE_V1 table uses EMP_NUM and EMP_SPOUSE to refer to an
employee. The MARRIED_V1 table uses the same synonyms.
쐌
Both solutions are likely to produce inconsistent data. For example, it is possible to enter employee 345 as
married to employee 347 and to enter employee 348 as married to employee 345.
쐌
Both solutions allow data entries to show one employee married to several other employees. For example, it
is possible to have data pairs such as 345,347 and 348,347 and 349,347, none of which will violate entity
integrity requirements, because they are all unique.
A third approach would be to have two new entities, MARRIAGE and MARPART, in a 1:M relationship. MARPART
contains the EMP_NUM foreign key to EMPLOYEE. (See the relational diagram in Figure 4.38.) But even this
approach has issues. It requires the collection of additional data regarding the employees’ marriage—the marriage
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date. If the business users do not need this data, then requiring them to collect it would be inappropriate. To ensure
that an employee occurs only once in any given marriage, you would have to create a unique index on the EMP_NUM
attribute in the MARPART table. Another potential problem with this solution is that the database implementation will
allow more than two employees to “participate” in the same marriage.
As you can see, a recursive 1:1 relationship yields many different solutions with varying degrees of effectiveness and
adherence to basic design principles. Any of the above solutions would likely involve the creation of program code to
help ensure the integrity and consistency of the data. In a later chapter, we will examine the creation of database
triggers that can do exactly that. Your job as a database designer is to use your professional judgment to yield a solution
that meets the requirements imposed by business rules, processing requirements, and basic design principles.
Finally, document, document, and document! Put all design activities in writing. Then review what you’ve written.
Documentation not only helps you stay on track during the design process, but also enables you (or those following
you) to pick up the design thread when the time comes to modify the design. Although the need for documentation
should be obvious, one of the most vexing problems in database and systems analysis work is that the “put it in writing”
rule is often not observed in all of the design and implementation stages. The development of organizational
documentation standards is a very important aspect of ensuring data compatibility and coherence.
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S u m m a r y
◗
◗
◗
◗
◗
◗
The ERM uses ERDs to represent the conceptual database as viewed by the end user. The ERM’s main components
are entities, relationships, and attributes. The ERD also includes connectivity and cardinality notations. An ERD can
also show relationship strength, relationship participation (optional or mandatory), and degree of relationship
(unary, binary, ternary, etc.).
Connectivity describes the relationship classification (1:1, 1:M, or M:N). Cardinality expresses the specific number
of entity occurrences associated with an occurrence of a related entity. Connectivities and cardinalities are usually
based on business rules.
In the ERM, an M:N relationship is valid at the conceptual level. However, when implementing the ERM in a
relational database, the M:N relationship must be mapped to a set of 1:M relationships through a composite entity.
ERDs may be based on many different ERMs. However, regardless of which model is selected, the modeling logic
remains the same. Because no ERM can accurately portray all real-world data and action constraints, application
software must be used to augment the implementation of at least some of the business rules.
Unified Modeling Language (UML) class diagrams are used to represent the static data structures in a data model.
The symbols used in the UML class and ER diagrams are very similar. The UML class diagrams can be used to
depict data models at the conceptual or implementation abstraction levels.
Database designers, no matter how well they are able to produce designs that conform to all applicable modeling
conventions, are often forced to make design compromises. Those compromises are required when end users have
vital transaction-speed and/or information requirements that prevent the use of “perfect” modeling logic and
adherence to all modeling conventions. Therefore, database designers must use their professional judgment to
determine how and to what extent the modeling conventions are subject to modification. To ensure that their
professional judgments are sound, database designers must have detailed and in-depth knowledge of data-modeling
conventions. It is also important to document the design process from beginning to end, which helps keep the
design process on track and allows for easy modifications down the road.
K e y
T e r m s
binary relationship, 116
mandatory participation, 113
required attribute, 101
cardinality, 107
multivalued attributes, 103
simple attribute, 103
composite attribute, 103
non-identifying relationship, 109
single-valued attribute, 103
composite identifier, 102
optional attribute, 101
strong entity, 108
connectivity, 107
optional participation, 113
strong relationship, 110
derived attribute, 105
participants, 105
ternary relationship, 116
existence-dependent, 108
recursive relationship, 116
unary relationship, 116
existence-independent, 108
regular entity, 108
weak entity, 110
identifiers, 101
relationship degree, 116
weak relationship, 109
identifying relationship, 110
iterative process, 123
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Online Content
Answers to selected Review Questions and Problems for this chapter are contained in the Premium Website for
this book.
R e v i e w
Q u e s t i o n s
1. What two conditions must be met before an entity can be classified as a weak entity? Give an example of a weak
entity.
2. What is a strong (or identifying) relationship, and how is it depicted in a Crow’s Foot ERD?
3. Given the business rule “an employee may have many degrees,” discuss its effect on attributes, entities, and
relationships. (Hint: Remember what a multivalued attribute is and how it might be implemented.)
4. What is a composite entity, and when is it used?
5. Suppose you are working within the framework of the conceptual model in Figure Q4.5.
FIGURE
The conceptual model for Question 5
Q4.5
Given the conceptual model in Figure Q4.5:
a. Write the business rules that are reflected in it.
b. Identify all of the cardinalities.
6. What is a recursive relationship? Give an example.
7. How would you (graphically) identify each of the following ERM components in a Crow’s Foot notation?
a. an entity
b. the cardinality (0,N)
c. a weak relationship
d. a strong relationship
8. Discuss the difference between a composite key and a composite attribute. How would each be indicated in
an ERD?
9. What two courses of action are available to a designer encountering a multivalued attribute?
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10. What is a derived attribute? Give an example.
11. How is a relationship between entities indicated in an ERD? Give an example, using the Crow’s Foot notation.
12. Discuss two ways in which the 1:M relationship between COURSE and CLASS can be implemented. (Hint:
Think about relationship strength.)
13. How is a composite entity represented in an ERD, and what is its function? Illustrate the Crow’s Foot notation.
14. What three (often conflicting) database requirements must be addressed in database design?
15. Briefly, but precisely, explain the difference between single-valued attributes and simple attributes. Give an
example of each.
16. What are multivalued attributes, and how can they be handled within the database design?
The next four questions are based on the ERD in Figure Q4.17.
FIGURE
The ERD for Questions 17–20
Q4.17
17. Write the 10 cardinalities that are appropriate for this ERD.
18. Write the business rules reflected in this ERD.
19. What two attributes must be contained in the composite entity between STORE and PRODUCT? Use proper
terminology in your answer.
20. Describe precisely the composition of the DEPENDENT weak entity’s primary key. Use proper terminology in
your answer.
21. The local city youth league needs a database system to help track children who sign up to play soccer. Data need
to be kept on each team and the children who will be playing on each team and their parents. Also, data need
to be kept on the coaches for each team.
Draw the data model described below.
Entities required: Team, Player, Coach, and Parent.
Attributes required:
Team: Team ID number, Team name, and Team colors.
Player: Player ID number, Player first name, Player last name, and Player age.
Coach: Coach ID number, Coach first name, Coach last name, and Coach home phone number.
Parent: Parent ID number, Parent last name, Parent first name, Home phone number, and Home Address
(Street, City, State, and Zip Code).
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The following relationships must be defined:
쐌
Team is related to Player.
쐌
Team is related to Coach.
쐌
Player is related to Parent.
Connectivities and participations are defined as follows:
쐌
A Team may or may not have a Player.
쐌
A Player must have a Team.
쐌
A Team may have many Players.
쐌
A Player has only one Team.
쐌
A Team may or may not have a Coach.
쐌
A Coach must have a Team.
쐌
A Team may have many Coaches.
쐌
A Coach has only one Team.
쐌
A Player must have a Parent.
쐌
A Parent must have a Player.
쐌
A Player may have many Parents.
쐌
A Parent may have many Players.
P r o b l e m s
1. Use the following business rules to create a Crow’s Foot ERD. Write all appropriate connectivities and
cardinalities in the ERD.
a. A department employs many employees, but each employee is employed by only one department.
b. Some employees, known as “rovers,” are not assigned to any department.
c. A division operates many departments, but each department is operated by only one division.
d. An employee may be assigned many projects, and a project may have many employees assigned to it.
e. A project must have at least one employee assigned to it.
f.
One of the employees manages each department, and each department is managed by only one employee.
g. One of the employees runs each division, and each division is run by only one employee.
2. The Jonesburgh County Basketball Conference (JCBC) is an amateur basketball association. Each city in the
county has one team as its representative. Each team has a maximum of 12 players and a minimum of 9 players.
Each team also has up to three coaches (offensive, defensive, and physical training coaches). During the season,
each team plays two games (home and visitor) against each of the other teams. Given those conditions, do the
following:
a. Identify the connectivity of each relationship.
b. Identify the type of dependency that exists between CITY and TEAM.
c. Identify the cardinality between teams and players and between teams and city.
d. Identify the dependency between coach and team and between team and player.
e. Draw the Chen and Crow’s Foot ERDs to represent the JCBC database.
f.
Draw the UML class diagram to depict the JCBC database.
3. Create an ERD based on the Crow’s Foot notation, using the following requirements:
a. An INVOICE is written by a SALESREP. Each sales representative can write many invoices, but each invoice
is written by a single sales representative.
b. The INVOICE is written for a single CUSTOMER. However, each customer can have many invoices.
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c. An INVOICE can include many detail lines (LINE), each of which describes one product bought by the
customer.
d. The product information is stored in a PRODUCT entity.
e. The product’s vendor information is found in a VENDOR entity.
4. The Hudson Engineering Group (HEG) has contacted you to create a conceptual model whose application will
meet the expected database requirements for the company’s training program. The HEG administrator gives you
the description (see below) of the training group’s operating environment. (Hint: Some of the following sentences
identify the volume of data rather than cardinalities. Can you tell which ones?)
The HEG has 12 instructors and can handle up to 30 trainees per class. HEG offers five Advanced Technology
courses, each of which may generate several classes. If a class has fewer than 10 trainees, it will be canceled.
Therefore, it is possible for a course not to generate any classes. Each class is taught by one instructor. Each
instructor may teach up to two classes or may be assigned to do research only. Each trainee may take up to two
classes per year.
Given that information, do the following:
a. Define all of the entities and relationships. (Use Table 4.4 as your guide.)
b. Describe the relationship between instructor and class in terms of connectivity, cardinality, and existence
dependence.
5. Automata, Inc. produces specialty vehicles by contract. The company operates several departments, each of
which builds a particular vehicle, such as a limousine, a truck, a van, or an RV.
쐌
Before a new vehicle is built, the department places an order with the purchasing department to request
specific components. Automata’s purchasing department is interested in creating a database to keep track
of orders and to accelerate the process of delivering materials.
쐌
The order received by the purchasing department may contain several different items. An inventory is
maintained so that the most frequently requested items are delivered almost immediately. When an order
comes in, it is checked to determine whether the requested item is in inventory. If an item is not in inventory,
it must be ordered from a supplier. Each item may have several suppliers.
Given that functional description of the processes encountered at Automata’s purchasing department, do the
following:
a. Identify all of the main entities.
b. Identify all of the relations and connectivities among entities.
c. Identify the type of existence dependence in all the relationships.
d. Give at least two examples of the types of reports that can be obtained from the database.
6. United Helpers is a nonprofit organization that provides aid to people after natural disasters. Based on the
following brief description of operations, create the appropriate fully labeled Crow’s Foot ERD.
쐌
Individuals volunteer their time to carry out the tasks of the organization. The name, address, and telephone
number for each voluteer are tracked. Each volunteer may be assigned to several tasks during the time that
he or she is doing volunteer work, and some tasks require many volunteers. It is possible for a volunteer to
be in the system without having been assigned a task yet. It is possible to have tasks that no one has been
assigned. When a volunteer is assigned to a task, the system should track the start time and end time of that
assignment.
쐌
For each task, there is a task code, task description, task type, and task status. For example, there may be
a task with task code “101,” a description of “answer the telephone,” a type of “recurring,” and a status of
“ongoing.” There could be another task with a code of “102,” a description of “prepare 5000 packages of
basic medical supplies,” a type of “packing,” and a status of “open.”
E N T I T Y
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쐌
For all tasks of type “packing,” there is a packing list that specifies the contents of the packages. There are
many different packing lists to produce different packages, such as basic medical packages, child-care
packages, food packages, etc. Each packing list has a packing list ID number, a packing list name, and a
packing list description, which describes the items that ideally go into making that type of package. Every
packing task is associated with only one packing list. A packing list may not be associated with any tasks,
or may be associated with many tasks. Tasks that are not packing tasks are not associated with any
packing list.
쐌
Packing tasks result in the creation of packages. Each individual package of supplies that is produced by the
organization is tracked. Each package is assigned an ID number. The date the package was created and the
total weight of the package are recorded. A given package is associated with only one task. Some tasks (e.g.,
“answer the phones”) will not have produced any packages, while other tasks (e.g., “prepare 5000 packages
of basic medical supplies”) will be associated with many packages.
쐌
The packing list describes the ideal contents of each package, but it is not always possible to include the ideal
number of each item. Therefore, the actual items included in each package should be tracked. A package
can contain many different items, and a given item can be used in many different packages.
쐌
For each item that the organization provides, there is an item ID number, item description, item value, and
item quantity on hand stored in the system. Along with tracking the actual items that are placed in each
package, the quantity of each item placed in the package must be tracked too. For example, a packing list
may state that basic medical packages should include 100 bandages, 4 bottles of iodine, and 4 bottles of
hydrogen peroxide. However, because of the limited supply of items, a given package may include only 10
bandages, 1 bottle of iodine, and no hydrogen peroxide. The fact that this package includes bandages and
iodine needs to be recorded along with the quantity of each that is included. It is possible for the organization
to have items donated that have not been included in any package yet, but every package will contain at least
one item.
7. Using the Crow’s Foot notation, create an ERD that can be implemented for a medical clinic, using the following
business rules:
쐌
A patient can make many appointments with one or more doctors in the clinic, and a doctor can accept
appointments with many patients. However, each appointment is made with only one doctor and one
patient.
쐌
Emergency cases do not require an appointment. However, for appointment management purposes, an
emergency is entered in the appointment book as “unscheduled.”
쐌
If kept, an appointment yields a visit with the doctor specified in the appointment. The visit yields a diagnosis
and, when appropriate, treatment.
쐌
With each visit, the patient’s records are updated to provide a medical history.
쐌
Each patient visit creates a bill. Each patient visit is billed by one doctor, and each doctor can bill many
patients.
쐌
Each bill must be paid. However, a bill may be paid in many installments, and a payment may cover more
than one bill.
쐌
A patient may pay the bill directly, or the bill may be the basis for a claim submitted to an insurance
company.
쐌
If the bill is paid by an insurance company, the deductible is submitted to the patient for payment.
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Note
The following cases and additional problems from the Instructor Online Companion may be used as the basis
for class projects. These problems illustrate the challenge of translating a description of operations into a set of
business rules that will define the components for an ERD that can be successfully implemented. These
problems can also be used as the basis for discussions about the components and contents of a proper
description of operations. One of the things you must learn if you want to create databases that can be
successfully implemented is to separate the generic background material from the details that directly affect
database design. You must also keep in mind that many constraints cannot be incorporated into the database
design; instead, such constraints are handled by the applications software.
C a s e s
8. The administrators of Tiny College are so pleased with your design and implementation of their student
registration/tracking system that they want you to expand the design to include the database for their motor
vehicle pool. A brief description of operations follows:
쐌
Faculty members may use the vehicles owned by Tiny College for officially sanctioned travel. For example,
the vehicles may be used by faculty members to travel to off-campus learning centers, to travel to locations
at which research papers are presented, to transport students to officially sanctioned locations, and to travel
for public service purposes. The vehicles used for such purposes are managed by Tiny College’s Travel Far
But Slowly (TFBS) Center.
쐌
Using reservation forms, each department can reserve vehicles for its faculty, who are responsible for filling
out the appropriate trip completion form at the end of a trip. The reservation form includes the expected
departure date, vehicle type required, destination, and name of the authorized faculty member. The faculty
member arriving to pick up a vehicle must sign a checkout form to log out the vehicle and pick up a trip
completion form. (The TFBS employee who releases the vehicle for use also signs the checkout form.) The
faculty member’s trip completion form includes the faculty member’s identification code, the vehicle’s
identification, the odometer readings at the start and end of the trip, maintenance complaints (if any), gallons
of fuel purchased (if any), and the Tiny College credit card number used to pay for the fuel. If fuel is
purchased, the credit card receipt must be stapled to the trip completion form. Upon receipt of the faculty
trip completion form, the faculty member’s department is billed at a mileage rate based on the vehicle type
(sedan, station wagon, panel truck, minivan, or minibus) used. (Hint: Do not use more entities than are
necessary. Remember the difference between attributes and entities!)
쐌
All vehicle maintenance is performed by TFBS. Each time a vehicle requires maintenance, a maintenance log
entry is completed on a prenumbered maintenance log form. The maintenance log form includes the vehicle
identification, a brief description of the type of maintenance required, the initial log entry date, the date on
which the maintenance was completed, and the identification of the mechanic who released the vehicle back
into service. (Only mechanics who have an inspection authorization may release the vehicle back into service.)
쐌
As soon as the log form has been initiated, the log form’s number is transferred to a maintenance detail
form; the log form’s number is also forwarded to the parts department manager, who fills out a parts usage
form on which the maintenance log number is recorded. The maintenance detail form contains separate
lines for each maintenance item performed, for the parts used, and for identification of the mechanic who
performed the maintenance item. When all maintenance items have been completed, the maintenance detail
form is stapled to the maintenance log form, the maintenance log form’s completion date is filled out, and
the mechanic who releases the vehicle back into service signs the form. The stapled forms are then filed, to
be used later as the source for various maintenance reports.
E N T I T Y
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M O D E L I N G
쐌
TFBS maintains a parts inventory, including oil, oil filters, air filters, and belts of various types. The parts
inventory is checked daily to monitor parts usage and to reorder parts that reach the “minimum quantity on
hand” level. To track parts usage, the parts manager requires each mechanic to sign out the parts that are
used to perform each vehicle’s maintenance; the parts manager records the maintenance log number under
which the part is used.
쐌
Each month TFBS issues a set of reports. The reports include the mileage driven by vehicle, by department,
and by faculty members within a department. In addition, various revenue reports are generated by vehicle
and department. A detailed parts usage report is also filed each month. Finally, a vehicle maintenance
summary is created each month.
Given that brief summary of operations, draw the appropriate (and fully labeled) ERD. Use the Chen
methodology to indicate entities, relationships, connectivities, and cardinalities.
9. During peak periods, Temporary Employment Corporation (TEC) places temporary workers in companies.
TEC’s manager gives you the following description of the business:
쐌
TEC has a file of candidates who are willing to work.
쐌
If the candidate has worked before, that candidate has a specific job history. (Naturally, no job history exists
if the candidate has never worked.) Each time the candidate works, one additional job history record is
created.
쐌
Each candidate has earned several qualifications. Each qualification may be earned by more than one
candidate. (For example, it is possible for more than one candidate to have earned a Bachelor of Business
Administration degree or a Microsoft Network Certification. And clearly, a candidate may have earned both
a BBA and a Microsoft Network Certification.)
쐌
TEC offers courses to help candidates improve their qualifications.
쐌
Every course develops one specific qualification; however, TEC does not offer a course for every
qualification. Some qualifications have multiple courses that develop that qualification.
쐌
Some courses cover advanced topics that require specific qualifications as prerequisites. Some courses cover
basic topics that do not require any prerequisite qualifications. A course can have several prerequisites. A
qualification can be a prerequisite for more than one course.
쐌
Courses are taught during training sessions. A training session is the presentation of a single course. Over
time, TEC will offer many training sessions for each course; however, new courses may not have any training
sessions scheduled right away.
쐌
Candidates can pay a fee to attend a training session. A training session can accommodate several
candidates, although new training sessions will not have any candidates registered at first.
쐌
TEC also has a list of companies that request temporaries.
쐌
Each time a company requests a temporary employee, TEC makes an entry in the Openings folder. That
folder contains an opening number, a company name, required qualifications, a starting date, an anticipated
ending date, and hourly pay.
쐌
Each opening requires only one specific or main qualification.
쐌
When a candidate matches the qualification, the job is assigned, and an entry is made in the Placement
Record folder. That folder contains an opening number, a candidate number, the total hours worked, etc. In
addition, an entry is made in the job history for the candidate.
쐌
An opening can be filled by many candidates, and a candidate can fill many openings.
쐌
TEC uses special codes to describe a candidate’s qualifications for an opening. The list of codes is shown in
Table P4.9.
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TABLE
P4.9
CODE
SEC-45
SEC-60
CLERK
PRG-VB
PRG-C++
DBA-ORA
DBA-DB2
DBA-SQLSERV
SYS-1
SYS-2
NW-NOV
WD-CF
DESCRIPTION
Secretarial work, at least 45 words per minute
Secretarial work, at least 60 words per minute
General clerking work
Programmer, Visual Basic
Programmer, C++
Database Administrator, Oracle
Database Administrator, IBM DB2
Database Administrator, MS SQL Server
Systems Analyst, level 1
Systems Analyst, level 2
Network Administrator, Novell experience
Web Developer, ColdFusion
TEC’s management wants to keep track of the following entities:
COMPANY, OPENING, QUALIFICATION, CANDIDATE, JOB_HISTORY, PLACEMENT, COURSE, and
SESSION. Given that information, do the following:
a. Draw the Crow’s Foot ERDs for this enterprise.
b. Identify all necessary relationships.
c. Identify the connectivity for each relationship.
d. Identify the mandatory/optional dependencies for the relationships.
e. Resolve all M:N relationships.
10. Use the following description of the operations of the RC_Charter2 Company to complete this exercise.
쐌
The RC_Charter2 Company operates a fleet of aircraft under the Federal Air Regulations (FAR) Part 135
(air taxi or charter) certificate, enforced by the FAA. The aircraft are available for air taxi (charter) operations
within the United States and Canada.
쐌
Charter companies provide so-called “unscheduled” operations—that is, charter flights take place only after
a customer reserves the use of an aircraft to fly at a customer-designated date and time to one or more
customer-designated destinations, transporting passengers, cargo, or some combination of passengers and
cargo. A customer can, of course, reserve many different charter flights (trips) during any time frame.
However, for billing purposes, each charter trip is reserved by one and only one customer. Some of
RC_Charter2’s customers do not use the company’s charter operations; instead, they purchase fuel, use
maintenance services, or use other RC_Charter2 services. However, this database design will focus on the
charter operations only.
쐌
Each charter trip yields revenue for the RC_Charter2 Company. This revenue is generated by the charges
a customer pays upon the completion of a flight. The charter flight charges are a function of aircraft model
used, distance flown, waiting time, special customer requirements, and crew expenses. The distance flown
charges are computed by multiplying the round-trip miles by the model’s charge per mile. Round-trip miles
are based on the actual navigational path flown. The sample route traced in Figure P4.10 illustrates the
procedure. Note that the number of round-trip miles is calculated to be 130 + 200 + 180 + 390 = 900.
Depending on whether a customer has RC_Charter2 credit authorization, the customer may:
쐌
Pay the entire charter bill upon the completion of the charter flight.
쐌
Pay a part of the charter bill and charge the remainder to the account. The charge amount may not exceed
the available credit.
쐌
Charge the entire charter bill to the account. The charge amount may not exceed the available credit.
쐌
Customers may pay all or part of the existing balance for previous charter trips. Such payments may be
E N T I T Y
FIGURE
R E L A T I O N S H I P
( E R )
M O D E L I N G
Round-trip mile determination
P4.10
Destination
180 miles
Intermediate Stop
200 miles
390 miles
Pax Pickup
130 miles
Home Base
made at any time and are not necessarily tied to a specific charter trip. The charter mileage charge includes
the expense of the pilot(s) and other crew required by FAR 135. However, if customers request additional
crew not required by FAR 135, those customers are charged for the crew members on an hourly basis. The
hourly crew-member charge is based on each crew member’s qualifications.
쐌
The database must be able to handle crew assignments. Each charter trip requires the use of an aircraft, and
a crew flies each aircraft. The smaller piston-engine-powered charter aircraft require a crew consisting of
only a single pilot. Larger aircraft (i.e., aircraft having a gross takeoff weight of 12,500 pounds or more) and
jet-powered aircraft require a pilot and a copilot, while some of the larger aircraft used to transport
passengers may require flight attendants as part of the crew. Some of the older aircraft require the
assignment of a flight engineer, and larger cargo-carrying aircraft require the assignment of a loadmaster. In
short, a crew can consist of more than one person, and not all crew members are pilots.
쐌
The charter flight’s aircraft waiting charges are computed by multiplying the hours waited by the model’s
hourly waiting charge. Crew expenses are limited to meals, lodging, and ground transportation.
The RC_Charter2 database must be designed to generate a monthly summary of all charter trips, expenses, and
revenues derived from the charter records. Such records are based on the data that each pilot in command is
required to record for each charter trip: trip date(s) and time(s), destination(s), aircraft number, pilot (and other
crew) data, distance flown, fuel usage, and other data pertinent to the charter flight. Such charter data are then
used to generate monthly reports that detail revenue and operating cost information for customers, aircraft, and
pilots. All pilots and other crew members are RC_Charter2 Company employees; that is, the company does not
use contract pilots and crew.
FAR Part 135 operations are conducted under a strict set of requirements that govern the licensing and training
of crew members. For example, pilots must have earned either a commercial license or an Airline Transport Pilot
(ATP) license. Both licenses require appropriate ratings. Ratings are specific competency requirements. For
example:
쐌
To operate a multiengine aircraft designed for takeoffs and landings on land only, the appropriate rating is
MEL, or Multiengine Landplane. When a multiengine aircraft can take off and land on water, the appropriate
rating is MES, or Multiengine Seaplane.
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4
쐌
The instrument rating is based on a demonstrated ability to conduct all flight operations with sole reference
to cockpit instrumentation. The instrument rating is required to operate an aircraft under Instrument
Meteorological Conditions (IMC), and all such operations are governed under FAR-specified Instrument
Flight Rules (IFR). In contrast, operations conducted under “good weather” or visual flight conditions are
based on the FAR Visual Flight Rules (VFR).
쐌
The type rating is required for all aircraft with a takeoff weight of more than 12,500 pounds or for aircraft
that are purely jet-powered. If an aircraft uses jet engines to drive propellers, that aircraft is said to be
turboprop-powered. A turboprop—that is, a turbo-propeller-powered aircraft—does not require a type rating
unless it meets the 12,500-pound weight limitation.
쐌
Although pilot licenses and ratings are not time-limited, exercising the privilege of the license and ratings
under Part 135 requires both a current medical certificate and a current Part 135 checkride. The
following distinctions are important:
쐌
The medical certificate may be Class I or Class II. The Class I medical is more stringent than the Class II, and
it must be renewed every six months. The Class II medical must be renewed yearly. If the Class I medical is
not renewed during the six-month period, it automatically reverts to a Class II certificate. If the Class II
medical is not renewed within the specified period, it automatically reverts to a Class III medical, which is not
valid for commercial flight operations.
쐌
A Part 135 checkride is a practical flight examination that must be successfully completed every six months.
The checkride includes all flight maneuvers and procedures specified in Part 135.
Nonpilot crew members must also have the proper certificates in order to meet specific job requirements. For
example, loadmasters need an appropriate certificate, as do flight attendants. In addition, crew members such as
loadmasters and flight attendants, who may be required in operations that involve large aircraft (more than a
12,500-pound takeoff weight and passenger configurations over 19) are also required periodically to pass a
written and practical exam. The RC_Charter2 Company is required to keep a complete record of all test types,
dates, and results for each crew member, as well as pilot medical certificate examination dates.
In addition, all flight crew members are required to submit to periodic drug testing; the results must be tracked,
too. (Note that nonpilot crew members are not required to take pilot-specific tests such as Part 135 checkrides.
Nor are pilots required to take crew tests such as loadmaster and flight attendant practical exams.) However,
many crew members have licenses and/or certifications in several areas. For example, a pilot may have an ATP
and a loadmaster certificate. If that pilot is assigned to be a loadmaster on a given charter flight, the loadmaster
certificate is required. Similarly, a flight attendant may have earned a commercial pilot’s license. Sample data
formats are shown in Table P4.10.
Pilots and other crew members must receive recurrency training appropriate to their work assignments.
Recurrency training is based on an FAA-approved curriculum that is job-specific. For example, pilot recurrency
training includes a review of all applicable Part 135 flight rules and regulations, weather data interpretation,
company flight operations requirements, and specified flight procedures. The RC_Charter2 Company is required
to keep a complete record of all recurrency training for each crew member subject to the training.
The RC_Charter2 Company is required to maintain a detailed record of all crew credentials and all training mandated
by Part 135. The company must keep a complete record of each requirement and of all compliance data.
To conduct a charter flight, the company must have a properly maintained aircraft available. A pilot who meets
all of the FAA’s licensing and currency requirements must fly the aircraft as Pilot in Command (PIC). For those
aircraft that are powered by piston engines or turboprops and have a gross takeoff weight under 12,500 pounds,
single-pilot operations are permitted under Part 135 as long as a properly maintained autopilot is available.
However, even if FAR Part 135 permits single-pilot operations, many customers require the presence of a copilot
who is capable of conducting the flight operations under Part 135.
The RC_Charter2 operations manager anticipates the lease of turbojet-powered aircraft, and those aircraft are
required to have a crew consisting of a pilot and copilot. Both pilot and copilot must meet the same Part 135
licensing, ratings, and training requirements.
E N T I T Y
R E L A T I O N S H I P
( E R )
M O D E L I N G
TABLE
P4.10
PART A TESTS
TEST CODE
1
2
3
4
5
6
7
PART B RESULTS
EMPLOYEE
101
103
112
103
112
101
101
125
TEST DESCRIPTION
Part 135 Flight Check
Medical, Class 1
Medical, Class 2
Loadmaster Practical
Flight Attendant Practical
Drug test
Operations, written exam
TEST CODE
1
6
4
7
7
7
6
2
TEST FREQUENCY
6 months
6 months
12 months
12 months
12 months
Random
6 months
TEST DATE
12-Nov-09
23-Dec-09
23-Dec-09
11-Jan-10
16-Jan-10
16-Jan-10
11-Feb-10
15-Feb-10
PART C LICENSES AND CERTIFICATIONS
LICENSE OR CERTIFICATE
ATP
Comm
Med-1
Med-2
Instr
MEL
LM
FA
EMPLOYEE
101
101
101
103
112
103
112
TEST RESULT
Pass-1
Pass-1
Pass-2
Pass-1
Pass-1
Pass-1
Pass-2
Pass-1
LICENSE OR CERTIFICATE DESCRIPTION
Airline Transport Pilot
Commercial license
Medical certificate, Class I
Medical certificate, Class II
Instrument rating
Multiengine Land aircraft rating
Loadmaster
Flight Attendant
LICENSE OR CERTIFICATE
Comm
Instr
MEL
Comm
FA
Instr
LM
DATE EARNED
12-Nov-93
28-Jun-94
9-Aug-94
21-Dec-95
23-Jun-02
18-Jan-96
27-Nov-05
The company also leases larger aircraft that exceed the 12,500-pound gross takeoff weight. Those aircraft can
carry the number of passengers that requires the presence of one or more flight attendants. If those aircraft carry
cargo weighing over 12,500 pounds, a loadmaster must be assigned as a crew member to supervise the loading
and securing of the cargo. The database must be designed to meet the anticipated additional charter crew
assignment capability.
a. Given this incomplete description of operations, write all applicable business rules to establish entities,
relationships, optionalities, connectivities, and cardinalities. (Hint: Use the following five business rules as
examples, writing the remaining business rules in the same format.)
쐌
A customer may request many charter trips.
쐌
Each charter trip is requested by only one customer.
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쐌
Some customers have not (yet) requested a charter trip.
쐌
An employee may be assigned to serve as a crew member on many charter trips.
쐌
Each charter trip may have many employees assigned to it to serve as crew members.
b. Draw the fully labeled and implementable Crow’s Foot ERD based on the business rules you wrote in Part (a)
of this problem. Include all entities, relationships, optionalities, connectivities, and cardinalities.
In this chapter, you will learn:
쐍 About the extended entity relationship (EER) model
쐍 How entity clusters are used to represent multiple entities and relationships
쐍 The characteristics of good primary keys and how to select them
쐍 How to use flexible solutions for special data-modeling cases
In the previous two chapters, you learned how to use entity relationship diagrams (ERDs)
to properly create a data model. In this chapter, you will learn about the extended entity
relationship (EER) model.The EER model builds on ER concepts and adds support for entity
P
review
supertypes, subtypes, and entity clustering.
Most current database implementations are based on relational databases. Because the
relational model uses keys to create associations among tables, it is essential to learn
the characteristics of good primary keys and how to select them. Selecting a good primary
key is too important to be left to chance, so in this chapter we cover the critical aspects
of primary key identification and placement.
Focusing on practical database design, this chapter also illustrates some special design cases that
highlight the importance of flexible designs, which can be adapted to meet the demands of
changing data and information requirements. Data modeling is a vital step in the development
of databases that in turn provide a good foundation for successful application development.
Remember that good database applications cannot be based on bad database designs, and no
amount of outstanding coding can overcome the limitations of poor database design.
5
F I V E
Advanced Data Modeling
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5.1 THE EXTENDED ENTITY RELATIONSHIP MODEL
As the complexity of the data structures being modeled has increased and as application software requirements have
become more stringent, there has been an increasing need to capture more information in the data model. The extended
entity relationship model (EERM), sometimes referred to as the enhanced entity relationship model, is the result of
adding more semantic constructs to the original entity relationship (ER) model. As you might expect, a diagram using
this model is called an EER diagram (EERD). In the following sections, you will learn about the main EER model
constructs—entity supertypes, entity subtypes, and entity clustering—and see how they are represented in ERDs.
5.1.1 Entity Supertypes and Subtypes
Because most employees possess a wide range of skills and special qualifications, data modelers must find a variety of
ways to group employees based on employee characteristics. For instance, a retail company could group employees
as salaried and hourly employees, while a university could group employees as faculty, staff, and administrators.
The grouping of employees to create various types of employees provides two important benefits:
쐌
It avoids unnecessary nulls in the employee attributes when some employees have characteristics that are not
shared by other employees.
쐌
It enables a particular employee type to participate in relationships that are unique to that employee type.
To illustrate those benefits, let’s explore the case of an aviation business. The aviation business employs pilots,
mechanics, secretaries, accountants, database managers, and many other types of employees. Figure 5.1 illustrates
how pilots share certain characteristics with other employees, such as a last name (EMP_LNAME) and hire date
(EMP_HIRE_DATE). On the other hand, many pilot characteristics are not shared by other employees. For example,
unlike other employees, pilots must meet special requirements such as flight hour restrictions, flight checks, and
periodic training. Therefore, if all employee characteristics and special qualifications were stored in a single
EMPLOYEE entity, you would have a lot of nulls or you would have to make a lot of needless dummy entries. In this
case, special pilot characteristics such as EMP_LICENSE, EMP_RATINGS, and EMP_MED_TYPE will generate nulls
for employees who are not pilots. In addition, pilots participate in some relationships that are unique to their
qualifications. For example, not all employees can fly airplanes; only employees who are pilots can participate in the
“employee flies airplane” relationship.
FIGURE
Nulls created by unique attributes
5.1
Based on the preceding discussion, you would correctly deduce that the PILOT entity stores only those attributes that
are unique to pilots, and that the EMPLOYEE entity stores attributes that are common to all employees. Based on that
hierarchy, you can conclude that PILOT is a subtype of EMPLOYEE, and that EMPLOYEE is the supertype of PILOT.
In modeling terms, an entity supertype is a generic entity type that is related to one or more entity subtypes, where
A D V A N C E D
D A T A
M O D E L I N G
the entity supertype contains the common characteristics, and the entity subtypes contain the unique characteristics of
each entity subtype.
There are two criteria that help the designer determine when to use subtypes and supertypes:
쐌
There must be different, identifiable kinds or types of the entity in the user’s environment.
쐌
The different kinds or types of instances should have one or more attributes that are unique to that kind or type
of instance.
In the preceding example, because pilots meet both criteria of being an identifiable kind of employee and having unique
attributes that other employees do not possess, it is appropriate to create PILOT as a subtype of EMPLOYEE. Let us
assume that mechanics and accountants also have attributes that are unique to them, respectively, and that clerks do
not. In that case, MECHANIC and ACCOUNTANT would also be legitimate subtypes of EMPLOYEE because they are
identifiable kinds of employees and they have unique attributes. CLERK would not be an acceptable subtype of
EMPLOYEE because it only satisfies one of the criteria—it is an identifiable kind of employee—but there are not any
attributes that are unique to just clerks. In the next section, you will learn how entity supertypes and subtypes are
related in a specialization hierarchy.
5.1.2 Specialization Hierarchy
Entity supertypes and subtypes are organized in a specialization hierarchy, which depicts the arrangement of
higher-level entity supertypes (parent entities) and lower-level entity subtypes (child entities). Figure 5.2 shows the
specialization hierarchy formed by an EMPLOYEE supertype and three entity subtypes—PILOT, MECHANIC, and
ACCOUNTANT. The specialization hierarchy reflects the 1:1 relationship between EMPLOYEE and its subtypes. For
example, a PILOT subtype occurrence is related to one instance of the EMPLOYEE supertype, and a MECHANIC
subtype occurrence is related to one instance of the EMPLOYEE supertype. The terminology and symbols in
Figure 5.2 are explained throughout this chapter.
The relationships depicted within the specialization hierarchy are sometimes described in terms of “is-a” relationships.
For example, a pilot is an employee, a mechanic is an employee, and an accountant is an employee. It is important
to understand that within a specialization hierarchy, a subtype can exist only within the context of a supertype, and
every subtype can have only one supertype to which it is directly related. However, a specialization hierarchy can have
many levels of supertype/subtype relationships—that is, you can have a specialization hierarchy in which a supertype
has many subtypes; in turn, one of the subtypes is the supertype to other lower-level subtypes.
Online Content
This chapter covers only specialization hierarchies. The EER model also supports specialization lattices, where
a subtype can have multiple parents (supertypes). However, those concepts are better covered under the
object-oriented model in Appendix G, Object-Oriented Databases. The appendix is available in the Premium
Website for this book.
As you can see in Figure 5.2, the arrangement of entity supertypes and subtypes in a specialization hierarchy is more
than a cosmetic convenience. Specialization hierarchies enable the data model to capture additional semantic content
(meaning) into the ERD. A specialization hierarchy provides the means to:
쐌
Support attribute inheritance.
쐌
Define a special supertype attribute known as the subtype discriminator.
쐌
Define disjoint/overlapping constraints and complete/partial constraints.
The following sections cover such characteristics and constraints in more detail.
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5
A specialization hierarchy
5.2
5.1.3 Inheritance
The property of inheritance enables an entity subtype to inherit the attributes and relationships of the supertype. As
discussed earlier, a supertype contains those attributes that are common to all of its subtypes. In contrast, subtypes
contain only the attributes that are unique to the subtype. For example, Figure 5.2 illustrates that pilots, mechanics,
and accountants all inherit the employee number, last name, first name, middle initial, hire date, and so on from the
EMPLOYEE entity. However, Figure 5.2 also illustrates that pilots have attributes that are unique; the same is true for
mechanics and accountants. One important inheritance characteristic is that all entity subtypes inherit their
primary key attribute from their supertype. Note in Figure 5.2 that the EMP_NUM attribute is the primary key for
each of the subtypes.
At the implementation level, the supertype and its subtype(s) depicted in the specialization hierarchy maintain a
1:1 relationship. For example, the specialization hierarchy lets you replace the undesirable EMPLOYEE table structure
in Figure 5.1 with two tables—one representing the supertype EMPLOYEE and the other representing the subtype
PILOT. (See Figure 5.3.)
Entity subtypes inherit all relationships in which the supertype entity participates. For example, Figure 5.2 shows the
EMPLOYEE entity supertype participating in a 1:M relationship with a DEPENDENT entity. Through inheritance, all
subtypes also participate in that relationship. In specialization hierarchies with multiple levels of supertype/subtypes,
a lower-level subtype inherits all of the attributes and relationships from all of its upper-level supertypes.
A D V A N C E D
FIGURE
D A T A
M O D E L I N G
The EMPLOYEE-PILOT supertype-subtype relationship
5.3
Table Name: EMPLOYEE
Table Name: PILOT
5.1.4 Subtype Discriminator
A subtype discriminator is the attribute in the supertype entity that determines to which subtype the supertype
occurrence is related. As seen in Figure 5.2, the subtype discriminator is the employee type (EMP_TYPE).
It is common practice to show the subtype discriminator and its value for each subtype in the ER diagram, as seen in
Figure 5.2. However, not all ER modeling tools follow that practice. For example, MS Visio shows the subtype
discriminator, but not its value. In Figure 5.2, the Visio text tool was used to manually add the discriminator value above
the entity subtype, close to the connector line. Using Figure 5.2 as your guide, note that the supertype is related to
a PILOT subtype if the EMP_TYPE has a value of “P.” If the EMP_TYPE value is “M,” the supertype is related to a
MECHANIC subtype. And if the EMP_TYPE value is “A,” the supertype is related to the ACCOUNTANT subtype.
Note that the default comparison condition for the subtype discriminator attribute is the equality comparison. However,
there may be situations in which the subtype discriminator is not necessarily based on an equality comparison. For
example, based on business requirements, you might create two new pilot subtypes, pilot-in-command (PIC)-qualified
and copilot-qualified only. A PIC-qualified pilot will be anyone with more than 1,500 PIC flight hours. In this case, the
subtype discriminator would be “Flight_Hours,” and the criteria would be > 1,500 or <= 1,500, respectively.
Note
In Visio, you select the subtype discriminator when creating a category using the Category shape from the
available shapes. The Category shape is a small circle with a horizontal line under it that connects the supertype
to its subtypes.
Online Content
For a tutorial on using MS Visio to create a specialization hierarchy, see Appendix A, Designing Databases with
Visio Professional: A Tutorial, in the Premium Website for this book.
5.1.5 Disjoint and Overlapping Constraints
An entity supertype can have disjoint or overlapping entity subtypes. For example, in the aviation example, an employee
can be a pilot or a mechanic or an accountant. Assume that one of the business rules dictates that an employee cannot
belong to more than one subtype at a time; that is, an employee cannot be a pilot and a mechanic at the same time.
Disjoint subtypes, also known as nonoverlapping subtypes, are subtypes that contain a unique subset of the
supertype entity set; in other words, each entity instance of the supertype can appear in only one of the subtypes. For
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example, in Figure 5.2, an employee (supertype) who is a pilot (subtype) can appear only in the PILOT subtype, not
in any of the other subtypes. In Visio, such disjoint subtypes are indicated by the letter d inside the category shape.
On the other hand, if the business rule specifies that employees can have multiple classifications, the EMPLOYEE
supertype may contain overlapping job classification subtypes. Overlapping subtypes are subtypes that contain
nonunique subsets of the supertype entity set; that is, each entity instance of the supertype may appear in more than
one subtype. For example, in a university environment, a person may be an employee or a student or both. In turn,
an employee may be a professor as well as an administrator. Because an employee may also be a student, STUDENT
and EMPLOYEE are overlapping subtypes of the supertype PERSON, just as PROFESSOR and ADMINISTRATOR
are overlapping subtypes of the supertype EMPLOYEE. Figure 5.4 illustrates overlapping subtypes with the use of the
letter o inside the category shape.
FIGURE
Specialization hierarchy with overlapping subtypes
5.4
It is common practice to show the disjoint/overlapping symbols in the ERD. (See Figure 5.2 and Figure 5.4.) However,
not all ER modeling tools follow that practice. For example, by default, Visio shows only the subtype discriminator
(using the Category shape) but not the disjoint/overlapping symbol. Therefore, the Visio text tool was used to manually
add the d and o symbols in Figures 5.2 and 5.4.
Note
Alternative notations exist for representing disjoint/overlapping subtypes. For example, Toby J. Teorey popularized the use of G and Gs to indicate disjoint and overlapping subtypes.
A D V A N C E D
D A T A
M O D E L I N G
As you learned earlier in this section, the implementation of disjoint subtypes is based on the value of the subtype
discriminator attribute in the supertype. However, implementing overlapping subtypes requires the use of one
discriminator attribute for each subtype. For example, in the case of the Tiny College database design you saw in
Chapter 4, Entity Relationship (ER) Modeling, a professor can also be an administrator. Therefore, the EMPLOYEE
supertype would have the subtype discriminator attributes and values shown in Table 5.1.
TABLE
5.1
Discriminator Attributes with Overlapping Subtypes
DISCRIMINATOR ATTRIBUTES
Professor
Administrator
“Y”
“N”
“N”
“Y”
“Y”
“Y”
COMMENT
The Employee is a member of the Professor subtype.
The Employee is a member of the Administrator subtype.
The Employee is both a Professor and an Administrator.
5.1.6 Completeness Constraint
The completeness constraint specifies whether each entity supertype occurrence must also be a member of at least
one subtype. The completeness constraint can be partial or total. Partial completeness (symbolized by a circle over
a single line) means that not every supertype occurrence is a member of a subtype; that is, there may be some
supertype occurrences that are not members of any subtype. Total completeness (symbolized by a circle over a
double line) means that every supertype occurrence must be a member of at least one subtype.
The ERDs in Figures 5.2 and 5.4 represent the completeness constraint based on the Visio Category shape. A single
horizontal line under the circle represents a partial completeness constraint; a double horizontal line under the circle
represents a total completeness constraint.
Note
Alternative notations exist to represent the completeness constraint. For example, some notations use a single
line (partial) or double line (total) to connect the supertype to the Category shape.
Given the disjoint/overlapping subtypes and completeness constraints, it’s possible to have the specialization hierarchy
constraint scenarios shown in Table 5.2.
TABLE
5.2
Specialization Hierarchy Constraint Scenarios
TYPE
Partial
DISJOINT CONSTRAINT
Supertype has optional subtypes.
Subtype discriminator can be null.
Subtype sets are unique.
OVERLAPPING CONSTRAINT
Supertype has optional subtypes.
Subtype discriminators can be null.
Subtype sets are not unique.
Total
Every supertype occurrence is a member of a
(at least one) subtype.
Subtype discriminator cannot be null.
Subtype sets are unique.
Every supertype occurrence is a member of a
(at least one) subtype.
Subtype discriminators cannot be null.
Subtype sets are not unique.
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5.1.7 Specialization and Generalization
You can use various approaches to develop entity supertypes and subtypes. For example, you can first identify a regular
entity, and then identify all entity subtypes based on their distinguishing characteristics. You can also start by identifying
multiple entity types and then later extract the common characteristics of those entities to create a higher-level
supertype entity.
Specialization is the top-down process of identifying lower-level, more specific entity subtypes from a higher-level
entity supertype. Specialization is based on grouping unique characteristics and relationships of the subtypes. In the
aviation example, you used specialization to identify multiple entity subtypes from the original employee supertype.
Generalization is the bottom-up process of identifying a higher-level, more generic entity supertype from lower-level
entity subtypes. Generalization is based on grouping common characteristics and relationships of the subtypes. For
example, you might identify multiple types of musical instruments: piano, violin, and guitar. Using the generalization
approach, you could identify a “string instrument” entity supertype to hold the common characteristics of the multiple
subtypes.
5.2 ENTITY CLUSTERING
Developing an ER diagram entails the discovery of possibly hundreds of entity types and their respective relationships.
Generally, the data modeler will develop an initial ERD containing a few entities. As the design approaches completion,
the ERD will contain hundreds of entities and relationships that crowd the diagram to the point of making it unreadable
and inefficient as a communication tool. In those cases, you can use entity clusters to minimize the number of entities
shown in the ERD.
An entity cluster is a “virtual” entity type used to represent multiple entities and relationships in the ERD. An entity
cluster is formed by combining multiple interrelated entities into a single abstract entity object. An entity cluster is
considered “virtual” or “abstract” in the sense that it is not actually an entity in the final ERD. Instead, it is a temporary
entity used to represent multiple entities and relationships, with the purpose of simplifying the ERD and thus enhancing
its readability.
Figure 5.5 illustrates the use of entity clusters based on the Tiny College example in Chapter 4. Note that the ERD
contains two entity clusters:
쐌
OFFERING, which groups the COURSE and CLASS entities and relationships.
쐌
LOCATION, which groups the ROOM and BUILDING entities and relationships.
Note also that the ERD in Figure 5.5 does not show attributes for the entities. When using entity clusters, the key attributes
of the combined entities are no longer available. Without the key attributes, primary key inheritance rules change. In turn,
the change in the inheritance rules can have undesirable consequences, such as changes in relationships—from identifying
to nonidentifying or vice versa—and the loss of foreign key attributes from some entities. To eliminate those problems, the
general rule is to avoid the display of attributes when entity clusters are used.
A D V A N C E D
FIGURE
D A T A
M O D E L I N G
Tiny College ERD using entity clusters
5.5
5.3 ENTITY INTEGRITY: SELECTING PRIMARY KEYS
Arguably, the most important characteristic of an entity is its primary key (a single attribute or some combination of
attributes), which uniquely identifies each entity instance. The primary key’s function is to guarantee entity integrity.
Furthermore, primary keys and foreign keys work together to implement relationships in the relational model.
Therefore, the importance of properly selecting the primary key has a direct bearing on the efficiency and effectiveness
of database implementation.
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5.3.1 Natural Keys and Primary Keys
The concept of a unique identifier is commonly encountered in the real world. For example, you use class (or section)
numbers to register for classes, invoice numbers to identify specific invoices, account numbers to identify credit cards,
and so on. Those examples illustrate natural identifiers or keys. A natural key or natural identifier is a real-world,
generally accepted identifier used to distinguish—that is, uniquely identify—real-world objects. As its name implies, a
natural key is familiar to end users and forms part of their day-to-day business vocabulary.
Usually, if an entity has a natural identifier, a data modeler uses that as the primary key of the entity being modeled.
Generally, most natural keys make acceptable primary key identifiers. The next section presents some basic guidelines
for selecting primary keys.
5.3.2 Primary Key Guidelines
A primary key is the attribute or combination of attributes that uniquely identifies entity instances in an entity set.
However, can the primary key be based on, say, 12 attributes? And just how long can a primary key be? In previous
examples, why was EMP_NUM selected as a primary key of EMPLOYEE and not a combination of EMP_LNAME,
EMP_FNAME, EMP_INITIAL, and EMP_DOB? Can a single 256-byte text attribute be a good primary key? There is
no single answer to those questions, but there is a body of practice that database experts have built over the years. This
section examines that body of documented practices.
First, you should understand the function of a primary key. The primary key’s main function is to uniquely identify an
entity instance or row within a table. In particular, given a primary key value—that is, the determinant—the relational
model can determine the value of all dependent attributes that “describe” the entity. Note that identification and
description are separate semantic constructs in the model. The function of the primary key is to guarantee entity
integrity, not to “describe” the entity.
Second, primary keys and foreign keys are used to implement relationships among entities. However, the implementation of such relationships is done mostly behind the scenes, hidden from end users. In the real world, end users
identify objects based on the characteristics they know about the objects. For example, when shopping at a grocery
store, you select products by taking them from a store display shelf and reading the labels, not by looking at the stock
number. It’s wise for database applications to mimic the human selection process as much as possible. Therefore,
database applications should let the end user choose among multiple descriptive narratives of different objects, while
using primary key values behind the scenes. Keeping those concepts in mind, look at Table 5.3, which summarizes
desirable primary key characteristics.
TABLE
5.3
Desirable Primary Key Characteristics
PK CHARACTERISTIC
Unique values
Nonintelligent
No change over time
RATIONALE
The PK must uniquely identify each entity instance. A primary key must be able
to guarantee unique values. It cannot contain nulls.
The PK should not have embedded semantic meaning other than to uniquely
identify each entity instance. An attribute with embedded semantic meaning is
probably better used as a descriptive characteristic of the entity than as an
identifier. For example, a student ID of 650973 would be preferred over Smith,
Martha L. as a primary key identifier.
If an attribute has semantic meaning, it might be subject to updates. This is why
names do not make good primary keys. If you have Vickie Smith as the primary
key, what happens if she changes her name when she gets married? If a primary
key is subject to change, the foreign key values must be updated, thus adding to
the database work load. Furthermore, changing a primary key value means that
you are basically changing the identity of an entity. In short, the PK should be
permanent and unchangeable.
A D V A N C E D
TABLE
5.3
D A T A
M O D E L I N G
Desirable Primary Key Characteristics (continued)
PK CHARACTERISTIC
Preferably single-attribute
Preferably numeric
Security-compliant
RATIONALE
A primary key should have the minimum number of attributes possible
(irreducible). Single-attribute primary keys are desirable but not required. Singleattribute primary keys simplify the implementation of foreign keys. Having
multiple-attribute primary keys can cause primary keys of related entities to grow
through the possible addition of many attributes, thus adding to the database
work load and making (application) coding more cumbersome.
Unique values can be better managed when they are numeric, because the
database can use internal routines to implement a counter-style attribute that
automatically increments values with the addition of each new row. In fact, most
database systems include the ability to use special constructs, such as Autonumber in Microsoft Access, to support self-incrementing primary key attributes.
The selected primary key must not be composed of any attribute(s) that might be
considered a security risk or violation. For example, using a Social Security number as a PK in an EMPLOYEE table is not a good idea.
5.3.3 When to Use Composite Primary Keys
In the previous section, you learned about the desirable characteristics of primary keys. For example, you learned that
the primary key should use the minimum number of attributes possible. However, that does not mean that composite
primary keys are not permitted in a model. In fact, composite primary keys are particularly useful in two cases:
쐌
As identifiers of composite entities, where each primary key combination is allowed only once in the
M:N relationship.
쐌
As identifiers of weak entities, where the weak entity has a strong identifying relationship with the parent entity.
To illustrate the first case, assume that you have a STUDENT entity set and a CLASS entity set. In addition, assume that
those two sets are related in an M:N relationship via an ENROLL entity set in which each student/class combination may
appear only once in the composite entity. Figure 5.6 shows the ERD to represent such a relationship.
As shown in Figure 5.6, the composite primary key automatically provides the benefit of ensuring that there cannot
be duplicate values—that is, it ensures that the same student cannot enroll more than once in the same class.
In the second case, a weak entity in a strong identifying relationship with a parent entity is normally used to represent
one of two situations:
1.
A real-world object that is existence-dependent on another real-world object. Those types of objects are
distinguishable in the real world. A dependent and an employee are two separate people who exist
independently of each other. However, such objects can exist in the model only when they relate to each other
in a strong identifying relationship. For example, the relationship between EMPLOYEE and DEPENDENT is
one of existence dependency in which the primary key of the dependent entity is a composite key that contains
the key of the parent entity.
2.
A real-world object that is represented in the data model as two separate entities in a strong identifying
relationship. For example, the real-world invoice object is represented by two entities in a data model:
INVOICE and LINE. Clearly, the LINE entity does not exist in the real world as an independent object, but
rather as part of an INVOICE.
In both situations, having a strong identifying relationship ensures that the dependent entity can exist only when it is
related to the parent entity. In summary, the selection of a composite primary key for composite and weak entity types
provides benefits that enhance the integrity and consistency of the model.
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FIGURE
5
The M:N relationship between STUDENT and CLASS
5.6
Database name: Ch06_Tinycollege
Table name: STUDENT
(first four fields)
Table name: CLASS
(first three fields)
Table name: ENROLL
5.3.4 When to Use Surrogate Primary Keys
There are some instances when a primary key doesn’t exist in the real world or when the existing natural key might
not be a suitable primary key. In these cases, it is standard practice to create a surrogate key. A surrogate key is a
primary key created by the database designer to simplify the identification of entity instances. The surrogate key has
no meaning in the user’s environment—it exists only to distinguish one entity instance from another. One practical
advantage of a surrogate key is that since it has no intrinsic meaning, values for it can be generated by the DBMS to
ensure that unique values are always provided.
For example, consider the case of a park recreation facility that rents rooms for small parties. The manager of the
facility keeps track of all events, using a folder with the format shown in Table 5.4.
TABLE
5.4
DATE
6/17/2010
6/17/2010
6/17/2010
6/17/2010
6/18/2010
6/18/2010
6/18/2010
Data Used to Keep Track of Events
TIME_START
11:00AM
11:00AM
3:00PM
3:30PM
1:00PM
11:00AM
11:00AM
TIME_END
2:00PM
2:00PM
5:30PM
5:30PM
3:00PM
2:00PM
12:30PM
ROOM
Allure
Bonanza
Allure
Bonanza
Bonanza
Allure
Bonanza
EVENT_NAME
Burton Wedding
Adams Office
Smith Family
Adams Office
Boy Scouts
March of Dimes
Smith Family
Given the data shown in Table 5.4, you would model the EVENT entity as:
EVENT (DATE, TIME_START, TIME_END, ROOM, EVENT_NAME, PARTY_OF)
PARTY_OF
60
12
15
12
33
25
12
A D V A N C E D
D A T A
M O D E L I N G
What primary key would you suggest? In this case, there is no simple natural key that could be used as a primary key
in the model. Based on the primary key concepts you learned about in previous chapters, you might suggest one of
these options:
(DATE, TIME_START, ROOM) or (DATE, TIME_END, ROOM)
Assume you select the composite primary key (DATE, TIME_START, ROOM) for the EVENT entity. Next, you
determine that one EVENT may use many RESOURCEs (such as tables, projectors, PCs, and stands), and that the
same RESOURCE may be used for many EVENTs. The RESOURCE entity would be represented by the following
attributes:
RESOURCE (RSC_ID, RSC_DESCRIPTION, RSC_TYPE, RSC_QTY, RSC_PRICE)
Given the business rules, the M:N relationship between RESOURCE and EVENT would be represented via the
EVNTRSC composite entity with a composite primary key as follows:
EVNTRSC (DATE, TIME_START, ROOM, RSC_ID, QTY_USED)
You now have a lengthy four-attribute composite primary key. What would happen if the EVNTRSC entity’s primary
key were inherited by another existence-dependent entity? At this point, you can see that the composite primary key
could make the implementation of the database and program coding unnecessarily complex.
As a data modeler, you probably noticed that the EVENT entity’s selected primary key might not fare well, given the
primary key guidelines in Table 5.3. In this case, the EVENT entity’s selected primary key contains embedded semantic
information and is formed by a combination of date, time, and text data columns. In addition, the selected primary key
would cause lengthy primary keys for existence-dependent entities. The preferred alternative is to use a numeric
single-attribute surrogate primary key.
Surrogate primary keys are accepted practice in today’s complex data environments. They are especially helpful when
there is no natural key, when the selected candidate key has embedded semantic contents, or when the selected candidate
key is too long or cumbersome. However, there is a trade-off: if you use a surrogate key, you must ensure that the
candidate key of the entity in question performs properly through the use of “unique index” and “not null” constraints.
5.4 DESIGN CASES: LEARNING FLEXIBLE DATABASE DESIGN
Data modeling and database design require skills that are acquired through experience. In turn, experience is acquired
through practice—regular and frequent repetition, applying the concepts learned to specific and different design
problems. This section presents four special design cases that highlight the importance of flexible designs, proper
identification of primary keys, and placement of foreign keys.
Note
In describing the various modeling concepts throughout this book, the focus is on relational models. Also, given
the focus on the practical nature of database design, all design issues are addressed with the implementation
goal in mind. Therefore, there is no sharp line of demarcation between design and implementation.
At the pure conceptual stage of the design, foreign keys are not part of an ER diagram. The ERD displays only
entities and relationships. Entities are identified by identifiers that may become primary keys. During design, the
modeler attempts to understand and define the entities and relationships. Foreign keys are the mechanism through
which the relationship designed in an ERD is implemented in a relational model. If you use Visio Professional as your
modeling tool, you will discover that this book’s methodology is reflected in the Visio modeling practice.
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5.4.1 Design Case #1: Implementing 1:1 Relationships
Foreign keys work with primary keys to properly implement relationships in the relational model. The basic rule is very
simple: put the primary key of the “one” side (the parent entity) on the “many” side (the dependent entity) as a foreign
key. However, where do you place the foreign key when you are working with a 1:1 relationship? For example, take
the case of a 1:1 relationship between EMPLOYEE and DEPARTMENT based on the business rule “one EMPLOYEE
is the manager of one DEPARTMENT, and one DEPARTMENT is managed by one EMPLOYEE.” In that case, there
are two options for selecting and placing the foreign key:
1.
Place a foreign key in both entities. This option is derived from the basic rule you learned in Chapter 4. Place
EMP_NUM as a foreign key in DEPARTMENT, and place DEPT_ID as a foreign key in EMPLOYEE. However,
this solution is not recommended, as it would create duplicated work, and it could conflict with other existing
relationships. (Remember that DEPARTMENT and EMPLOYEE also participate in a 1:M relationship—one
department employs many employees.)
2.
Place a foreign key in one of the entities. In that case, the primary key of one of the two entities appears
as a foreign key in the other entity. That is the preferred solution, but there is a remaining question: which
primary key should be used as a foreign key? The answer to that question is found in Table 5.5. Table 5.5
shows the rationale for selecting the foreign key in a 1:1 relationship based on the relationship properties in
the ERD.
TABLE
5.5
CASE
I
Selection of Foreign Key in a 1:1 Relationship
II
ER RELATIONSHIP CONSTRAINTS
One side is mandatory and the other
side is optional.
Both sides are optional.
III
Both sides are mandatory.
ACTION
Place the PK of the entity on the mandatory side in the entity
on the optional side as a FK, and make the FK mandatory.
Select the FK that causes the fewest nulls, or place the FK in
the entity in which the (relationship) role is played.
See Case II, or consider revising your model to ensure that
the two entities do not belong together in a single entity.
Figure 5.7 illustrates the “EMPLOYEE manages DEPARTMENT” relationship. Note that in this case, EMPLOYEE is
mandatory to DEPARTMENT. Therefore, EMP_NUM is placed as the foreign key in DEPARTMENT. Alternatively, you
might also argue that the “manager” role is played by the EMPLOYEE in the DEPARTMENT.
FIGURE
The 1:1 relationship between DEPARTMENT and EMPLOYEE
5.7
As a designer, you must recognize that 1:1 relationships exist in the real world, and therefore, they should be supported
in the data model. In fact, a 1:1 relationship is used to ensure that two entity sets are not placed in the same table.
A D V A N C E D
D A T A
M O D E L I N G
In other words, EMPLOYEE and DEPARTMENT are clearly separate and unique entity types that do not belong
together in a single entity. If you grouped them together in one entity, what would be the name of that entity?
5.4.2 Design Case #2: Maintaining History of Time-Variant Data
Company managers generally realize that good decision making is based on the information that is generated through
the data stored in databases. Such data reflect current as well as past events. Company managers use the data stored
in databases to answer questions such as: “How do the current company profits compare to those of previous years?”
and “What are XYZ product’s sales trends?” In other words, the data stored on databases reflect not only current data,
but also historic data.
Normally, data changes are managed by replacing the existing attribute value with the new value, without regard to the
previous value. However, there are situations in which the history of values for a given attribute must be preserved.
From a data-modeling point of view, time-variant data refer to data whose values change over time and for which
you must keep a history of the data changes. You could argue that all data in a database are subject to change over
time and are, therefore, time variant. However, some attribute values, such as your date of birth or your Social Security
number, are not time variant. On the other hand, attributes such as your student GPA or your bank account balance
are subject to change over time. Sometimes the data changes are externally originated and event driven, such as a
product price change. On other occasions, changes are based on well-defined schedules, such as the daily stock quote
“open” and “close” values.
In any case, keeping the history of time-variant data is equivalent to having a multivalued attribute in your entity. To
model time-variant data, you must create a new entity in a 1:M relationship with the original entity. This new entity
will contain the new value, the date of the change, and whatever other attribute is pertinent to the event being
modeled. For example, if you want to keep track of the current manager as well as the history of all department
managers, you can create the model shown in Figure 5.8.
FIGURE
5.8
Maintaining manager history
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Note that in Figure 5.8, the MGR_HIST entity has a 1:M relationship with EMPLOYEE and a 1:M relationship with
DEPARTMENT to reflect the fact that, over time, an employee could be the manager of many different departments,
and a department could have many different employee managers. Because you are recording time-variant data, you
must store the DATE_ASSIGN attribute in the MGR_HIST entity to provide the date on which the employee
(EMP_NUM) became the manager of the department. The primary key of MGR_HIST permits the same employee to
be the manager of the same department, but on different dates. If that scenario is not the case in your environment—if,
for example, an employee is the manager of a department only once—you could make DATE_ASSIGN a nonprime
attribute in the MGR_HIST entity.
Note in Figure 5.8 that the “manages” relationship is optional in theory and redundant in practice. At any time, you
could find out who the manager of a department is by retrieving the most recent DATE_ASSIGN date from MGR_HIST
for a given department. On the other hand, the ERD in Figure 5.8 differentiates between current data and historic data.
The current manager relationship is implemented by the “manages” relationship between EMPLOYEE and
DEPARTMENT. Additionally, the historic data are managed through EMP_MGR_HIST and DEPT_MGR_HIST. The
trade-off with that model is that each time a new manager is assigned to a department, there will be two data
modifications: one update in the DEPARTMENT entity and one insert in the MGR_HIST entity.
The flexibility of the model proposed in Figure 5.8 becomes more apparent when you add the 1:M “one department
employs many employees” relationship. In that case, the PK of the “1” side (DEPT_ID) appears in the “many” side
(EMPLOYEE) as a foreign key. Now suppose you would like to keep track of the job history for each of the company’s
employees—you’d probably want to store the department, the job code, the date assigned, and the salary. To
accomplish that task, you would modify the model in Figure 5.8 by adding a JOB_HIST entity. Figure 5.9 shows the
use of the new JOB_HIST entity to maintain the employee’s history.
Again, it’s worth emphasizing that the “manages” and “employs” relationships are theoretically optional and redundant
in practice. You can always find out where each employee works by looking at the job history and selecting only the
most current data row for each employee. However, as you will discover in Chapter 7, Introduction to Structured Query
Language (SQL), and in Chapter 8, Advanced SQL, finding where each employee works is not a trivial task. Therefore,
the model represented in Figure 5.9 includes the admittedly redundant but unquestionably useful “manages” and
“employs” relationships to separate current data from historic data.
5.4.3 Design Case #3: Fan Traps
Creating a data model requires proper identification of the data relationships among entities. However, due to
miscommunication or incomplete understanding of the business rules or processes, it is not uncommon to misidentify
relationships among entities. Under those circumstances, the ERD may contain a design trap. A design trap occurs
when a relationship is improperly or incompletely identified and is therefore represented in a way that is not consistent
with the real world. The most common design trap is known as a fan trap.
A fan trap occurs when you have one entity in two 1:M relationships to other entities, thus producing an association
among the other entities that is not expressed in the model. For example, assume the JCB basketball league has many
divisions. Each division has many players, and each division has many teams. Given those “incomplete” business rules,
you might create an ERD that looks like the one in Figure 5.10.
As you can see in Figure 5.10, DIVISION is in a 1:M relationship with TEAM and in a 1:M relationship with PLAYER.
Although that representation is semantically correct, the relationships are not properly identified. For example, there
is no way to identify which players belong to which team. Figure 5.10 also shows a sample instance relationship
representation for the ERD. Note that the relationship lines for the DIVISION instances fan out to the TEAM and
PLAYER entity instances—thus the “fan trap” label.
Figure 5.11 shows the correct ERD after the fan trap has been eliminated. Note that, in this case, DIVISION is in a
1:M relationship with TEAM. In turn, TEAM is in a 1:M relationship with PLAYER. Figure 5.11 also shows the
instance relationship representation after eliminating the fan trap.
A D V A N C E D
FIGURE
D A T A
M O D E L I N G
Maintaining job history
5.9
FIGURE
Incorrect ERD with fan trap problem
5.10
Given the design in Figure 5.11, note how easy it is to see which players play for which team. However, to find out
which players play in which division, you first need to see what teams belong to each division; then you need to find
out which players play on each team. In other words, there is a transitive relationship between DIVISION and PLAYER
via the TEAM entity.
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FIGURE
5
Corrected ERD after removal of the fan trap
5.11
5.4.4 Design Case #4: Redundant Relationships
Although redundancy is often a good thing to have in computer environments (multiple backups in multiple places, for
example), redundancy is seldom a good thing in the database environment. (As you learned in Chapter 3, The
Relational Database Model, redundancies can cause data anomalies in a database.) Redundant relationships occur
when there are multiple relationship paths between related entities. The main concern with redundant relationships is
that they remain consistent across the model. However, it’s important to note that some designs use redundant
relationships as a way to simplify the design.
An example of redundant relationships was first introduced in Figure 5.8 during the discussion on maintaining a history
of time-variant data. However, the use of the redundant “manages” and “employs” relationships was justified by the
fact that such relationships were dealing with current data rather than historic data. Another more specific example of
a redundant relationship is represented in Figure 5.12.
FIGURE
A redundant relationship
5.12
In Figure 5.12, note the transitive 1:M relationship between DIVISION and PLAYER through the TEAM entity set.
Therefore, the relationship that connects DIVISION and PLAYER is, for all practical purposes, redundant. In that case,
the relationship could be safely deleted without losing any information-generation capabilities in the model.
A D V A N C E D
D A T A
M O D E L I N G
S u m m a r y
◗
◗
◗
◗
◗
◗
◗
◗
◗
The extended entity relationship (EER) model adds semantics to the ER model via entity supertypes, subtypes, and
clusters. An entity supertype is a generic entity type that is related to one or more entity subtypes.
A specialization hierarchy depicts the arrangement and relationships between entity supertypes and entity subtypes.
Inheritance means that an entity subtype inherits the attributes and relationships of the supertype. Subtypes can be
disjoint or overlapping. A subtype discriminator is used to determine to which entity subtype the supertype
occurrence is related. The subtypes can exhibit partial or total completeness. There are basically two approaches
to developing a specialization hierarchy of entity supertypes and subtypes: specialization and generalization.
An entity cluster is a “virtual” entity type used to represent multiple entities and relationships in the ERD. An entity
cluster is formed by combining multiple interrelated entities and relationships into a single, abstract entity object.
Natural keys are identifiers that exist in the real world. Natural keys sometimes make good primary keys, but this is not
necessarily true. Primary keys should have these characteristics: they must have unique values, they should be
nonintelligent, they must not change over time, and they are preferably numeric and composed of a single attribute.
Composite keys are useful to represent M:N relationships and weak (strong identifying) entities.
Surrogate primary keys are useful when there is no natural key that makes a suitable primary key, when the primary
key is a composite primary key with multiple different data types, or when the primary key is too long to be usable.
In a 1:1 relationship, place the PK of the mandatory entity as a foreign key in the optional entity, as an FK in the
entity that causes the least number of nulls, or as an FK where the role is played.
Time-variant data refers to data whose values change over time and whose requirements mandate that you keep
a history of data changes. To maintain the history of time-variant data, you must create an entity containing the
new value, the date of change, and any other time-relevant data. This entity maintains a 1:M relationship with the
entity for which the history is to be maintained.
A fan trap occurs when you have one entity in two 1:M relationships to other entities and there is an association
among the other entities that is not expressed in the model. Redundant relationships occur when there are multiple
relationship paths between related entities. The main concern with redundant relationships is that they remain
consistent across the model.
K e y
T e r m s
extended entity relationship
model (EERM), 148
partial completeness, 153
disjoint subtype (nonoverlapping
subtype), 151
fan trap, 162
specialization hierarchy, 149
generalization, 154
subtype discriminator, 151
EER diagram (EERD), 148
inheritance, 150
surrogate key, 158
entity cluster, 154
natural key (natural identifier), 156
time-variant data, 161
entity subtype, 148
overlapping subtype, 152
total completeness, 153
completeness constraint, 153
design trap, 162
entity supertype, 148
specialization, 154
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Online Content
Answers to selected Review Questions and Problems for this chapter are contained in the Premium Website for
this book.
R e v i e w
Q u e s t i o n s
1. What is an entity supertype, and why is it used?
2. What kinds of data would you store in an entity subtype?
3. What is a specialization hierarchy?
4. What is a subtype discriminator? Give an example of its use.
5. What is an overlapping subtype? Give an example.
6. What is the difference between partial completeness and total completeness?
For questions 7–9, refer to Figure Q5.7.
FIGURE
The PRODUCT data model
Q5.7
7. List all of the attributes of a movie.
8. According to the data model, is it required that every entity instance in the PRODUCT table be associated with
an entity instance in the CD table? Why, or why not?
9. Is it possible for a book to appear in the BOOK table without appearing in the PRODUCT table? Why, or why not?
10. What is an entity cluster, and what advantages are derived from its use?
11. What primary key characteristics are considered desirable? Explain why each characteristic is considered desirable.
12. Under what circumstances are composite primary keys appropriate?
13. What is a surrogate primary key, and when would you use one?
14. When implementing a 1:1 relationship, where should you place the foreign key if one side is mandatory and one
side is optional? Should the foreign key be mandatory or optional?
A D V A N C E D
D A T A
M O D E L I N G
15. What are time-variant data, and how would you deal with such data from a database design point of view?
16. What is the most common design trap, and how does it occur?
P r o b l e m s
1. Given the following business scenario, create a Crow’s Foot ERD using a specialization hierarchy if appropriate.
Two-Bit Drilling Company keeps information on employees and their insurance dependents. Each employee has
an employee number, name, date of hire, and title. If an employee is an inspector, then the date of certification
and the renewal date for that certification should also be recorded in the system. For all employees, the Social
Security number and dependent names should be kept. All dependents must be associated with one and only one
employee. Some employees will not have dependents, while others will have many dependents.
2. Given the following business scenario, create a Crow’s Foot ERD using a specialization hierarchy if appropriate.
Tiny Hospital keeps information on patients and hospital rooms. The system assigns each patient a patient ID
number. In addition, the patient’s name and date of birth are recorded. Some patients are resident patients (they
spend at least one night in the hospital) and others are outpatients (they are treated and released). Resident patients
are assigned to a room. Each room is identified by a room number. The system also stores the room type (private
or semiprivate) and room fee. Over time, each room will have many patients who stay in it. Each resident patient will
stay in only one room. Every room must have had a patient, and every resident patient must have a room.
3. Given the following business scenario, create a Crow’s Foot ERD using a specialization hierarchy if appropriate.
Granite Sales Company keeps information on employees and the departments that they work in. For each
department, the department name, internal mail box number, and office phone extension are kept. A department
can have many assigned employees, and each employee is assigned to only one department. Employees can be
salaried employees, hourly employees, or contract employees. All employees are assigned an employee number.
This is kept along with the employee’s name and address. For hourly employees, hourly wage and target weekly
work hours are stored (e.g., the company may target 40 hours/week for some, 32 hours/week for others, and
20 hours/week for others). Some salaried employees are salespeople who can earn a commission in addition to
their base salary. For all salaried employees, the yearly salary amount is recorded in the system. For salespeople,
their commission percentage on sales and commission percentage on profit are stored in the system. For
example, John is a salesperson with a base salary of $50,000 per year plus 2% commission on the sales price
for all sales he makes, plus another 5% of the profit on each of those sales. For contract employees, the
beginning date and end date of their contract are stored along with the billing rate for their hours.
4. In Chapter 4, you saw the creation of the Tiny College database design. That design reflected such business rules
as “a professor may advise many students” and “a professor may chair one department.” Modify the design
shown in Figure 4.36 to include these business rules:
쐌
An employee could be staff or a professor or an administrator.
쐌
A professor may also be an administrator.
쐌
Staff employees have a work level classification, such as Level I and Level II.
쐌
Only professors can chair a department. A department is chaired by only one professor.
쐌
Only professors can serve as the dean of a college. Each of the university’s colleges is served by one dean.
쐌
A professor can teach many classes.
쐌
Administrators have a position title.
Given that information, create the complete ERD containing all primary keys, foreign keys, and main attributes.
5. Tiny College wants to keep track of the history of all administrative appointments (date of appointment and date
of termination). (Hint: Time-variant data are at work.) The Tiny College chancellor may want to know how many
deans worked in the College of Business between January 1, 1960, and January 1, 2010, or who the dean of
the College of Education was in 1990. Given that information, create the complete ERD containing all primary
keys, foreign keys, and main attributes.
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6. Some Tiny College staff employees are information technology (IT) personnel. Some IT personnel provide
technology support for academic programs. Some IT personnel provide technology infrastructure support. Some
IT personnel provide technology support for academic programs and technology infrastructure support. IT
personnel are not professors. IT personnel are required to take periodic training to retain their technical
expertise. Tiny College tracks all IT personnel training by date, type, and results (completed vs. not completed).
Given that information, create the complete ERD containing all primary keys, foreign keys, and main attributes.
7. The FlyRight Aircraft Maintenance (FRAM) division of the FlyRight Company (FRC) performs all maintenance for
FRC’s aircraft. Produce a data model segment that reflects the following business rules:
쐌
All mechanics are FRC employees. Not all employees are mechanics.
쐌
Some mechanics are specialized in engine (EN) maintenance. Some mechanics are specialized in airframe
(AF) maintenance. Some mechanics are specialized in avionics (AV) maintenance. (Avionics are the
electronic components of an aircraft that are used in communication and navigation.) All mechanics take
periodic refresher courses to stay current in their areas of expertise. FRC tracks all courses taken by each
mechanic—date, course type, certification (Y/N), and performance.
쐌
FRC keeps a history of the employment of all mechanics. The history includes the date hired, date promoted,
date terminated, and so on. (Note: The “and so on” component is, of course, not a real-world requirement.
Instead, it has been used here to limit the number of attributes you will show in your design.)
Given those requirements, create the Crow’s Foot ERD segment.
C a s e s
8. “Martial Arts R Us” (MARU) needs a database. MARU is a martial arts school with hundreds of students. It is
necessary to keep track of all the different classes that are being offered, who is assigned to teach each class, and
which students attend each class. Also, it is important to track the progress of each student as they advance.
Create a complete Crow’s Foot ERD for these requirements:
쐌
Students are given a student number when they join the school. This is stored along with their name, date
of birth, and the date they joined the school.
쐌
All instructors are also students, but clearly, not all students are instructors. In addition to the normal student
information, for all instructors, the date that they start working as an instructor must be recorded, along with
their instructor status (compensated or volunteer).
쐌
An instructor may be assigned to teach any number of classes, but each class has one and only one assigned
instructor. Some instructors, especially volunteer instructors, may not be assigned to any class.
쐌
A class is offered for a specific level at a specific time, day of the week, and location. For example, one class
taught on Mondays at 5:00 p.m. in Room #1 is an intermediate-level class. Another class taught on Mondays
at 6:00 p.m. in Room #1 is a beginner-level class. A third class taught on Tuesdays at 5:00 p.m. in
Room #2 is an advanced-level class.
쐌
Students may attend any class of the appropriate level during each week, so there is no expectation that any
particular student will attend any particular class session. Therefore, the actual attendance of students at each
individual class meeting must be tracked.
쐌
A student will attend many different class meetings, and each class meeting is normally attended by many
students. Some class meetings may have no students show up for that meeting. New students may not have
attended any class meetings yet.
쐌
At any given meeting of a class, instructors other than the assigned instructor may show up to help.
Therefore, a given class meeting may have several instructors (a head instructor and many assistant
instructors), but it will always have at least the one instructor who is assigned to that class. For each class
meeting, the date that the class was taught and the instructors’ roles (head instructor or assistant instructor)
need to be recorded. For example, Mr. Jones is assigned to teach the Monday, 5:00 p.m., intermediate class
in Room #1. During one particular meeting of that class, Mr. Jones was present as the head instructor and
Ms. Chen came to help as an assistant instructor.
A D V A N C E D
D A T A
M O D E L I N G
쐌
Each student holds a rank in the martial arts. The rank name, belt color, and rank requirements are stored.
Each rank will have numerous rank requirements. Each requirement is considered a requirement just for the
rank at which the requirement is introduced. Every requirement is associated with a particular rank. All ranks
except white belt have at least one requirement.
쐌
A given rank may be held by many students. While it is customary to think of a student as having a single
rank, it is necessary to track each student’s progress through the ranks. Therefore, every rank that a student
attains is kept in the system. New students joining the school are automatically given a white belt rank. The
date that a student is awarded each rank should be kept in the system. All ranks have at least one student
who has achieved that rank at some time.
9. The Journal of E-commerce Research Knowledge is a prestigious information systems research journal. It uses
a peer-review process to select manuscripts for publication. Only about 10 percent of the manuscripts submitted
to the journal are accepted for publication. A new issue of the journal is published each quarter. Create a
complete ERD to support the business needs described below.
쐌
Unsolicited manuscripts are submitted by authors. When a manuscript is received, the editor will assign the
manuscript a number, and record some basic information about it in the system. The title of the manuscript,
the date it was received, and a manuscript status of “received” are entered. Information about the author(s) is
also recorded. For each author, the author’s name, mailing address, e-mail address, and affiliation (school or
company for which the author works) are recorded. Every manuscript must have an author. Only authors who
have submitted manuscripts are kept in the system. It is typical for a manuscript to have several authors. A single
author may have submitted many different manuscripts to the journal. Additionally, when a manuscript has
multiple authors, it is important to record the order in which the authors are listed in the manuscript credits.
쐌
At her earliest convenience, the editor will briefly review the topic of the manuscript to ensure that the
manuscript’s contents fall within the scope of the journal. If the content is not within the scope of the journal,
the manuscript’s status is changed to “rejected” and the author is notified via e-mail. If the content is within
the scope of the journal, then the editor selects three or more reviewers to review the manuscript. Reviewers
work for other companies or universities and read manuscripts to ensure the scientific validity of the
manuscripts. For each reviewer, the system records a reviewer number, name, e-mail address, affiliation, and
areas of interest. Areas of interest are predefined areas of expertise that the reviewer has specified. An area
of interest is identified by an IS code and includes a description (e.g., IS2003 is the code for “database
modeling”). A reviewer can have many areas of interest, and an area of interest can be associated with many
reviewers. All reviewers must specify at least one area of interest. It is unusual, but it is possible, to have an
area of interest for which the journal has no reviewers. The editor will change the status of the manuscript
to “under review” and record which reviewers the manuscript was sent to and the date on which it was sent
to each reviewer. A reviewer will typically receive several manuscripts to review each year, although new
reviewers may not have received any manuscripts yet.
쐌
The reviewers will read the manuscript at their earliest convenience and provide feedback to the editor regarding
the manuscript. The feedback from each reviewer includes rating the manuscript on a 10-point scale for
appropriateness, clarity, methodology, and contribution to the field, as well as a recommendation for publication
(accept or reject). The editor will record all of this information in the system for each review received from each
reviewer and the date that the feedback was received. Once all of the reviewers have provided their evaluation
of the manuscript, the editor will decide whether or not to publish the manuscript. If the editor decides to publish
the manuscript, the manuscript’s status is changed to “accepted,” and the date of acceptance for the manuscript
is recorded. If the manuscript is not to be published, the status is changed to “rejected.”
쐌
Once a manuscript has been accepted for publication, it must be scheduled. For each issue of the journal,
the publication period (fall, winter, spring, or summer), publication year, volume, and number are recorded.
An issue will contain many manuscripts, although the issue may be created in the system before it is known
which manuscripts will go in that issue. An accepted manuscript appears in only one issue of the journal.
Each manuscript goes through a typesetting process that formats the content (font, font size, line spacing,
justification, etc.). Once the manuscript has been typeset, the number of pages that the manuscript will
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5
occupy is recorded in the system. The editor will then make decisions about which issue each accepted
manuscript will appear in and the order of manuscripts within each issue. The order and the beginning page
number for each manuscript must be stored in the system. Once the manuscript has been scheduled for an
issue, the status of the manuscript is changed to “scheduled.” Once an issue is published, the print date for
the issue is recorded, and the statuses of all of the manuscripts in that issue are changed to “published.”
10. Global Computer Solutions (GCS) is an information technology consulting company with many offices located
throughout the United States. The company’s success is based on its ability to maximize its resources—that is,
its ability to match highly skilled employees with projects according to region. To better manage its projects, GCS
has contacted you to design a database so that GCS managers can keep track of their customers, employees,
projects, project schedules, assignments, and invoices.
The GCS database must support all of GCS’s operations and information requirements. A basic description of
the main entities follows:
쐌
The employees working for GCS have an employee ID, an employee last name, a middle initial, a first
name, a region, and a date of hire.
쐌
Valid regions are as follows: Northwest (NW), Southwest (SW), Midwest North (MN), Midwest South (MS),
Northeast (NE), and Southeast (SE).
쐌
Each employee has many skills, and many employees have the same skill.
쐌
Each skill has a skill ID, description, and rate of pay. Valid skills are as follows: data entry I, data entry II,
systems analyst I, systems analyst II, database designer I, database designer II, Cobol I, Cobol II, C++ I,
C++ II, VB I, VB II, ColdFusion I, ColdFusion II, ASP I, ASP II, Oracle DBA, MS SQL Server DBA, network
engineer I, network engineer II, Web administrator, technical writer, and project manager. Table P5.10a
shows an example of the Skills Inventory.
TABLE
P5.10a
SKILL
Data Entry I
Data Entry II
Systems Analyst I
Systems Analyst II
DB Designer I
DB Designer II
Cobol I
Cobol II
C++ I
C++ II
VB I
VB II
ColdFusion I
ColdFusion II
ASP I
ASP II
Oracle DBA
SQL Server DBA
Network Engineer I
Network Engineer II
Web Administrator
Technical Writer
Project Manager
EMPLOYEE
Seaton Amy; Williams Josh; Underwood Trish
Williams Josh; Seaton Amy
Craig Brett; Sewell Beth; Robbins Erin; Bush Emily; Zebras Steve
Chandler Joseph; Burklow Shane; Robbins Erin
Yarbrough Peter; Smith Mary
Yarbrough Peter; Pascoe Jonathan
Kattan Chris; Ephanor Victor; Summers Anna; Ellis Maria
Kattan Chris; Ephanor Victor; Batts Melissa
Smith Jose; Rogers Adam; Cope Leslie
Rogers Adam; Bible Hanah
Zebras Steve; Ellis Maria
Zebras Steve; Newton Christopher
Duarte Miriam; Bush Emily
Bush Emily; Newton Christopher
Duarte Miriam; Bush Emily
Duarte Miriam; Newton Christopher
Smith Jose; Pascoe Jonathan
Yarbrough Peter; Smith Jose
Bush Emily; Smith Mary
Bush Emily; Smith Mary
Bush Emily; Smith Mary; Newton Christopher
Kilby Surgena; Bender Larry
Paine Brad; Mudd Roger; Kenyon Tiffany; Connor Sean
A D V A N C E D
D A T A
M O D E L I N G
쐌
GCS has many customers. Each customer has a customer ID, customer name, phone number, and region.
쐌
GCS works by projects. A project is based on a contract between the customer and GCS to design, develop,
and implement a computerized solution. Each project has specific characteristics such as the project ID, the
customer to which the project belongs, a brief description, a project date (that is, the date on which the
project’s contract was signed), a project start date (an estimate), a project end date (also an estimate), a
project budget (total estimated cost of the project), an actual start date, an actual end date, an actual cost,
and one employee assigned as the manager of the project.
쐌
The actual cost of the project is updated each Friday by adding that week’s cost (computed by multiplying
the hours each employee worked by the rate of pay for that skill) to the actual cost.
쐌
The employee who is the manager of the project must complete a project schedule, which is, in effect, a
design and development plan. In the project schedule (or plan), the manager must determine the tasks that
will be performed to take the project from beginning to end. Each task has a task ID, a brief task description,
the task’s starting and ending dates, the types of skills needed, and the number of employees (with the
required skills) required to complete the task. General tasks are initial interview, database and system design,
implementation, coding, testing, and final evaluation and sign-off. For example, GCS might have the project
schedule shown in Table P5.10b.
TABLE
P5.10b
PROJECT ID:
COMPANY :
START DATE:
START
DATE
3/1/10
1
SEE ROCKS
3/1/2010
END DATE
DESCRIPTION: SALES MANAGEMENT SYSTEM
CONTRACT DATE: 2/12/2010
REGION:
END DATE: 7/1/2010
BUDGET:
TASK DESCRIPTION
SKILL(S) REQUIRED
3/6/10
Initial Interview
3/11/10
3/11/10
3/15/10
4/12/10
Database Design
System Design
3/18/10
3/25/10
3/22/10
5/20/10
Database Implementation
System Coding & Testing
3/25/10
6/10/10
6/7/10
6/14/10
System Documentation
Final Evaluation
6/17/10
6/21/10
On-Site System Online and
Data Loading
7/1/10
7/1/10
Sign-Off
쐌
Project Manager
Systems Analyst II
DB Designer I
DB Designer I
Systems Analyst II
Systems Analyst I
Oracle DBA
Cobol I
Cobol II
Oracle DBA
Technical Writer
Project Manager
Systems Analyst II
DB Designer I
Cobol II
Project Manager
Systems Analyst II
DB Designer I
Cobol II
Project Manager
NW
$15,500
QUANTITY
REQUIRED
1
1
1
1
1
2
1
2
1
1
1
1
1
1
1
1
1
1
1
1
Assignments: GCS pools all of its employees by region, and from this pool, employees are assigned to a specific
task scheduled by the project manager. For example, for the first project’s schedule, you know that for the period
3/1/10 to 3/6/10, a Systems Analyst II, a Database Designer I, and a Project Manager are needed. (The
project manager is assigned when the project is created and remains for the duration of the project.) Using that
information, GCS searches the employees who are located in the same region as the customer, matching the
skills required and assigning them to the project task.
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5
쐌
Each project schedule task can have many employees assigned to it, and a given employee can work on
multiple project tasks. However, an employee can work on only one project task at a time. For example, if
an employee is already assigned to work on a project task from 2/20/10 to 3/3/10, (s)he cannot work on
another task until the current assignment is closed (ends). The date on which an assignment is closed does
not necessarily match the ending date of the project schedule task, because a task can be completed ahead
of or behind schedule.
쐌
Given all of the preceding information, you can see that the assignment associates an employee with a
project task, using the project schedule. Therefore, to keep track of the assignment, you require at least the
following information: assignment ID, employee, project schedule task, date assignment starts, and date
assignment ends (which could be any date, as some projects run ahead of or behind schedule). Table P5.10c
shows a sample assignment form.
TABLE
P5.10c
PROJECT ID: 1
DESCRIPTION: SALES MANAGEMENT SYSTEM
COMPANY: SEE ROCKS
CONTRACT DATE: 2/12/2010
AS OF: 03/29/10
SCHEDULED
ACTUAL ASSIGNMENTS
Project Task
Start
End Date
Skill
Employee
Start
End Date
Date
Date
3/6/10
3/1/10
101—Connor S.
Initial Interview
3/1/10
3/6/10
Project Mgr.
3/6/10
3/1/10
102—Burklow S.
Sys. Analyst II
3/6/10
3/1/10
103—Smith M.
DB Designer I
Database Design
3/11/10
3/15/10
DB Designer I
104—Smith M.
3/11/10
3/14/10
3/11/10
105—Burklow S.
System Design
3/11/10
4/12/10
Sys. Analyst II
3/11/10
106—Bush E.
Sys. Analyst I
3/11/10
107—Zebras S.
Sys. Analyst I
3/22/10
Oracle DBA
108—Smith J.
3/15/10
3/19/10
Database
3/18/10
Implementation
3/21/10
109—Summers A.
System Coding & 3/25/10
5/20/10
Cobol I
3/21/10
110—Ellis M.
Testing
Cobol I
3/21/10
111—Ephanor V.
Cobol II
3/21/10
112—Smith J.
Oracle DBA
System
3/25/10
6/7/10
Tech. Writer
113—Kilby S.
3/25/10
Documentation
Final Evaluation
6/10/10
6/14/10
Project Mgr.
Sys. Analyst II
DB Designer I
Cobol II
6/17/10
6/21/10
Project Mgr.
On-Site System
Sys. Analyst II
Online and Data
DB Designer I
Loading
Cobol II
7/1/10
Project Mgr.
Sign-Off
7/1/10
(Note: The assignment number is shown as a prefix of the employee name; for example, 101, 102.) Assume that
the assignments shown previously are the only ones existing as of the date of this design. The assignment number
can be whatever number matches your database design.
쐌
The hours an employee works are kept in a work log containing a record of the actual hours worked by an
employee on a given assignment. The work log is a weekly form that the employee fills out at the end of
each week (Friday) or at the end of each month. The form contains the date (of each Friday of the month
A D V A N C E D
D A T A
M O D E L I N G
or the last workday of the month, if it doesn’t fall on a Friday), the assignment ID, the total hours worked
that week (or up to the end of the month), and the number of the bill to which the work-log entry is charged.
Obviously, each work-log entry can be related to only one bill. A sample list of the current work-log entries
for the first sample project is shown in Table P5.10d.
TABLE
P5.10d
EMPLOYEE NAME
Burklow S.
Connor S.
Smith M.
Burklow S.
Connor S.
Smith M.
Burklow S.
Bush E.
Smith J.
Smith M.
Zebras S.
Burklow S.
Bush E.
Ellis M.
Ephanor V.
Smith J.
Smith J.
Summers A.
Zebras S.
Burklow S.
Bush E.
Ellis M.
Ephanor V.
Kilby S.
Smith J.
Summers A.
Zebras S.
Note: xxx represents the
쐌
WEEK ENDING
ASSIGNMENT NUMBER
HOURS WORKED
3/1/10
1-102
4
3/1/10
1-101
4
3/1/10
1-103
4
3/8/10
1-102
24
3/8/10
1-101
24
3/8/10
1-103
24
3/15/10
1-105
40
3/15/10
1-106
40
3/15/10
1-108
6
3/15/10
1-104
32
3/15/10
1-107
35
3/22/10
1-105
40
3/22/10
1-106
40
3/22/10
1-110
12
3/22/10
1-111
12
3/22/10
1-108
12
3/22/10
1-112
12
3/22/10
1-109
12
1-107
35
3/22/10
3/29/10
1-105
40
3/29/10
1-106
40
3/29/10
1-110
35
3/29/10
1-111
35
3/29/10
1-113
40
3/29/10
1-112
35
3/29/10
1-109
35
3/29/10
1-107
35
bill ID. Use the one that matches the bill number in your database.
BILL NUMBER
xxx
xxx
xxx
xxx
xxx
xxx
xxx
xxx
xxx
xxx
xxx
Finally, every 15 days, a bill is written and sent to the customer, totaling the hours worked on the project
that period. When GCS generates a bill, it uses the bill number to update the work-log entries that are part
of that bill. In summary, a bill can refer to many work-log entries, and each work-log entry can be related
to only one bill. GCS sent one bill on 3/15/10 for the first project (SEE ROCKS), totaling the hours worked
between 3/1/10 and 3/15/10. Therefore, you can safely assume that there is only one bill in this table and
that that bill covers the work-log entries shown in the above form.
Your assignment is to create a database that will fulfill the operations described in this problem. The minimum
required entities are employee, skill, customer, region, project, project schedule, assignment, work log, and bill.
(There are additional required entities that are not listed.)
쐌
Create all of the required tables and all of the required relationships.
쐌
Create the required indexes to maintain entity integrity when using surrogate primary keys.
쐌
Populate the tables as needed (as indicated in the sample data and forms).
173
S I X
6
Normalization of Database Tables
In this chapter, you will learn:
쐍 What normalization is and what role it plays in the database design process
쐍 About the normal forms 1NF, 2NF, 3NF, BCNF, and 4NF
쐍 How normal forms can be transformed from lower normal forms to higher normal forms
쐍 That normalization and ER modeling are used concurrently to produce a good database
design
쐍 That some situations require denormalization to generate information efficiently
Good database design must be matched to good table structures. In this chapter, you will
learn to evaluate and design good table structures to control data redundancies, thereby
avoiding data anomalies. The process that yields such desirable results is known as
normalization.
In order to recognize and appreciate the characteristics of a good table structure, it is useful
to examine a poor one.Therefore, the chapter begins by examining the characteristics of a
poor table structure and the problems it creates. You then learn how to correct a poor
table structure. This methodology will yield important dividends: you will know how to
design a good table structure and how to repair an existing poor one.
You will discover not only that data anomalies can be eliminated through normalization, but
also that a properly normalized set of table structures is actually less complicated to use
than an unnormalized set. In addition, you will learn that the normalized set of table
structures more faithfully reflects an organization’s real operations.
P
review
N O R M A L I Z A T I O N
O F
D A T A B A S E
TA B L E S
6.1 DATABASE TABLES AND NORMALIZATION
Having good relational database software is not enough to avoid the data redundancy discussed in Chapter 1, Database
Systems. If the database tables are treated as though they are files in a file system, the relational database management
system (RDBMS) never has a chance to demonstrate its superior data-handling capabilities.
The table is the basic building block of database design. Consequently, the table’s structure is of great interest. Ideally,
the database design process explored in Chapter 4, Entity Relationship (ER) Modeling, yields good table structures. Yet
it is possible to create poor table structures even in a good database design. So how do you recognize a poor table
structure, and how do you produce a good table? The answer to both questions involves normalization.
Normalization is a process for evaluating and correcting table structures to minimize data redundancies, thereby
reducing the likelihood of data anomalies. The normalization process involves assigning attributes to tables based on
the concept of determination you learned about in Chapter 3, The Relational Database Model.
Normalization works through a series of stages called normal forms. The first three stages are described as first normal
form (1NF), second normal form (2NF), and third normal form (3NF). From a structural point of view, 2NF is better
than 1NF, and 3NF is better than 2NF. For most purposes in business database design, 3NF is as high as you need
to go in the normalization process. However, you will discover that properly designed 3NF structures also meet the
requirements of fourth normal form (4NF).
Although normalization is a very important database design ingredient, you should not assume that the highest level
of normalization is always the most desirable. Generally, the higher the normal form, the more relational join
operations are required to produce a specified output and the more resources are required by the database system to
respond to end-user queries. A successful design must also consider end-user demand for fast performance. Therefore,
you will occasionally be expected to denormalize some portions of a database design in order to meet performance
requirements. Denormalization produces a lower normal form; that is, a 3NF will be converted to a 2NF through
denormalization. However, the price you pay for increased performance through denormalization is greater data
redundancy.
Note
Although the word table is used throughout this chapter, formally, normalization is concerned with relations. In
Chapter 3 you learned that the terms table and relation are frequently used interchangeably. In fact, you can say
that a table is the implementation view of a logical relation that meets some specific conditions. (See Table 3.1.)
However, being more rigorous, the mathematical relation does not allow duplicate tuples, whereas duplicate
tuples could exist in tables (see Section 6.5). Also, in normalization terminology, any attribute that is at least part
of a key is known as a prime attribute instead of the more common term key attribute, which was introduced
earlier. Conversely, a nonprime attribute, or a nonkey attribute, is not part of any candidate key.
6.2 THE NEED FOR NORMALIZATION
Normalization is typically used in conjunction with the entity relationship modeling that you learned in the previous
chapters. There are two common situations in which database designers use normalization. When designing a new
database structure based on the business requirements of the end users, the database designer will construct a data
model using a technique such as Crow’s Foot notation ERDs. After the initial design is complete, the designer can use
normalization to analyze the relationships that exist among the attributes within each entity, to determine if the
structure can be improved through normalization. Alternatively, database designers are often asked to modify existing
data structures that can be in the form of flat files, spreadsheets, or older database structures. Again, through an
analysis of the relationships among the attributes or fields in the data structure, the database designer can use the
normalization process to improve the existing data structure to create an appropriate database design. Whether
designing a new database structure or modifying an existing one, the normalization process is the same.
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C H A P T E R
6
To get a better idea of the normalization process, consider the simplified database activities of a construction company
that manages several building projects. Each project has its own project number, name, employees assigned to it, and
so on. Each employee has an employee number, name, and job classification, such as engineer or computer
technician.
The company charges its clients by billing the hours spent on each contract. The hourly billing rate is dependent on
the employee’s position. For example, one hour of computer technician time is billed at a different rate than one hour
of engineer time. Periodically, a report is generated that contains the information displayed in Table 6.1.
The total charge in Table 6.1 is a derived attribute and, at this point, is not stored in the table.
The easiest short-term way to generate the required report might seem to be a table whose contents correspond to
the reporting requirements. (See Figure 6.1.)
FIGURE
6.1
Tabular representation of the report format
Table name: RPT_FORMAT
Database name: Ch06_ConstructCo
Online Content
The databases used to illustrate the material in this chapter are found in the Premium Website for this book.
Note that the data in Figure 6.1 reflect the assignment of employees to projects. Apparently, an employee can be
assigned to more than one project. For example, Darlene Smithson (EMP_NUM = 112) has been assigned to two
projects: Amber Wave and Starflight. Given the structure of the dataset, each project includes only a single occurrence
of any one employee. Therefore, knowing the PROJ_NUM and EMP_NUM value will let you find the job classification
and its hourly charge. In addition, you will know the total number of hours each employee worked on each project.
(The total charge—a derived attribute whose value can be computed by multiplying the hours billed and the charge per
hour—has not been included in Figure 6.1. No structural harm is done if this derived attribute is included.)
EMPLOYEE
NUMBER
103
101
105
106
102
114
118
104
112
PROJECT
NAME
Evergreen
Amber Wave
Rolling Tide
Starflight
PROJECT
NUMBER
15
18
22
25
Maria D. Alonzo
Travis B. Bawangi
John G. News *
Annelise Jones
Ralph B. Washington
James J. Frommer
Darlene M. Smithson
Alice K. Johnson
Anne K. Ramoras
Delbert K. Joenbrood
Geoff B. Wabash
William Smithfield
Annelise Jones
James J. Frommer
Anne K. Ramoras *
Darlene M. Smithson
EMPLOYEE
NAME
June E. Arbough
John G. News
Alice K. Johnson *
William Smithfield
David H. Senior
Total
JOB
CLASS
Elec. Engineer
Database Designer
Database Designer
Programmer
Systems Analyst
Subtotal
Applications Designer
General Support
Systems Analyst
DSS Analyst
Subtotal
Database Designer
Systems Analyst
Applications Designer
Clerical Support
Programmer
Subtotal
Programmer
Systems Analyst
Database Designer
Applications Designer
Systems Analyst
General Support
DSS Analyst
Subtotal
HOURS
BILLED
23.8
19.4
35.7
12.6
23.8
25.6
45.3
32.4
45.0
65.7
48.4
23.6
22.0
12.8
25.6
45.8
56.3
33.1
23.6
30.5
41.4
CHARGE/
HOUR
$ 85.50
$105.00
$105.00
$ 35.75
$ 96.75
48.10
18.36
96.75
45.95
$
$
$
$
$105.00
$ 96.75
$ 48.10
$ 26.87
$ 35.75
$ 35.75
$ 96.75
$105.00
$ 48.10
$ 96.75
$ 18.36
$ 45.95
$49,199.69
TOTAL
CHARGE
$ 2,034.90
$ 2,037.00
$ 3,748.50
$ 450.45
$ 2,302.65
$10,573.50
$ 1,183.26
$ 831.71
$ 3,134.70
$ 2,067.75
$ 7,265.52
$ 6,998.50
$ 4,682.70
$ 1,135.16
$ 591.14
$ 457.60
$13,765.10
$ 915.20
$ 4,431.15
$ 5,911.50
$ 1,592.11
$ 2,283.30
$ 559.98
$ 1,902.33
$17,595.57
O F
Note: * indicates project leader
107
115
101
114
108
118
112
105
104
113
111
106
A Sample Report Layout
6.1
TABLE
N O R M A L I Z A T I O N
D A T A B A S E
TA B L E S
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Unfortunately, the structure of the dataset in Figure 6.1 does not conform to the requirements discussed in Chapter 3,
nor does it handle data very well. Consider the following deficiencies:
1.
The project number (PROJ_NUM) is apparently intended to be a primary key or at least a part of a PK, but it
contains nulls. (Given the preceding discussion, you know that PROJ_NUM + EMP_NUM will define each row.)
2.
The table entries invite data inconsistencies. For example, the JOB_CLASS value “Elect. Engineer” might be
entered as “Elect.Eng.” in some cases, “El. Eng.” in others, and “EE” in still others.
3.
The table displays data redundancies. Those data redundancies yield the following anomalies:
a. Update anomalies. Modifying the JOB_CLASS for employee number 105 requires (potentially) many
alterations, one for each EMP_NUM = 105.
b. Insertion anomalies. Just to complete a row definition, a new employee must be assigned to a project. If
the employee is not yet assigned, a phantom project must be created to complete the employee data entry.
c. Deletion anomalies. Suppose that only one employee is associated with a given project. If that employee
leaves the company and the employee data are deleted, the project information will also be deleted. To
prevent the loss of the project information, a fictitious employee must be created just to save the project
information.
In spite of those structural deficiencies, the table structure appears to work; the report is generated with ease.
Unfortunately, the report might yield varying results depending on what data anomaly has occurred. For example, if
you want to print a report to show the total “hours worked” value by the job classification “Database Designer,” that
report will not include data for “DB Design” and “Database Design” data entries. Such reporting anomalies cause a
multitude of problems for managers—and cannot be fixed through applications programming.
Even if very careful data-entry auditing can eliminate most of the reporting problems (at a high cost), it is easy to
demonstrate that even a simple data entry becomes inefficient. Given the existence of update anomalies, suppose
Darlene M. Smithson is assigned to work on the Evergreen project. The data-entry clerk must update the PROJECT
file with the entry:
15
Evergreen
112
Darlene M Smithson
DSS Analyst
$45.95
0.0
to match the attributes PROJ_NUM, PROJ_NAME, EMP_NUM, EMP_NAME, JOB_CLASS, CHG_HOUR, and
HOURS. (When Ms. Smithson has just been assigned to the project, she has not yet worked, so the total number of
hours worked is 0.0.)
Note
Remember that the naming convention makes it easy to see what each attribute stands for and what its likely
origin is. For example, PROJ_NAME uses the prefix PROJ to indicate that the attribute is associated with the
PROJECT table, while the NAME component is self-documenting, too. However, keep in mind that name length
is also an issue, especially in the prefix designation. For that reason, the prefix CHG was used rather than
CHARGE. (Given the database’s context, it is not likely that that prefix will be misunderstood.)
Each time another employee is assigned to a project, some data entries (such as PROJ_NAME, EMP_NAME, and
CHG_HOUR) are unnecessarily repeated. Imagine the data-entry chore when 200 or 300 table entries must be made!
Note that the entry of the employee number should be sufficient to identify Darlene M. Smithson, her job description,
and her hourly charge. Because there is only one person identified by the number 112, that person’s characteristics
(name, job classification, and so on) should not have to be typed in each time the main file is updated. Unfortunately,
the structure displayed in Figure 6.1 does not make allowances for that possibility.
N O R M A L I Z A T I O N
O F
D A T A B A S E
TA B L E S
The data redundancy evident in Figure 6.1 leads to wasted disk space. What’s more, data redundancy produces data
anomalies. For example, suppose the data-entry clerk had entered the data as:
15
Evergeen
112
Darla Smithson
DCS Analyst
$45.95
0.0
At first glance, the data entry appears to be correct. But is Evergeen the same project as Evergreen? And is DCS
Analyst supposed to be DSS Analyst? Is Darla Smithson the same person as Darlene M. Smithson? Such confusion
is a data integrity problem that was caused because the data entry failed to conform to the rule that all copies of
redundant data must be identical.
The possibility of introducing data integrity problems caused by data redundancy must be considered when a database
is designed. The relational database environment is especially well suited to help the designer overcome those
problems.
6.3 THE NORMALIZATION PROCESS
In this section, you will learn how to use normalization to produce a set of normalized tables to store the data that will
be used to generate the required information. The objective of normalization is to ensure that each table conforms to
the concept of well-formed relations—that is, tables that have the following characteristics:
쐌
Each table represents a single subject. For example, a course table will contain only data that directly pertain
to courses. Similarly, a student table will contain only student data.
쐌
No data item will be unnecessarily stored in more than one table (in short, tables have minimum controlled
redundancy). The reason for this requirement is to ensure that the data are updated in only one place.
쐌
All nonprime attributes in a table are dependent on the primary key—the entire primary key and nothing but
the primary key. The reason for this requirement is to ensure that the data are uniquely identifiable by a primary
key value.
쐌
Each table is void of insertion, update, or deletion anomalies. This is to ensure the integrity and consistency
of the data.
To accomplish the objective, the normalization process takes you through the steps that lead to successively higher
normal forms. The most common normal forms and their basic characteristic are listed in Table 6.2. You will learn the
details of these normal forms in the indicated sections.
TABLE
6.2
Normal Forms
NORMAL FORM
First normal form (1NF)
Second normal form (2NF)
Third normal form (3NF)
Boyce-Codd normal form (BCNF)
Fourth normal form (4NF)
CHARACTERISTIC
Table format, no repeating groups, and PK identified
1NF and no partial dependencies
2NF and no transitive dependencies
Every determinant is a candidate key (special case of 3NF)
3NF and no independent multivalued dependencies
SECTION
6.3.1
6.3.2
6.3.3
6.6.1
6.6.2
The concept of keys is central to the discussion of normalization. Recall from Chapter 3 that a candidate key is a
minimal (irreducible) superkey. The primary key is the candidate key that is selected to be the primary means used to
identify the rows in the table. Although normalization is typically presented from the perspective of candidate keys, for
the sake of simplicity while initially explaining the normalization process, we will make the assumption that for each
table there is only one candidate key, and therefore, that candidate key is the primary key.
From the data modeler’s point of view, the objective of normalization is to ensure that all tables are at least in third
normal form (3NF). Even higher-level normal forms exist. However, normal forms such as the fifth normal form (5NF)
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and domain-key normal form (DKNF) are not likely to be encountered in a business environment and are mainly of
theoretical interest. More often than not, such higher normal forms increase joins (slowing performance) without
adding any value in the elimination of data redundancy. Some very specialized applications, such as statistical research,
might require normalization beyond the 4NF, but those applications fall outside the scope of most business operations.
Because this book focuses on practical applications of database techniques, the higher-level normal forms are not
covered.
Functional Dependence
Before outlining the normalization process, it’s a good idea to review the concepts of determination and functional
dependence that were covered in detail in Chapter 3. Table 6.3 summarizes the main concepts.
TABLE
6.3
Functional Dependence Concepts
CONCEPT
Functional dependence
Functional dependence
(generalized definition)
Fully functional dependence
(composite key)
DEFINITION
The attribute B is fully functionally dependent on the attribute A if each value of A
determines one and only one value of B.
Example: PROJ_NUM → PROJ_NAME
(read as “PROJ_NUM functionally determines PROJ_NAME”)
In this case, the attribute PROJ_NUM is known as the “determinant” attribute,
and the attribute PROJ_NAME is known as the “dependent” attribute.
Attribute A determines attribute B (that is, B is functionally dependent on A) if all
of the rows in the table that agree in value for attribute A also agree in value for
attribute B.
If attribute B is functionally dependent on a composite key A but not on any subset of that composite key, the attribute B is fully functionally dependent on A.
It is crucial to understand these concepts because they are used to derive the set of functional dependencies for a given
relation. The normalization process works one relation at a time, identifying the dependencies on that relation and
normalizing the relation. As you will see in the following sections, normalization starts by identifying the dependencies
of a given relation and progressively breaking up the relation (table) into a set of new relations (tables) based on the
identified dependencies.
Two types of functional dependencies that are of special interest in normalization are partial dependencies and
transitive dependencies. A partial dependency exists when there is a functional dependence in which the
determinant is only part of the primary key (remember we are assuming there is only one candidate key). For example,
if (A, B) → (C,D), B → C, and (A, B) is the primary key, then the functional dependence B → C is a partial dependency
because only part of the primary key (B) is needed to determine the value of C. Partial dependencies tend to be rather
straightforward and easy to identify.
A transitive dependency exists when there are functional dependencies such that X → Y, Y → Z, and X is the
primary key. In that case, the dependency X → Z is a transitive dependency because X determines the value of Z via
Y. Unlike partial dependencies, transitive dependencies are more difficult to identify among a set of data. Fortunately,
there is an easier way to identify transitive dependencies. A transitive dependency will occur only when a functional
dependence exists among nonprime attributes. In the previous example, the actual transitive dependency is X → Z.
However, the dependency Y → Z signals that a transitive dependency exists. Hence, throughout the discussion of the
normalization process, the existence of a functional dependence among nonprime attributes will be considered a sign
of a transitive dependency. To address the problems related to transitive dependencies, changes to the table structure
are made based on the functional dependence that signals the transitive dependency’s existence. Therefore, to simplify
the description of normalization, from this point forward we will refer to the signaling dependency as the transitive
dependency.
N O R M A L I Z A T I O N
O F
D A T A B A S E
TA B L E S
6.3.1 Conversion to First Normal Form
Because the relational model views data as part of a table or a collection of tables in which all key values must be
identified, the data depicted in Figure 6.1 might not be stored as shown. Note that Figure 6.1 contains what is known
as repeating groups. A repeating group derives its name from the fact that a group of multiple entries of the same
type can exist for any single key attribute occurrence. In Figure 6.1, note that each single project number
(PROJ_NUM) occurrence can reference a group of related data entries. For example, the Evergreen project
(PROJ_NUM = 15) shows five entries at this point—and those entries are related because they each share the
PROJ_NUM = 15 characteristic. Each time a new record is entered for the Evergreen project, the number of entries
in the group grows by one.
A relational table must not contain repeating groups. The existence of repeating groups provides evidence that the
RPT_FORMAT table in Figure 6.1 fails to meet even the lowest normal form requirements, thus reflecting data
redundancies.
Normalizing the table structure will reduce the data redundancies. If repeating groups do exist, they must be eliminated
by making sure that each row defines a single entity. In addition, the dependencies must be identified to diagnose the
normal form. Identification of the normal form will let you know where you are in the normalization process. The
normalization process starts with a simple three-step procedure.
Step 1: Eliminate the Repeating Groups
Start by presenting the data in a tabular format, where each cell has a single value and there are no repeating groups.
To eliminate the repeating groups, eliminate the nulls by making sure that each repeating group attribute contains an
appropriate data value. That change converts the table in Figure 6.1 to 1NF in Figure 6.2.
FIGURE
6.2
A table in first normal form
Table name: DATA_ORG_1NF
Database name: Ch06_ConstructCo
Step 2: Identify the Primary Key
The layout in Figure 6.2 represents more than a mere cosmetic change. Even a casual observer will note that
PROJ_NUM is not an adequate primary key because the project number does not uniquely identify all of the remaining
entity (row) attributes. For example, the PROJ_NUM value 15 can identify any one of five employees. To maintain a
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proper primary key that will uniquely identify any attribute value, the new key must be composed of a combination
of PROJ_NUM and EMP_NUM. For example, using the data shown in Figure 6.2, if you know that PROJ_NUM =
15 and EMP_NUM = 103, the entries for the attributes PROJ_NAME, EMP_NAME, JOB_CLASS, CHG_HOUR, and
HOURS must be Evergreen, June E. Arbough, Elect. Engineer, $84.50, and 23.8, respectively.
Step 3: Identify All Dependencies
The identification of the PK in Step 2 means that you have already identified the following dependency:
PROJ_NUM, EMP_NUM → PROJ_NAME, EMP_NAME, JOB_CLASS, CHG_HOUR, HOURS
That is, the PROJ_NAME, EMP_NAME, JOB_CLASS, CHG_HOUR, and HOURS values are all dependent on—that
is, they are determined by—the combination of PROJ_NUM and EMP_NUM. There are additional dependencies. For
example, the project number identifies (determines) the project name. In other words, the project name is dependent
on the project number. You can write that dependency as:
PROJ_NUM → PROJ_NAME
Also, if you know an employee number, you also know that employee’s name, that employee’s job classification, and
that employee’s charge per hour. Therefore, you can identify the dependency shown next:
EMP_NUM → EMP_NAME, JOB_CLASS, CHG_HOUR
However, given the previous dependency components, you can see that knowing the job classification means knowing
the charge per hour for that job classification. In other words, you can identify one last dependency:
JOB_CLASS → CHG_HOUR
This dependency exists between two nonprime attributes; therefore it is a signal that a transitive dependency exists,
and we will refer to it as a transitive dependency. The dependencies you have just examined can also be depicted with
the help of the diagram shown in Figure 6.3. Because such a diagram depicts all dependencies found within a given
table structure, it is known as a dependency diagram. Dependency diagrams are very helpful in getting a bird’s-eye
view of all of the relationships among a table’s attributes, and their use makes it less likely that you will overlook an
important dependency.
As you examine Figure 6.3, note the following dependency diagram features:
1.
The primary key attributes are bold, underlined, and shaded in a different color.
2.
The arrows above the attributes indicate all desirable dependencies, that is, dependencies that are based on the
primary key. In this case, note that the entity’s attributes are dependent on the combination of PROJ_NUM
and EMP_NUM.
3.
The arrows below the dependency diagram indicate less desirable dependencies. Two types of such
dependencies exist:
a. Partial dependencies. You need to know only the PROJ_NUM to determine the PROJ_NAME; that is, the
PROJ_NAME is dependent on only part of the primary key. And you need to know only the EMP_NUM
to find the EMP_NAME, the JOB_CLASS, and the CHG_HOUR. A dependency based on only a part of
a composite primary key is a partial dependency.
b. Transitive dependencies. Note that CHG_HOUR is dependent on JOB_CLASS. Because neither
CHG_HOUR nor JOB_CLASS is a prime attribute—that is, neither attribute is at least part of a key—the
condition is a transitive dependency. In other words, a transitive dependency is a dependency of one
nonprime attribute on another nonprime attribute. The problem with transitive dependencies is that they
still yield data anomalies.
N O R M A L I Z A T I O N
FIGURE
O F
D A T A B A S E
TA B L E S
First normal form (1NF) dependency diagram
6.3
PROJ_NUM
PROJ_NAME
EMP_NUM
EMP_NAME
JOB_CLASS
Partial dependency
CHG_HOUR
HOURS
Transitive
dependency
Partial dependencies
1NF (PROJ_NUM, EMP_NUM, PROJ_NAME, EMP_NAME, JOB_CLASS, CHG_HOURS, HOURS)
PARTIAL DEPENDENCIES:
(PROJ_NUM
PROJ_NAME)
(EMP_NUM
EMP_NAME, JOB_CLASS, CHG_HOUR)
TRANSITIVE DEPENDENCY:
(JOB CLASS
CHG_HOUR)
Note that Figure 6.3 includes the relational schema for the table in 1NF and a textual notation for each identified
dependency.
Note
The term first normal form (1NF) describes the tabular format in which:
• All of the key attributes are defined.
• There are no repeating groups in the table. In other words, each row/column intersection contains one and
only one value, not a set of values.
• All attributes are dependent on the primary key.
All relational tables satisfy the 1NF requirements. The problem with the 1NF table structure shown in Figure 6.3 is that
it contains partial dependencies—that is, dependencies based on only a part of the primary key.
While partial dependencies are sometimes used for performance reasons, they should be used with caution. (If the
information requirements seem to dictate the use of partial dependencies, it is time to evaluate the need for a data
warehouse design, discussed in Chapter 13, Business Intelligence and Data Warehouses.) Such caution is warranted
because a table that contains partial dependencies is still subject to data redundancies, and therefore, to various
anomalies. The data redundancies occur because every row entry requires duplication of data. For example, if Alice
K. Johnson submits her work log, then the user would have to make multiple entries during the course of a day. For
each entry, the EMP_NAME, JOB_CLASS, and CHG_HOUR must be entered each time, even though the attribute
values are identical for each row entered. Such duplication of effort is very inefficient. What’s more, the duplication
of effort helps create data anomalies; nothing prevents the user from typing slightly different versions of the employee
name, the position, or the hourly pay. For instance, the employee name for EMP_NUM = 102 might be entered as
Dave Senior or D. Senior. The project name might also be entered correctly as Evergreen or misspelled as Evergeen.
Such data anomalies violate the relational database’s integrity and consistency rules.
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6.3.2 Conversion to Second Normal Form
Converting to 2NF is done only when the 1NF has a composite primary key. If the 1NF has a single-attribute primary
key, then the table is automatically in 2NF. The 1NF-to-2NF conversion is simple. Starting with the 1NF format
displayed in Figure 6.3, you do the following:
Step 1: Make New Tables to Eliminate Partial Dependencies
For each component of the primary key that acts as a determinant in a partial dependency, create a new table with
a copy of that component as the primary key. While these components are placed in the new tables, it is important
that they also remain in the original table as well. It is important that the determinants remain in the original table
because they will be the foreign keys for the relationships that are needed to relate these new tables to the original
table. For the construction of our revised dependency diagram, write each key component on a separate line; then
write the original (composite) key on the last line. For example:
PROJ_NUM
EMP_NUM
PROJ_NUM EMP_NUM
Each component will become the key in a new table. In other words, the original table is now divided into three tables
(PROJECT, EMPLOYEE, and ASSIGNMENT).
Step 2: Reassign Corresponding Dependent Attributes
Use Figure 6.3 to determine those attributes that are dependent in the partial dependencies. The dependencies for the
original key components are found by examining the arrows below the dependency diagram shown in Figure 6.3. The
attributes that are dependent in a partial dependency are removed from the original table and placed in the new table
with its determinant. Any attributes that are not dependent in a partial dependency will remain in the original table.
In other words, the three tables that result from the conversion to 2NF are given appropriate names (PROJECT,
EMPLOYEE, and ASSIGNMENT) and are described by the following relational schemas:
PROJECT (PROJ_NUM, PROJ_NAME)
EMPLOYEE (EMP_NUM, EMP_NAME, JOB_CLASS, CHG_HOUR)
ASSIGNMENT (PROJ_NUM, EMP_NUM, ASSIGN_HOURS)
Because the number of hours spent on each project by each employee is dependent on both PROJ_NUM and
EMP_NUM in the ASSIGNMENT table, you leave those hours in the ASSIGNMENT table as ASSIGN_HOURS.
Notice that the ASSIGNMENT table contains a composite primary key composed of the attributes PROJ_NUM and
EMP_NUM. Notice that by leaving the determinants in the original table as well as making them the primary keys of
the new tables, primary key/foreign key relationships have been created. For example, in the EMPLOYEE table,
EMP_NUM is the primary key. In the ASSIGNMENT table, EMP_NUM is part of the composite primary key
(PROJ_NUM, EMP_NUM) and is a foreign key relating the EMPLOYEE table to the ASSIGNMENT table.
The results of Steps 1 and 2 are displayed in Figure 6.4. At this point, most of the anomalies discussed earlier have
been eliminated. For example, if you now want to add, change, or delete a PROJECT record, you need to go only to
the PROJECT table and make the change to only one row.
Because a partial dependency can exist only when a table’s primary key is composed of several attributes, a table
whose primary key consists of only a single attribute is automatically in 2NF once it is in 1NF.
Figure 6.4 still shows a transitive dependency, which can generate anomalies. For example, if the charge per hour
changes for a job classification held by many employees, that change must be made for each of those employees. If
N O R M A L I Z A T I O N
FIGURE
O F
D A T A B A S E
TA B L E S
Second normal form (2NF) conversion results
6.4
PROJECT (PROJ_NUM, PROJ_NAME)
Table name: PROJECT
PROJ_NUM
PROJ_NAME
Table name: EMPLOYEE
EMPLOYEE (EMP_NUM, EMP_NAME, JOB_CLASS, CHG_HOUR)
TRANSITIVE DEPENDENCY
(JOB_CLASS
CHG_HOUR)
EMP_NUM
EMP_NAME
JOB_CLASS
CHG_HOUR
Transitive
dependency
Table name: ASSIGNMENT
PROJ_NUM
EMP_NUM
ASSIGNMENT (PROJ_NUM, EMP_NUM, ASSIGN_HOURS)
ASSIGN_HOURS
you forget to update some of the employee records that are affected by the charge per hour change, different
employees with the same job description will generate different hourly charges.
Note
A table is in second normal form (2NF) when:
• It is in 1NF.
and
• It includes no partial dependencies; that is, no attribute is dependent on only a portion of the primary key.
Note that it is still possible for a table in 2NF to exhibit transitive dependency; that is, the primary key may
rely on one or more nonprime attributes to functionally determine other nonprime attributes, as is indicated by
a functional dependence among the nonprime attributes.
6.3.3 Conversion to Third Normal Form
The data anomalies created by the database organization shown in Figure 6.4 are easily eliminated by completing the
following two steps:
Step 1: Make New Tables to Eliminate Transitive Dependencies
For every transitive dependency, write a copy of its determinant as a primary key for a new table. A determinant is
any attribute whose value determines other values within a row. If you have three different transitive dependencies, you
will have three different determinants. As with the conversion to 2NF, it is important that the determinant remain in
the original table to serve as a foreign key. Figure 6.4 shows only one table that contains a transitive dependency.
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Therefore, write the determinant for this transitive dependency as:
JOB_CLASS
Step 2: Reassign Corresponding Dependent Attributes
Using Figure 6.4, identify the attributes that are dependent on each determinant identified in Step 1. Place the
dependent attributes in the new tables with their determinants and remove them from their original tables. In this
example, eliminate CHG_HOUR from the EMPLOYEE table shown in Figure 6.4 to leave the EMPLOYEE table
dependency definition as:
EMP_NUM → EMP_NAME, JOB_CLASS
Draw a new dependency diagram to show all of the tables you have defined in Steps 1 and 2. Name the table to reflect
its contents and function. In this case, JOB seems appropriate. Check all of the tables to make sure that each table
has a determinant and that no table contains inappropriate dependencies. When you have completed these steps, you
will see the results in Figure 6.5.
FIGURE
Third normal form (3NF) conversion results
6.5
PROJ_NUM
PROJ_NAME
EMP_NUM
EMP_NAME
JOB_CLASS
Table name: PROJECT
Table name: EMPLOYEE
PROJECT (PROJ_NUM, PROJ_NAME)
EMPLOYEE (EMP_NUM, EMP_NAME, JOB_CLASS)
JOB_CLASS
CHG_HOUR
PROJ_NUM
EMP_NUM
ASSIGN_HOURS
Table name: JOB
Table name: ASSIGNMENT
JOB (JOB_CLASS, CHG_HOUR)
ASSIGNMENT (PROJ_NUM, EMP_NUM, ASSIGN_HOURS)
In other words, after the 3NF conversion has been completed, your database will contain four tables:
PROJECT (PROJ_NUM, PROJ_NAME)
EMPLOYEE (EMP_NUM, EMP_NAME, JOB_CLASS)
JOB (JOB_CLASS, CHG_HOUR)
ASSIGNMENT (PROJ_NUM, EMP_NUM, ASSIGN_HOURS)
Note that this conversion has eliminated the original EMPLOYEE table’s transitive dependency; the tables are now said
to be in third normal form (3NF).
N O R M A L I Z A T I O N
O F
D A T A B A S E
TA B L E S
Note
A table is in third normal form (3NF) when:
• It is in 2NF.
and
• It contains no transitive dependencies.
It is interesting to note the similarities between resolving 2NF and 3NF problems. To convert a table from 1NF to 2NF,
it is necessary to remove the partial dependencies. To convert a table from 2NF to 3NF, it is necessary to remove the
transitive dependencies. No matter whether the “problem” dependency is a partial dependency or a transitive
dependency, the solution is the same. Create a new table for each problem dependency. The determinant of the
problem dependency remains in the original table and is placed as the primary key of the new table. The dependents
of the problem dependency are removed from the original table and placed as nonprime attributes in the new table.
Be aware, however, that while the technique is the same, it is imperative that 2NF be achieved before moving on to
3NF; be certain to resolve the partial dependencies before resolving the transitive dependencies. Recall, however, the
assumption that was made at the beginning of the discussion of the normalization process—that each table has only
one candidate key, which is the primary key. If a table has multiple candidate keys, then the overall process remains
the same, but there are additional considerations.
For example, if a table has multiple candidate keys and one of those candidate keys is a composite key, the table can
have partial dependencies based on this composite candidate key, even when the primary key chosen is a single
attribute. In those cases, following the process described above, those dependencies would be perceived as transitive
dependencies and would not be resolved until 3NF. The simplified process described above will allow the designer to
achieve the correct result, but through practice, you should recognize all candidate keys and their dependencies as
such, and resolve them appropriately. The existence of multiple candidate keys can also influence the identification of
transitive dependencies. Previously, a transitive dependency was defined to exist when one nonprime attribute
determined another nonprime attribute. In the presence of multiple candidate keys, the definition of a nonprime
attribute as an attribute that is not a part of any candidate key is critical. If the determinant of a functional dependence
is not the primary key but is a part of another candidate key, then it is not a nonprime attribute and does not signal
the presence of a transitive dependency.
6.4 IMPROVING THE DESIGN
The table structures are cleaned up to eliminate the troublesome partial and transitive dependencies. You can now
focus on improving the database’s ability to provide information and on enhancing its operational characteristics. In
the next few paragraphs, you will learn about the various types of issues you need to address to produce a good
normalized set of tables. Please note that for space issues, each section presents just one example—the designer must
apply the principle to all remaining tables in the design. Remember that normalization cannot, by itself, be relied on
to make good designs. Instead, normalization is valuable because its use helps eliminate data redundancies.
Evaluate PK Assignments
Each time a new employee is entered into the EMPLOYEE table, a JOB_CLASS value must be entered. Unfortunately,
it is too easy to make data-entry errors that lead to referential integrity violations. For example, entering DB Designer
instead of Database Designer for the JOB_CLASS attribute in the EMPLOYEE table will trigger such a violation.
Therefore, it would be better to add a JOB_CODE attribute to create a unique identifier. The addition of a JOB_CODE
attribute produces the dependency:
JOB_CODE → JOB_CLASS, CHG_HOUR
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If you assume that the JOB_CODE is a proper primary key, this new attribute does produce the dependency:
JOB_CLASS → CHG_HOUR
However, this dependency is not a transitive dependency because the determinant is a candidate key. Further, the
presence of JOB_CODE greatly decreases the likelihood of referential integrity violations. Note that the new JOB table
now has two candidate keys—JOB_CODE and JOB_CLASS. In this case, JOB_CODE is the chosen primary key as
well as a surrogate key. A surrogate key, as you should recall, is an artificial PK introduced by the designer with the
purpose of simplifying the assignment of primary keys to tables. Surrogate keys are usually numeric, they are often
automatically generated by the DBMS, they are free of semantic content (they have no special meaning), and they are
usually hidden from the end users.
Evaluate Naming Conventions
It is best to adhere to the naming conventions outlined in Chapter 2, Data Models. Therefore, CHG_HOUR will be
changed to JOB_CHG_HOUR to indicate its association with the JOB table. In addition, the attribute name JOB_CLASS
does not quite describe entries such as Systems Analyst, Database Designer, and so on; the label JOB_DESCRIPTION
fits the entries better. Also, you might have noticed that HOURS was changed to ASSIGN_HOURS in the conversion
from 1NF to 2NF. That change lets you associate the hours worked with the ASSIGNMENT table.
Refine Attribute Atomicity
It is generally good practice to pay attention to the atomicity requirement. An atomic attribute is one that cannot
be further subdivided. Such an attribute is said to display atomicity. Clearly, the use of the EMP_NAME in the
EMPLOYEE table is not atomic because EMP_NAME can be decomposed into a last name, a first name, and an initial.
By improving the degree of atomicity, you also gain querying flexibility. For example, if you use EMP_LNAME,
EMP_FNAME, and EMP_INITIAL, you can easily generate phone lists by sorting last names, first names, and initials.
Such a task would be very difficult if the name components were within a single attribute. In general, designers prefer
to use simple, single-valued attributes as indicated by the business rules and processing requirements.
Identify New Attributes
If the EMPLOYEE table were used in a real-world environment, several other attributes would have to be added. For
example, year-to-date gross salary payments, Social Security payments, and Medicare payments would be desirable.
An employee hire date attribute (EMP_HIREDATE) could be used to track an employee’s job longevity and serve as
a basis for awarding bonuses to long-term employees and for other morale-enhancing measures. The same principle
must be applied to all other tables in your design.
Identify New Relationships
According to the original report, the users need to track which employee is acting as the manager of each project. This
can be implemented as a relationship between EMPLOYEE and PROJECT. From the original report, it is clear that
each project has only one manager. Therefore, the system’s ability to supply detailed information about each project’s
manager is ensured by using the EMP_NUM as a foreign key in PROJECT. That action ensures that you can access
the details of each PROJECT’s manager data without producing unnecessary and undesirable data duplication. The
designer must take care to place the right attributes in the right tables by using normalization principles.
Refine Primary Keys as Required for Data Granularity
Granularity refers to the level of detail represented by the values stored in a table’s row. Data stored at their lowest
level of granularity are said to be atomic data, as explained earlier. In Figure 6.5, the ASSIGNMENT table in 3NF uses
the ASSIGN_HOURS attribute to represent the hours worked by a given employee on a given project. However, are
those values recorded at their lowest level of granularity? In other words, does ASSIGN_HOURS represent the hourly
total, daily total, weekly total, monthly total, or yearly total? Clearly, ASSIGN_HOURS requires more careful
N O R M A L I Z A T I O N
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definition. In this case, the relevant question would be as follows: For what time frame—hour, day, week, month, and
so on—do you want to record the ASSIGN_HOURS data?
For example, assume that the combination of EMP_NUM and PROJ_NUM is an acceptable (composite) primary key
in the ASSIGNMENT table. That primary key is useful in representing only the total number of hours an employee
worked on a project since its start. Using a surrogate primary key such as ASSIGN_NUM provides lower granularity
and yields greater flexibility. For example, assume that the EMP_NUM and PROJ_NUM combination is used as the
primary key, and then an employee makes two “hours worked” entries in the ASSIGNMENT table. That action violates
the entity integrity requirement. Even if you add the ASSIGN_DATE as part of a composite PK, an entity integrity
violation is still generated if any employee makes two or more entries for the same project on the same day. (The
employee might have worked on the project a few hours in the morning and then worked on it again later in the day.)
The same data entry yields no problems when ASSIGN_NUM is used as the primary key.
Note
In an ideal (database design) world, the level of desired granularity is determined at the conceptual design or at
the requirements-gathering phase. However, as you have already seen in this chapter, many database designs
involve the refinement of existing data requirements, thus triggering design modifications. In a real-world
environment, changing granularity requirements might dictate changes in primary key selection, and those
changes might ultimately require the use of surrogate keys.
Maintain Historical Accuracy
Writing the job charge per hour into the ASSIGNMENT table is crucial to maintaining the historical accuracy of the data
in the ASSIGNMENT table. It would be appropriate to name this attribute ASSIGN_CHG_HOUR. Although this attribute
would appear to have the same value as JOB_CHG_HOUR, this is true only if the JOB_CHG_HOUR value remains
the same forever. However, it is reasonable to assume that the job charge per hour will change over time. But suppose
that the charges to each project were figured (and billed) by multiplying the hours worked on the project, found in the
ASSIGNMENT table, by the charge per hour, found in the JOB table. Those charges would always show the current
charge per hour stored in the JOB table, rather than the charge per hour that was in effect at the time of the
assignment.
Evaluate Using Derived Attributes
Finally, you can use a derived attribute in the ASSIGNMENT table to store the actual charge made to a project. That
derived attribute, to be named ASSIGN_CHARGE, is the result of multiplying ASSIGN_HOURS by ASSIGN_CHG_
HOUR. This creates a transitive dependency such that
(ASSIGN_CHARGE + ASSIGN_HOURS) → ASSIGN_CHG_HOUR.
From a strictly database point of view, such derived attribute values can be calculated when they are needed to write
reports or invoices. However, storing the derived attribute in the table makes it easy to write the application software
to produce the desired results. Also, if many transactions must be reported and/or summarized, the availability of the
derived attribute will save reporting time. (If the calculation is done at the time of data entry, it will be completed when
the end user presses the Enter key, thus speeding up the process.)
The enhancements described in the preceding sections are illustrated in the tables and dependency diagrams shown
in Figure 6.6.
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FIGURE
6
The completed database
6.6
Table name: PROJECT
Table name: JOB
PROJ_NUM
PROJ_NAME
EMP_NUM
JOB_CODE
JOB_DESCRIPTION
JOB_CHG_HOUR
Database name: Ch06_ConstructCo
Table name: PROJECT
Table name: JOB
Table name: ASSIGNMENT
ASSIGN_NUM
ASSIGN_DATE
PROJ_NUM
EMP_NUM
ASSIGN_HOURS
ASSIGN_CHG_HOUR ASSIGN_CHARGE
Table name: ASSIGNMENT
Figure 6.6 is a vast improvement over the original database design. If the application software is designed properly,
the most active table (ASSIGNMENT) requires the entry of only the PROJ_NUM, EMP_NUM, and ASSIGN_HOURS
N O R M A L I Z A T I O N
FIGURE
O F
D A T A B A S E
TA B L E S
The completed database (continued)
6.6
Table name: EMPLOYEE
EMP_NUM
EMP_LNAME
EMP_FNAME
EMP_INITIAL
EMP_HIREDATE
JOB_CODE
Table name: EMPLOYEE
values. The values for the attributes ASSIGN_NUM and ASSIGN_DATE can be generated by the application. For
example, the ASSIGN_NUM can be created by using a counter, and the ASSIGN_DATE can be the system date read
by the application and automatically entered into the ASSIGNMENT table. In addition, the application software can
automatically insert the correct ASSIGN_CHG_HOUR value by writing the appropriate JOB table’s JOB_CHG_
HOUR value into the ASSIGNMENT table. (The JOB and ASSIGNMENT tables are related through the JOB_CODE
attribute.) If the JOB table’s JOB_CHG_HOUR value changes, the next insertion of that value into the ASSIGNMENT
table will reflect the change automatically. The table structure thus minimizes the need for human intervention. In fact,
if the system requires the employees to enter their own work hours, they can scan their EMP_NUM into the
ASSIGNMENT table by using a magnetic card reader that enters their identity. Thus, the ASSIGNMENT table’s
structure can set the stage for maintaining some desired level of security.
6.5 SURROGATE KEY CONSIDERATIONS
Although this design meets the vital entity and referential integrity requirements, the designer must still address some
concerns. For example, a composite primary key might become too cumbersome to use as the number of attributes
grows. (It becomes difficult to create a suitable foreign key when the related table uses a composite primary key. In
addition, a composite primary key makes it more difficult to write search routines.) Or a primary key attribute might
simply have too much descriptive content to be usable—which is why the JOB_CODE attribute was added to the JOB
table to serve as that table’s primary key. When, for whatever reason, the primary key is considered to be unsuitable,
designers use surrogate keys, as discussed in the previous chapter.
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At the implementation level, a surrogate key is a system-defined attribute generally created and managed via the
DBMS. Usually, a system-defined surrogate key is numeric, and its value is automatically incremented for each new
row. For example, Microsoft Access uses an AutoNumber data type, Microsoft SQL Server uses an identity column,
and Oracle uses a sequence object.
Recall from Section 6.4 that the JOB_CODE attribute was designated to be the JOB table’s primary key. However,
remember that the JOB_CODE does not prevent duplicate entries from being made, as shown in the JOB table in
Table 6.4.
TABLE
6.4
JOB_CODE
511
512
Duplicate Entries in the Job Table
JOB_DESCRIPTION
Programmer
Programmer
JOB_CHG_HOUR
$35.75
$35.75
Clearly, the data entries in Table 6.4 are inappropriate because they duplicate existing records—yet there has been no
violation of either entity integrity or referential integrity. This “multiple duplicate records” problem was created when
the JOB_CODE attribute was added as the PK. (When the JOB_DESCRIPTION was initially designated to be the PK,
the DBMS would ensure unique values for all job description entries when it was asked to enforce entity integrity. But
that option created the problems that caused the use of the JOB_CODE attribute in the first place!) In any case, if
JOB_CODE is to be the surrogate PK, you still must ensure the existence of unique values in the JOB_DESCRIPTION
through the use of a unique index.
Note that all of the remaining tables (PROJECT, ASSIGNMENT, and EMPLOYEE) are subject to the same limitations.
For example, if you use the EMP_NUM attribute in the EMPLOYEE table as the PK, you can make multiple entries
for the same employee. To avoid that problem, you might create a unique index for EMP_LNAME, EMP_FNAME, and
EMP_INITIAL. But how would you then deal with two employees named Joe B. Smith? In that case, you might use
another (preferably externally defined) attribute to serve as the basis for a unique index.
It is worth repeating that database design often involves trade-offs and the exercise of professional judgment. In a
real-world environment, you must strike a balance between design integrity and flexibility. For example, you might
design the ASSIGNMENT table to use a unique index on PROJ_NUM, EMP_NUM, and ASSIGN_DATE if you want
to limit an employee to only one ASSIGN_HOURS entry per date. That limitation would ensure that employees
couldn’t enter the same hours multiple times for any given date. Unfortunately, that limitation is likely to be undesirable
from a managerial point of view. After all, if an employee works several different times on a project during any given
day, it must be possible to make multiple entries for that same employee and the same project during that day. In that
case, the best solution might be to add a new externally defined attribute—such as a stub, voucher, or ticket
number—to ensure uniqueness. In any case, frequent data audits would be appropriate.
6.6 HIGHER-LEVEL NORMAL FORMS
Tables in 3NF will perform suitably in business transactional databases. However, there are occasions when higher
normal forms are useful. In this section, you will learn about a special case of 3NF, known as Boyce-Codd normal form
(BCNF), and about fourth normal form (4NF).
6.6.1 The Boyce-Codd Normal Form (BCNF)
A table is in Boyce-Codd normal form (BCNF) when every determinant in the table is a candidate key. (Recall from
Chapter 3 that a candidate key has the same characteristics as a primary key, but for some reason, it was not chosen
to be the primary key.) Clearly, when a table contains only one candidate key, the 3NF and the BCNF are equivalent.
N O R M A L I Z A T I O N
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Putting that proposition another way, BCNF can be violated only when the table contains more than one
candidate key.
Note
A table is in Boyce-Codd normal form (BCNF) when every determinant in the table is a candidate key.
Most designers consider the BCNF to be a special case of the 3NF. In fact, if the techniques shown here are used, most
tables conform to the BCNF requirements once the 3NF is reached. So how can a table be in 3NF and not be in
BCNF? To answer that question, you must keep in mind that a transitive dependency exists when one nonprime
attribute is dependent on another nonprime attribute.
In other words, a table is in 3NF when it is in 2NF and there are no transitive dependencies. But what about a case
in which a nonkey attribute is the determinant of a key attribute? That condition does not violate 3NF, yet it fails to
meet the BCNF requirements because BCNF requires that every determinant in the table be a candidate key. The
situation just described (a 3NF table that fails to meet BCNF requirements) is shown in Figure 6.7.
FIGURE
6.7
A table that is in 3NF but not
in BCNF
Note these functional dependencies in Figure 6.7:
A + B → C, D
A + C → B, D
C→B
Notice that this structure has two candidate keys: (A + B) and
(A + C). The table structure shown in Figure 6.7 has no partial
dependencies, nor does it contain transitive dependencies.
(The condition C → B indicates that a nonkey attribute
determines part of the primary key—and that dependency
is not transitive or partial because the dependent is a prime
attribute!) Thus, the table structure in Figure 6.7 meets the
3NF requirements. Yet the condition C → B causes the table to fail to meet the BCNF requirements.
A
B
C
D
To convert the table structure in Figure 6.7 into table structures that are in 3NF and in BCNF, first change the primary
key to A + C. That is an appropriate action because the dependency C → B means that C is, in effect, a superset of
B. At this point, the table is in 1NF because it contains a partial dependency, C → B. Next, follow the standard
decomposition procedures to produce the results shown in Figure 6.8.
To see how this procedure can be applied to an actual problem, examine the sample data in Table 6.5.
TABLE
6.5
STU_ID
125
125
135
144
144
Sample Data for a BCNF Conversion
STAFF_ID
25
20
20
25
20
CLASS_CODE
21334
32456
28458
27563
32456
ENROLL_GRADE
A
C
B
C
B
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FIGURE
6
Decomposition to BCNF
6.8
3NF, but not BCNF
A
B
C
D
1NF
A
C
B
D
Partial dependency
A
C
D
3NF and BCNF
C
B
3NF and BCNF
Table 6.5 reflects the following conditions:
쐌
Each CLASS_CODE identifies a class uniquely. This condition illustrates the case in which a course might
generate many classes. For example, a course labeled INFS 420 might be taught in two classes (sections), each
identified by a unique code to facilitate registration. Thus, the CLASS_CODE 32456 might identify INFS 420,
class section 1, while the CLASS_CODE 32457 might identify INFS 420, class section 2. Or the
CLASS_CODE 28458 might identify QM 362, class section 5.
쐌
A student can take many classes. Note, for example, that student 125 has taken both 21334 and 32456,
earning the grades A and C, respectively.
쐌
A staff member can teach many classes, but each class is taught by only one staff member. Note that staff
member 20 teaches the classes identified as 32456 and 28458.
The structure shown in Table 6.5 is reflected in Panel A of Figure 6.9:
STU_ID + STAFF_ID → CLASS_CODE, ENROLL_GRADE
CLASS_CODE → STAFF_ID
Panel A of Figure 6.9 shows a structure that is clearly in 3NF, but the table represented by this structure has a major
problem, because it is trying to describe two things: staff assignments to classes and student enrollment information.
Such a dual-purpose table structure will cause anomalies. For example, if a different staff member is assigned to teach
class 32456, two rows will require updates, thus producing an update anomaly. And if student 135 drops class 28458,
N O R M A L I Z A T I O N
FIGURE
O F
D A T A B A S E
TA B L E S
Another BCNF decomposition
6.9
Panel A: 3NF, but not BCNF
STU_ID
STAFF_ID
CLASS_CODE
ENROLL_GRADE
Panel B: 3NF and BCNF
STU_ID
CLASS_CODE
ENROLL_GRADE
CLASS_CODE
STAFF_ID
information about who taught that class is lost, thus producing a deletion anomaly. The solution to the problem is to
decompose the table structure, following the procedure outlined earlier. Note that the decomposition of Panel B shown
in Figure 6.9 yields two table structures that conform to both 3NF and BCNF requirements.
Remember that a table is in BCNF when every determinant in that table is a candidate key. Therefore, when a table
contains only one candidate key, 3NF and BCNF are equivalent.
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6.6.2 Fourth Normal Form (4NF)
You might encounter poorly designed databases, or you might be asked to convert spreadsheets into a database format
in which multiple multivalued attributes exist. For example, consider the possibility that an employee can have multiple
assignments and can also be involved in multiple service organizations. Suppose employee 10123 does volunteer work
for the Red Cross and United Way. In addition, the same employee might be assigned to work on three projects: 1,
3, and 4. Figure 6.10 illustrates how that set of facts can be recorded in very different ways.
FIGURE
Tables with multivalued dependencies
6.10
Database name: Ch06_Service
Table name: VOLUNTEER_V1
Table name: VOLUNTEER_V2
Table name: VOLUNTEER_V3
There is a problem with the tables in Figure 6.10. The attributes ORG_CODE and ASSIGN_NUM each may have
many different values. In normalization terminology, this situation is referred to as a multivalued dependency. A
multivalued dependency occurs when one key determines multiple values of two other attributes and those attributes
are independent of each other. (One employee can have many service entries and many assignment entries. Therefore,
one EMP_NUM can determine multiple values of ORG_CODE and multiple values of ASSIGN_NUM; however,
ORG_CODE and ASSIGN_NUM are independent of each other.) The presence of a multivalued dependency means
that if versions 1 and 2 are implemented, the tables are likely to contain quite a few null values; in fact, the tables do
not even have a viable candidate key. (The EMP_NUM values are not unique, so they cannot be PKs. No combination
of the attributes in table versions 1 and 2 can be used to create a PK because some of them contain nulls.) Such a
condition is not desirable, especially when there are thousands of employees, many of whom may have multiple job
assignments and many service activities. Version 3 at least has a PK, but it is composed of all of the attributes in the
table. In fact, version 3 meets 3NF requirements, yet it contains many redundancies that are clearly undesirable.
The solution is to eliminate the problems caused by the multivalued dependency. You do this by creating new tables
for the components of the multivalued dependency. In this example, the multivalued dependency is resolved by creating
the ASSIGNMENT and SERVICE_V1 tables depicted in Figure 6.11. Note that in Figure 6.11, neither the
ASSIGNMENT nor the SERVICE_V1 table contains a multivalued dependency. Those tables are said to be in 4NF.
If you follow the proper design procedures illustrated in this book, you shouldn’t encounter the previously described
problem. Specifically, the discussion of 4NF is largely academic if you make sure that your tables conform to the
following two rules:
1.
All attributes must be dependent on the primary key, but they must be independent of each other.
2.
No row may contain two or more multivalued facts about an entity.
N O R M A L I Z A T I O N
FIGURE
O F
D A T A B A S E
TA B L E S
A set of tables in 4NF
6.11
Database name: CH06_Service
Table name: PROJECT
Table name: EMPLOYEE
Table name: ORGANIZATION
Table name: ASSIGNMENT
Table name: SERVICE_V1
The relational diagram
Note
A table is in fourth normal form (4NF) when it is in 3NF and has no multivalued dependencies.
6.7 NORMALIZATION AND DATABASE DESIGN
The tables shown in Figure 6.6 illustrate how normalization procedures can be used to produce good tables from poor
ones. You will likely have ample opportunity to put this skill into practice when you begin to work with real-world
databases. Normalization should be part of the design process. Therefore, make sure that proposed entities meet
the required normal form before the table structures are created. Keep in mind that if you follow the design procedures
discussed in Chapter 3 and Chapter 4, the likelihood of data anomalies will be small. But even the best database
designers are known to make occasional mistakes that come to light during normalization checks. However, many of
the real-world databases you encounter will have been improperly designed or burdened with anomalies if they were
improperly modified over the course of time. And that means you might be asked to redesign and modify existing
databases that are, in effect, anomaly traps. Therefore, you should be aware of good design principles and procedures
as well as normalization procedures.
First, an ERD is created through an iterative process. You begin by identifying relevant entities, their attributes, and
their relationships. Then you use the results to identify additional entities and attributes. The ERD provides the big
picture, or macro view, of an organization’s data requirements and operations.
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Second, normalization focuses on the characteristics of specific entities; that is, normalization represents a micro view
of the entities within the ERD. And as you learned in the previous sections of this chapter, the normalization process
might yield additional entities and attributes to be incorporated into the ERD. Therefore, it is difficult to separate the
normalization process from the ER modeling process; the two techniques are used in an iterative and incremental
process.
To illustrate the proper role of normalization in the design process, let’s reexamine the operations of the contracting
company whose tables were normalized in the preceding sections. Those operations can be summarized by using the
following business rules:
쐌
The company manages many projects.
쐌
Each project requires the services of many employees.
쐌
An employee may be assigned to several different projects.
쐌
Some employees are not assigned to a project and perform duties not specifically related to a project. Some
employees are part of a labor pool, to be shared by all project teams. For example, the company’s executive
secretary would not be assigned to any one particular project.
쐌
Each employee has a single primary job classification. That job classification determines the hourly billing rate.
쐌
Many employees can have the same job classification. For example, the company employs more than one
electrical engineer.
Given that simple description of the company’s operations, two entities and their attributes are initially defined:
쐌
PROJECT (PROJ_NUM, PROJ_NAME)
쐌
EMPLOYEE (EMP_NUM, EMP_LNAME, EMP_FNAME, EMP_INITIAL, JOB_DESCRIPTION,
JOB_CHG_HOUR)
Those two entities constitute the initial ERD shown in Figure 6.12.
FIGURE
6.12
Initial contracting company
ERD
After creating the initial ERD shown in Figure 6.12, the
normal forms are defined:
쐌
PROJECT is in 3NF and needs no modification at
this point.
쐌
EMPLOYEE requires additional scrutiny. The JOB_
DESCRIPTION attribute defines job classifications
such as Systems Analyst, Database Designer, and
Programmer. In turn, those classifications determine
the billing rate, JOB_CHG_HOUR. Therefore,
EMPLOYEE contains a transitive dependency.
The removal of EMPLOYEE’s transitive dependency yields three entities:
쐌
PROJECT (PROJ_NUM, PROJ_NAME)
쐌
EMPLOYEE (EMP_NUM, EMP_LNAME, EMP_FNAME, EMP_INITIAL, JOB_CODE)
쐌
JOB (JOB_CODE, JOB_DESCRIPTION, JOB_CHG_HOUR)
Because the normalization process yields an additional entity (JOB), the initial ERD is modified as shown in
Figure 6.13.
To represent the M:N relationship between EMPLOYEE and PROJECT, you might think that two 1:M relationships
could be used—an employee can be assigned to many projects, and each project can have many employees assigned
to it. (See Figure 6.14.) Unfortunately, that representation yields a design that cannot be correctly implemented.
Because the M:N relationship between EMPLOYEE and PROJECT cannot be implemented, the ERD in Figure 6.14
must be modified to include the ASSIGNMENT entity to track the assignment of employees to projects, thus yielding
N O R M A L I Z A T I O N
FIGURE
O F
D A T A B A S E
TA B L E S
Modified contracting company ERD
6.13
FIGURE
Incorrect M:N relationship representation
6.14
the ERD shown in Figure 6.15. The ASSIGNMENT entity in Figure 6.15 uses the primary keys from the entities
PROJECT and EMPLOYEE to serve as its foreign keys. However, note that in this implementation, the ASSIGNMENT
entity’s surrogate primary key is ASSIGN_NUM, to avoid the use of a composite primary key. Therefore, the “enters”
relationship between EMPLOYEE and ASSIGNMENT and the “requires” relationship between PROJECT and
ASSIGNMENT are shown as weak or nonidentifying.
Note that in Figure 6.15, the ASSIGN_HOURS attribute is assigned to the composite entity named ASSIGNMENT.
Because you will likely need detailed information about each project’s manager, the creation of a “manages”
relationship is useful. The “manages” relationship is implemented through the foreign key in PROJECT. Finally, some
additional attributes may be created to improve the system’s ability to generate additional information. For example,
199
you may want to include the date on which the employee was hired (EMP_HIREDATE) to keep track of worker
longevity. Based on this last modification, the model should include four entities and their attributes:
PROJECT (PROJ_NUM, PROJ_NAME, EMP_NUM)
EMPLOYEE (EMP_NUM, EMP_LNAME, EMP_FNAME, EMP_INITIAL, EMP_HIREDATE, JOB_CODE)
JOB (JOB_CODE, JOB_DESCRIPTION, JOB_CHG_HOUR)
ASSIGNMENT (ASSIGN_NUM, ASSIGN_DATE, PROJ_NUM, EMP_NUM, ASSIGN_HOURS, ASSIGN_CHG_
HOUR, ASSIGN_CHARGE)
The design process is now on the right track. The ERD represents the operations accurately, and the entities now
reflect their conformance to 3NF. The combination of normalization and ER modeling yields a useful ERD, whose
entities may now be translated into appropriate table structures. In Figure 6.15, note that PROJECT is optional to
EMPLOYEE in the “manages” relationship. This optionality exists because not all employees manage projects. The
final database contents are shown in Figure 6.16.
6.8 DENORMALIZATION
It’s important to remember that the optimal relational database implementation requires that all tables be at least in
third normal form (3NF). A good relational DBMS excels at managing normalized relations; that is, relations void of
any unnecessary redundancies that might cause data anomalies. Although the creation of normalized relations is an
important database design goal, it is only one of many such goals. Good database design also considers processing (or
reporting) requirements and processing speed. The problem with normalization is that as tables are decomposed to
conform to normalization requirements, the number of database tables expands. Therefore, in order to generate
information, data must be put together from various tables. Joining a large number of tables takes additional
input/output (I/O) operations and processing logic, thereby reducing system speed. Most relational database systems
are able to handle joins very efficiently. However, rare and occasional circumstances may allow some degree of
denormalization so processing speed can be increased.
N O R M A L I Z A T I O N
FIGURE
O F
D A T A B A S E
TA B L E S
The implemented database
6.16
Table name: EMPLOYEE
Database name: Ch06_ConstructCo
Table name: JOB
Table name: PROJECT
Table name: ASSIGNMENT
Keep in mind that the advantage of higher processing speed must be carefully weighed against the disadvantage of data
anomalies. On the other hand, some anomalies are of only theoretical interest. For example, should people in a
real-world database environment worry that a ZIP_CODE determines CITY in a CUSTOMER table whose primary key
is the customer number? Is it really practical to produce a separate table for
ZIP (ZIP_CODE, CITY)
to eliminate a transitive dependency from the CUSTOMER table? (Perhaps your answer to that question changes if you
are in the business of producing mailing lists.) As explained earlier, the problem with denormalized relations and
redundant data is that the data integrity could be compromised due to the possibility of data anomalies (insert, update,
and deletion anomalies). The advice is simple: use common sense during the normalization process.
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C H A P T E R
6
Furthermore, the database design process could, in some cases, introduce some small degree of redundant data in the
model (as seen in the previous example). This, in effect, creates “denormalized” relations. Table 6.6 shows some
common examples of data redundancy that are generally found in database implementations.
TABLE
6.6
Common Denormalization Examples
CASE
Redundant data
EXAMPLE
Storing ZIP and CITY attributes in the CUSTOMER
table when ZIP determines CITY. (See Table 1.4.)
Derived data
Storing STU_HRS and STU_CLASS (student classification) when STU_HRS determines STU_CLASS. (See
Figure 3.29.)
Preaggregated
data (also derived
data)
Storing the student grade point average (STU_GPA)
aggregate value in the STUDENT table when this can
be calculated from the ENROLL and COURSE tables.
(See Figure 3.29.)
Information
requirements
Using a temporary denormalized table to hold report
data. This is required when creating a tabular report in
which the columns represent data that are stored in
the table as rows. (See Figure 6.17 and Figure 6.18.)
RATIONALE AND CONTROLS
• Avoid extra joint operations.
• Program can validate city
(drop-down box) based on the
zip code.
• Avoid extra joint operations.
• Program can validate classification (lookup) based on the student hours.
• Avoid extra joint operations.
• Program computes the GPA
every time a grade is entered or
updated.
• STU_GPA can be updated only
via administrative routine.
• Impossible to generate the data
required by the report using plain
SQL.
• No need to maintain table.
Temporary table is deleted once
report is done.
• Processing speed is not an issue.
A more comprehensive example of the need for denormalization due to reporting requirements is the case of a faculty
evaluation report in which each row list the scores obtained during the last four semesters taught. (See Figure 6.17.)
FIGURE
The faculty evaluation report
6.17
Although this report seems simple enough, the problem arises from the fact that the data are stored in a normalized
table in which each row represents a different score for a given faculty member in a given semester. (See Figure 6.18.)
The difficulty of transposing multirow data to multicolumnar data is compounded by the fact that the last four semesters
taught are not necessarily the same for all faculty members (some might have taken sabbaticals, some might have had
N O R M A L I Z A T I O N
FIGURE
O F
D A T A B A S E
TA B L E S
The EVALDATA and FACHIST tables
6.18
Table name: EVALDATA
Table name: FACHIST
Denormalized
Database name: Ch06_EVAL
Repeating Group
Normalized
research appointments, some might be new faculty with only two semesters on the job, etc.). To generate this report,
the two tables you see in Figure 6.18 were used. The EVALDATA table is the master data table containing the
evaluation scores for each faculty member for each semester taught; this table is normalized. The FACHIST table
contains the last four data points—that is, evaluation score and semester—for each faculty member. The FACHIST
table is a temporary denormalized table created from the EVALDATA table via a series of queries. (The FACHIST table
is the basis for the report shown in Figure 6.17.)
As seen in the faculty evaluation report, the conflicts between design efficiency, information requirements, and
performance are often resolved through compromises that may include denormalization. In this case, and assuming
there is enough storage space, the designer’s choices could be narrowed down to:
쐌
Store the data in a permanent denormalized table. This is not the recommended solution, because the
denormalized table is subject to data anomalies (insert, update, and delete). This solution is viable only if
performance is an issue.
쐌
Create a temporary denormalized table from the permanent normalized table(s). Because the denormalized
table exists only as long as it takes to generate the report, it disappears after the report is produced. Therefore,
there are no data anomaly problems. This solution is practical only if performance is not an issue and there
are no other viable processing options.
As shown, normalization purity is often difficult to sustain in the modern database environment. You will learn
in Chapter 13, Business Intelligence and Data Warehouses, that lower normalization forms occur (and are even
required) in specialized databases known as data warehouses. Such specialized databases reflect the ever-growing
demand for greater scope and depth in the data on which decision support systems increasingly rely. You will discover
that the data warehouse routinely uses 2NF structures in its complex, multilevel, multisource data environment. In
short, although normalization is very important, especially in the so-called production database environment, 2NF is
no longer disregarded as it once was.
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C H A P T E R
6
Although 2NF tables cannot always be avoided, the problem of working with tables that contain partial and/or
transitive dependencies in a production database environment should not be minimized. Aside from the possibility of
troublesome data anomalies being created, unnormalized tables in a production database tend to suffer from these
defects:
쐌
Data updates are less efficient because programs that read and update tables must deal with larger tables.
쐌
Indexing is more cumbersome. It is simply not practical to build all of the indexes required for the many
attributes that might be located in a single unnormalized table.
쐌
Unnormalized tables yield no simple strategies for creating virtual tables known as views. You will learn how
to create and use views in Chapter 7, Introduction to Structured Query Language (SQL).
Remember that good design cannot be created in the application programs that use a database. Also keep in mind that
unnormalized database tables often lead to various data redundancy disasters in production databases such as the ones
examined thus far. In other words, use denormalization cautiously and make sure that you can explain why the
unnormalized tables are a better choice in certain situations than their normalized counterparts.
6.9 DATA-MODELING CHECKLIST
In the chapters of Part II, you have learned how data modeling translates a specific real-world environment into a data
model that represents the real-world data, users, processes, and interactions. The modeling techniques you have
learned thus far give you the tools needed to produce successful database designs. However, just as any good pilot uses
a checklist to ensure that all is in order for a successful flight, the data-modeling checklist shown in Table 6.7 will help
ensure that you perform data-modeling tasks successfully based on the concepts and tools you have learned in this text.
Note
You can also find this Data-Modeling Checklist on the inside back cover of this book for easy reference.
N O R M A L I Z A T I O N
TABLE
6.7
O F
D A T A B A S E
TA B L E S
Data-Modeling Checklist
DATA-MODELING CHECKLIST
BUSINESS RULES
• Properly document and verify all business rules with the end users.
• Ensure that all business rules are written precisely, clearly, and simply. The business rules must help identify
entities, attributes, relationships, and constraints.
• Identify the source of all business rules, and ensure that each business rule is justified, dated, and signed off by
an approving authority.
DATA MODELING
Naming Conventions: All names should be limited in length (database-dependent size).
• Entity Names:
• Should be nouns that are familiar to business and should be short and meaningful
• Should document abbreviations, synonyms, and aliases for each entity
• Should be unique within the model
• For composite entities, may include a combination of abbreviated names of the entities linked through the
composite entity
• Attribute Names:
• Should be unique within the entity
• Should use the entity abbreviation as a prefix
• Should be descriptive of the characteristic
• Should use suffixes such as _ID, _NUM, or _CODE for the PK attribute
• Should not be a reserved word
• Should not contain spaces or special characters such as @, !, or &
• Relationship Names:
• Should be active or passive verbs that clearly indicate the nature of the relationship
Entities:
• Each entity should represent a single subject.
• Each entity should represent a set of distinguishable entity instances.
• All entities should be in 3NF or higher. Any entities below 3NF should be justified.
• The granularity of the entity instance should be clearly defined.
• The PK should be clearly defined and support the selected data granularity.
Attributes:
• Should be simple and single-valued (atomic data)
• Should document default values, constraints, synonyms, and aliases
• Derived attributes should be clearly identified and include source(s)
• Should not be redundant unless this is required for transaction accuracy, performance, or
maintaining a history
• Nonkey attributes must be fully dependent on the PK attribute
Relationships:
• Should clearly identify relationship participants
• Should clearly define participation, connectivity, and document cardinality
ER Model:
• Should be validated against expected processes: inserts, updates, and deletes
• Should evaluate where, when, and how to maintain a history
• Should not contain redundant relationships except as required (see attributes)
• Should minimize data redundancy to ensure single-place updates
• Should conform to the minimal data rule: “All that is needed is there, and all that is there is needed.”
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206
C H A P T E R
6
S u m m a r y
◗
◗
◗
◗
◗
◗
◗
◗
◗
Normalization is a technique used to design tables in which data redundancies are minimized. The first three normal
forms (1NF, 2NF, and 3NF) are most commonly encountered. From a structural point of view, higher normal forms
are better than lower normal forms, because higher normal forms yield relatively fewer data redundancies in the
database. Almost all business designs use 3NF as the ideal normal form. A special, more restricted 3NF known as
Boyce-Codd normal form, or BCNF, is also used.
A table is in 1NF when all key attributes are defined and when all remaining attributes are dependent on the primary
key. However, a table in 1NF can still contain both partial and transitive dependencies. (A partial dependency is one
in which an attribute is functionally dependent on only a part of a multiattribute primary key. A transitive
dependency is one in which one attribute is functionally dependent on another nonkey attribute.) A table with a
single-attribute primary key cannot exhibit partial dependencies.
A table is in 2NF when it is in 1NF and contains no partial dependencies. Therefore, a 1NF table is automatically
in 2NF when its primary key is based on only a single attribute. A table in 2NF may still contain transitive
dependencies.
A table is in 3NF when it is in 2NF and contains no transitive dependencies. Given that definition of 3NF, the
Boyce-Codd normal form (BCNF) is merely a special 3NF case in which all determinant keys are candidate keys.
When a table has only a single candidate key, a 3NF table is automatically in BCNF.
A table that is not in 3NF may be split into new tables until all of the tables meet the 3NF requirements.
Normalization is an important part—but only a part—of the design process. As entities and attributes are defined
during the ER modeling process, subject each entity (set) to normalization checks and form new entity (sets) as
required. Incorporate the normalized entities into the ERD and continue the iterative ER process until all entities
and their attributes are defined and all equivalent tables are in 3NF.
A table in 3NF might contain multivalued dependencies that produce either numerous null values or redundant data.
Therefore, it might be necessary to convert a 3NF table to the fourth normal form (4NF) by splitting the table to
remove the multivalued dependencies. Thus, a table is in 4NF when it is in 3NF and contains no multivalued
dependencies.
The larger the number of tables, the more additional I/O operations and processing logic required to join them.
Therefore, tables are sometimes denormalized to yield less I/O in order to increase processing speed. Unfortunately, with larger tables, you pay for the increased processing speed by making the data updates less efficient, by
making indexing more cumbersome, and by introducing data redundancies that are likely to yield data anomalies.
In the design of production databases, use denormalization sparingly and cautiously.
The Data-Modeling Checklist provides a way for the designer to check that the ERD meets a set of minimum
requirements.
K e y
T e r m s
atomic attribute, 188
first normal form (1NF), 183
partial dependency, 180
atomicity, 188
fourth normal form (4NF), 197
prime attribute, 175
Boyce-Codd normal form
(BCNF), 192
granularity, 188
repeating group, 181
key attribute, 175
second normal form (2NF), 185
denormalization, 175
nonkey attribute, 175
third normal form (3NF), 187
dependency diagram, 182
nonprime attribute, 175
transitive dependency, 180
determinant, 185
normalization, 175
N O R M A L I Z A T I O N
O F
D A T A B A S E
TA B L E S
Online Content
Answers to selected Review Questions and Problems for this chapter are contained in the Premium Website for
this book.
R e v i e w
Q u e s t i o n s
1. What is normalization?
2. When is a table in 1NF?
3. When is a table in 2NF?
4. When is a table in 3NF?
5. When is a table in BCNF?
6. Given the dependency diagram shown in Figure Q6.6, answer Items 6a−6c.
FIGURE
Dependency diagram for Question 6
Q6.6
C1
C2
C3
C4
C5
a. Identify and discuss each of the indicated dependencies.
b. Create a database whose tables are at least in 2NF, showing the dependency diagrams for each table.
c. Create a database whose tables are at least in 3NF, showing the dependency diagrams for each table.
7. The dependency diagram in Figure Q6.7 indicates that authors are paid royalties for each book that they write
for a publisher. The amount of the royalty can vary by author, by book, and by edition of the book.
FIGURE
Book royalty dependency diagram
Q6.7
ISBN
BookTitle
Author_Num
LastName
Publisher
Royalty
Edition
207
208
C H A P T E R
6
a. Based on the dependency diagram, create a database whose tables are at least in 2NF, showing the
dependency diagram for each table.
b. Create a database whose tables are at least in 3NF, showing the dependency diagram for each table.
8. The dependency diagram in Figure Q6.8 indicates that a patient can receive many prescriptions for one or more
medicines over time. Based on the dependency diagram, create a database whose tables are in at least 2NF,
showing the dependency diagram for each table.
FIGURE
Prescription dependency diagram
Q6.8
MedName
PatientID
Date
RefillsAllowed PatientName
Dosage
ShelfLife
9. What is a partial dependency? With what normal form is it associated?
10. What three data anomalies are likely to be the result of data redundancy? How can such anomalies be eliminated?
11. Define and discuss the concept of transitive dependency.
12. What is a surrogate key, and when should you use one?
13. Why is a table whose primary key consists of a single attribute automatically in 2NF when it is in 1NF?
14. How would you describe a condition in which one attribute is dependent on another attribute, when neither
attribute is part of the primary key?
15. Suppose that someone tells you that an attribute that is part of a composite primary key is also a candidate key.
How would you respond to that statement?
16. A table is in
dependencies.
normal form when it is in
and there are no transitive
P r o b l e m s
1. Using the descriptions of the attributes given in the figure, convert the ERD shown in Figure P6.1 into a
dependency diagram that is in at least 3NF.
2. Using the descriptions of the attributes given in the figure, convert the ERD shown in Figure P6.2 into a
dependency diagram that is in at least 3NF.
3. Using the INVOICE table structure shown in Table P6.3, do the following:
a. Write the relational schema, draw its dependency diagram, and identify all dependencies, including all partial
and transitive dependencies. You can assume that the table does not contain repeating groups and that an
invoice number references more than one product. (Hint: This table uses a composite primary key.)
b. Remove all partial dependencies, write the relational schema, and draw the new dependency diagrams.
Identify the normal forms for each table structure you created.
c. Remove all transitive dependencies, write the relational schema, and draw the new dependency diagrams.
Also identify the normal forms for each table structure you created.
d. Draw the Crow’s Foot ERD.
N O R M A L I Z A T I O N
FIGURE
Appointment ERD for Problem 1
P6.1
FIGURE
P6.2
Presentation ERD for Problem 2
O F
D A T A B A S E
TA B L E S
209
210
C H A P T E R
6
TABLE
P6.3
ATTRIBUTE
NAME
INV_NUM
PROD_NUM
SALE_DATE
PROD_LABEL
VEND_CODE
VEND_NAME
QUANT_SOLD
PROD_PRICE
SAMPLE VALUE
SAMPLE VALUE
SAMPLE VALUE
SAMPLE VALUE
SAMPLE VALUE
211347
AA-E3422QW
15-Jan-2010
Rotary sander
211
NeverFail, Inc.
1
$49.95
211347
QD-300932X
15-Jan-2010
0.25-in. drill bit
211
NeverFail, Inc.
8
$3.45
211347
RU-995748G
15-Jan-2010
Band saw
309
BeGood, Inc.
1
$39.99
211348
AA-E3422QW
15-Jan-2010
Rotary sander
211
NeverFail, Inc.
2
$49.95
211349
GH-778345P
16-Jan-2010
Power drill
157
ToughGo, Inc.
1
$87.75
Note
You can assume that any given product is supplied by a single vendor, but a vendor can supply many products.
Therefore, it is proper to conclude that the following dependency exists:
PROD_NUM → PROD_DESCRIPTION, PROD_PRICE, VEND_CODE, VEND_NAME
(Hint: Your actions should produce three dependency diagrams.)
4. Using the STUDENT table structure shown in Table P6.4 do the following:
a. Write the relational schema and draw its dependency diagram. Identify all dependencies, including all
transitive dependencies.
b. Write the relational schema and draw the dependency diagram to meet the 3NF requirements to the greatest
practical extent possible. If you believe that practical considerations dictate using a 2NF structure, explain why
your decision to retain 2NF is appropriate. If necessary, add or modify attributes to create appropriate
determinants and to adhere to the naming conventions.
TABLE
P6.4
ATTRIBUTE NAME
STU_NUM
STU_LNAME
STU_MAJOR
DEPT_CODE
DEPT_NAME
DEPT_PHONE
COLLEGE_NAME
ADVISOR_LNAME
ADVISOR_OFFICE
ADVISOR_BLDG
ADVISOR_PHONE
STU_GPA
STU_HOURS
STU_CLASS
SAMPLE VALUE
211343
Stephanos
Accounting
ACCT
Accounting
4356
Business Admin
Grastrand
T201
Torre Building
2115
3.87
75
Junior
SAMPLE VALUE
200128
Smith
Accounting
ACCT
Accounting
4356
Business Admin
Grastrand
T201
Torre Building
2115
2.78
45
Sophomore
SAMPLE VALUE
199876
Jones
Marketing
MKTG
Marketing
4378
Business Admin
Gentry
T228
Torre Building
2123
2.31
117
Senior
SAMPLE VALUE
199876
Ortiz
Marketing
MKTG
Marketing
4378
Business Admin
Tillery
T356
Torre Building
2159
3.45
113
Senior
SAMPLE VALUE
223456
McKulski
Statistics
MATH
Mathematics
3420
Arts & Sciences
Chen
J331
Jones Building
3209
3.58
87
Junior
N O R M A L I Z A T I O N
O F
D A T A B A S E
TA B L E S
c. Write the relational schema and draw the dependency diagram to meet the 3NF requirements to the greatest
practical extent possible. If you believe that practical considerations dictate using a 2NF structure, explain why
your decision to retain 2NF is appropriate. If necessary, add or modify attributes to create appropriate
determinants and to adhere to the naming conventions.
d. Using the results of Problem 4, draw the Crow’s Foot ERD.
Note
Although the completed student hours (STU_HOURS) do determine the student classification (STU_CLASS),
this dependency is not as obvious as you might initially assume it to be. For example, a student is considered a
junior if that student has completed between 61 and 90 credit hours. Therefore, a student who is classified as
a junior may have completed 66, 72, or 87 hours or any other number of hours within the specified range of
61−90 hours. In short, any hour value within a specified range will define the classification.
5. To keep track of office furniture, computers, printers, and so on, the FOUNDIT Company uses the table structure
shown in Table P6.5.
TABLE
P6.5
ATTRIBUTE NAME
ITEM_ID
ITEM_LABEL
ROOM_NUMBER
BLDG_CODE
BLDG_NAME
BLDG_MANAGER
SAMPLE VALUE
231134-678
HP DeskJet 895Cse
325
NTC
Nottooclear
I. B. Rightonit
SAMPLE VALUE
342245-225
HP Toner
325
NTC
Nottoclear
I. B. Rightonit
SAMPLE VALUE
254668-449
DT Scanner
123
CSF
Canseefar
May B. Next
a. Given that information, write the relational schema and draw the dependency diagram. Make sure that you
label the transitive and/or partial dependencies.
b. Write the relational schema and create a set of dependency diagrams that meet 3NF requirements. Rename
attributes to meet the naming conventions and create new entities and attributes as necessary.
c. Draw the Crow’s Foot ERD.
211
212
C H A P T E R
6
6. The table structure shown in Table P6.6 contains many unsatisfactory components and characteristics. For example,
there are several multivalued attributes, naming conventions are violated, and some attributes are not atomic.
TABLE
P6.6
EMP_NUM
EMP_LNAME
EMP_EDUCATION
JOB_CLASS
EMP_DEPENDENTS
DEPT_CODE
DEPT_NAME
DEPT_MANAGER
EMP_TITLE
EMP_DOB
EMP_HIRE_DATE
EMP_TRAINING
EMP_BASE_SALARY
EMP_COMMISSION_RATE
1003
Willaker
BBA, MBA
SLS
Gerald (spouse),
Mary (daughter),
John (son)
MKTG
Marketing
Jill H. Martin
Sales Agent
23-Dec-1968
14-Oct-1997
L1, L2
$38,255.00
0.015
1018
Smith
BBA
SLS
1019
McGuire
MKTG
Marketing
Jill H. Martin
Sales Agent
28-Mar-1979
15-Jan-2006
L1
$30,500.00
0.010
SVC
General Service
Hank B. Jones
Janitor
18-May-1982
21-Apr-2003
L1
$19,750.00
JNT
JoAnne (spouse)
1023
McGuire
BS, MS, Ph.D.
DBA
George (spouse)
Jill (daughter)
INFS
Info. Systems
Carlos G. Ortez
DB Admin
20-Jul-1959
15-Jul-1999
L1, L3, L8, L15
$127,900.00
a. Given the structure shown in Table P6.6, write the relational schema and draw its dependency diagram. Label
all transitive and/or partial dependencies.
b. Draw the dependency diagrams that are in 3NF. (Hint: You might have to create a few new attributes. Also
make sure that the new dependency diagrams contain attributes that meet proper design criteria; that is,
make sure that there are no multivalued attributes, that the naming conventions are met, and so on.)
c. Draw the relational diagram.
d. Draw the Crow’s Foot ERD.
7. Suppose you are given the following business rules to form the basis for a database design. The database must
enable the manager of a company dinner club to mail invitations to the club’s members, to plan the meals, to
keep track of who attends the dinners, and so on.
쐌
Each dinner serves many members, and each member may attend many dinners.
쐌
A member receives many invitations, and each invitation is mailed to many members.
쐌
A dinner is based on a single entree, but an entree may be used as the basis for many dinners. For example,
a dinner may be composed of a fish entree, rice, and corn; or the dinner may be composed of a fish entree,
a baked potato, and string beans.
쐌
A member may attend many dinners, and each dinner may be attended by many members.
Because the manager is not a database expert, the first attempt at creating the database uses the structure shown
in Table P6.7.
a. Given the table structure illustrated in Table P6.7, write the relational schema and draw its dependency
diagram. Label all transitive and/or partial dependencies. (Hint: This structure uses a composite primary key.)
b. Break up the dependency diagram you drew in Problem 7a to produce dependency diagrams that are in 3NF,
and write the relational schema. (Hint: You might have to create a few new attributes. Also, make sure that
the new dependency diagrams contain attributes that meet proper design criteria; that is, make sure that there
are no multivalued attributes, that the naming conventions are met, and so on.)
c. Using the results of Problem 7b, draw the Crow’s Foot ERD.
N O R M A L I Z A T I O N
O F
D A T A B A S E
TA B L E S
TABLE
P6.7
ATTRIBUTE NAME
MEMBER_NUM
MEMBER_NAME
MEMBER_ADDRESS
MEMBER_CITY
MEMBER_ZIPCODE
INVITE_NUM
INVITE_DATE
ACCEPT_DATE
DINNER_DATE
DINNER_ATTENDED
DINNER_CODE
DINNER_DESCRIPTION
ENTREE_CODE
ENTREE_DESCRIPTION
DESSERT_CODE
DESSERT_DESCRIPTION
SAMPLE VALUE
214
Alice B. VanderVoort
325 Meadow Park
Murkywater
12345
8
23-Feb-2008
27-Feb-2008
15-Mar-2008
Yes
DI5
Glowing ea Delight
EN3
Stuffed crab
DE8
Chocolate mousse
with raspberry sauce
SAMPLE VALUE
235
Gerald M. Gallega
123 Rose Court
Highlight
12349
9
12-Mar-2008
15-Mar-2008
17-Mar-2008
Yes
DI5
Glowing Sea Delight
EN3
Stuffed crab
DE5
Cherries jubilee
SAMPLE VALUE
214
Alice B. VanderVoort
325 Meadow Park
Murkywater
12345
10
23-Feb-2008
27-Feb-2008
15-Mar-2008
No
DI2
Ranch Superb
EN5
Marinated steak
DE2
Apple pie with honey
crust
8. Use the dependency diagram shown in Figure P6.8 to work the following problems.
a. Break up the dependency diagram shown in Figure P6.8 to create two new dependency diagrams, one in
3NF and one in 2NF.
b. Modify the dependency diagrams you created in Problem 8a to produce a set of dependency diagrams that
are in 3NF. (Hint: One of your dependency diagrams will be in 3NF but not in BCNF.)
c. Modify the dependency diagrams you created in Problem 8b to produce a collection of dependency diagrams
that are in 3NF and BCNF.
FIGURE
Initial dependency diagram for Problem 8
P6.8
A
B
C
D
E
F
G
213
214
C H A P T E R
6
9. Suppose you have been given the table structure and data shown in Table P6.9, which was imported from an
Excel spreadsheet. The data reflect that a professor can have multiple advisees, can serve on multiple committees,
and can edit more than one journal.
TABLE
P6.9
ATTRIBUTE NAME
EMP_NUM
PROF_RANK
EMP_NAME
DEPT_CODE
DEPT_NAME
PROF_OFFICE
ADVISEE
COMMITTEE_CODE
JOURNAL_CODE
SAMPLE VALUE
123
Professor
Ghee
CIS
Computer Info.
Systems
KDD-567
1215, 2312, 3233,
2218, 2098
PROMO, TRAF,
APPL, DEV
JMIS, QED,
JMGT
SAMPLE VALUE
104
Asst. Professor
Rankin
CHEM
Chemistry
BLF-119
3102, 2782,
3311, 2008,
2876, 2222,
3745, 1783,
2378
DEV
SAMPLE VALUE
118
Assoc. Professor
Ortega
CIS
Computer Info.
Systems
KDD-562
2134, 2789, 3456,
2002, 2046, 2018,
2764
SAMPLE VALUE
SPR, TRAF
PROMO, SPR,
DEV
Assoc. Professor
Smith
ENG
English
PRT-345
2873, 2765, 2238,
2901, 2308
JCIS, JMGT
Given the information in Table P6.9:
a. Draw the dependency diagram.
b. Identify the multivalued dependencies.
c. Create the dependency diagrams to yield a set of table structures in 3NF.
d. Eliminate the multivalued dependencies by converting the affected table structures to 4NF.
e. Draw the Crow’s Foot ERD to reflect the dependency diagrams you drew in Problem 9c. (Note: You might
have to create additional attributes to define the proper PKs and FKs. Make sure that all of your attributes
conform to the naming conventions.)
10. The manager of a consulting firm has asked you to evaluate a database that contains the table structure shown
in Table P6.10.
Table P6.10 was created to enable the manager to match clients with consultants. The objective is to match a
client within a given region with a consultant in that region and to make sure that the client’s need for specific
consulting services is properly matched to the consultant’s expertise. For example, if the client needs help with
database design and is located in the Southeast, the objective is to make a match with a consultant who is located
in the Southeast and whose expertise is in database design. (Although the consulting company manager tries to
match consultant and client locations to minimize travel expense, it is not always possible to do so.) The following
basic business rules are maintained:
쐌
Each client is located in one region.
쐌
A region can contain many clients.
쐌
Each consultant can work on many contracts.
쐌
Each contract might require the services of many consultants.
쐌
A client can sign more than one contract, but each contract is signed by only one client.
쐌
Each contract might cover multiple consulting classifications. (For example, a contract may list consulting
services in database design and networking.)
쐌
Each consultant is located in one region.
N O R M A L I Z A T I O N
O F
D A T A B A S E
TA B L E S
TABLE
P6.10
ATTRIBUTE NAME
CLIENT_NUM
CLIENT_NAME
CLIENT_REGION
CONTRACT_DATE
CONTRACT_NUMBER
CONTRACT_AMOUNT
CONSULT_CLASS_1
CONSULT_CLASS_2
CONSULT_CLASS_3
CONSULT_CLASS_4
CONSULTANT_NUM_1
CONSULTANT_NAME_1
CONSULTANT_REGION_1
CONSULTANT_NUM_2
CONSULTANT_NAME_2
CONSULTANT_REGION_2
CONSULTANT_NUM_3
CONSULTANT_NAME_3
CONSULTANT_REGION_3
CONSULTANT_NUM_4
CONSULTANT_NAME_4
CONSULTANT_REGION_4
SAMPLE VALUE
298
Marianne R. Brown
Midwest
10-Feb-2010
5841
$2,985,000.00
Database Administration
Web Applications
SAMPLE VALUE
289
James D. Smith
Southeast
15-Feb-2010
5842
$670,300.00
Internet Services
SAMPLE VALUE
289
James D. Smith
Southeast
12-Mar-2010
5843
$1,250,000.00
Database Design
Database Administration
Network Installation
29
Rachel G. Carson
Midwest
56
Karl M. Spenser
Midwest
22
Julian H. Donatello
Midwest
34
Gerald K. Ricardo
Southeast
38
Anne T. Dimarco
Southeast
45
Geraldo J. Rivera
Southeast
18
Donald Chen
West
25
Angela M. Jamison
Southeast
34
Gerald K. Ricardo
Southeast
쐌
A region can contain many consultants.
쐌
Each consultant has one or more areas of expertise (class). For example, a consultant might be classified as
an expert in both database design and networking.
쐌
Each area of expertise (class) can have many consultants in it. For example, the consulting company might
employ many consultants who are networking experts.
a. Given this brief description of the requirements and the business rules, write the relational schema and draw
the dependency diagram for the preceding (and very poor) table structure. Label all transitive and/or partial
dependencies.
b. Break up the dependency diagram you drew in Problem 10a to produce dependency diagrams that are in
3NF and write the relational schema. (Hint: You might have to create a few new attributes. Also make sure
that the new dependency diagrams contain attributes that meet proper design criteria; that is, make sure that
there are no multivalued attributes, that the naming conventions are met, and so on.)
c. Using the results of Problem 10b, draw the Crow’s Foot ERD.
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6
11. Given the sample records in the CHARTER table shown in Table P6.11, do the following:
a. Write the relational schema and draw the dependency diagram for the table structure. Make sure that you
label all dependencies. CHAR_PAX indicates the number of passengers carried. The CHAR_MILES entry is
based on round-trip miles, including pickup points. (Hint: Look at the data values to determine the nature of
the relationships. For example, note that employee Melton has flown two charter trips as pilot and one trip
as copilot.)
b. Decompose the dependency diagram you drew to solve Problem 11a to create table structures that are in
3NF and write the relational schema.
c. Draw the Crow’s Foot ERD to reflect the properly decomposed dependency diagrams you created in
Problem 11b. Make sure that the ERD yields a database that can track all of the data shown in Problem 11.
Show all entities, relationships, connectivities, optionalities, and cardinalities.
TABLE
P6.11
ATTRIBUTE NAME
CHAR_TRIP
CHAR_DATE
CHAR_CITY
CHAR_MILES
CUST_NUM
CUST_LNAME
CHAR_PAX
CHAR_CARGO
PILOT
COPILOT
FLT_ENGINEER
LOAD_MASTER
AC_NUMBER
MODEL_CODE
MODEL_SEATS
MODEL_CHG_MILE
SAMPLE VALUE
10232
15-Jan-2008
STL
580
784
Brown
5
235 lbs.
Melton
1234Q
PA31-350
10
$2.79
SAMPLE VALUE
10233
15-Jan-2008
MIA
1,290
231
Hanson
12
18,940 lbs.
Chen
Henderson
O'Shaski
Benkasi
3456Y
CV-580
38
$23.36
SAMPLE VALUE
10234
16-Jan-2008
TYS
524
544
Bryana
2
348 lbs.
Henderson
Melton
SAMPLE VALUE
10235
17-Jan-2008
ATL
768
784
Brown
5
155 lbs.
Melton
1234Q
PA31-350
10
$2.79
2256W
PA31-350
10
$2.79
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PART
III
Advanced Design and
Implementation
Introduction to Structured Query
Language (SQL)
7
Advanced SQL
8
Database Design
9
The Many Benefits of BI
Since its inception in 1962, the financial service company Raymond James has distinguished itself by focusing on the financial needs of its clients. Raymond James was the first
financial service company to introduce a client bill of rights, and it invested heavily in tools
and strategies to help clients meet their financial goals. Today it is one of the largest
financial service companies in the United States, with annual revenues of $3.2 billion in
2008 and over 5,000 financial advisors worldwide.
When financial firms faced hard times following the crisis of 2008, Raymond James
responded by improving services though IT innovation. The company created a business
intelligence (BI) system that provided faster, more accurate information to help financial
advisors evaluate the performance of investment programs and client portfolios.
With increased regulations, rising acquisitions, and the credit crises, Raymond James
decided to expand its existing data warehouse and BI services. In 2003, Raymond James
had pioneered a data warehouse employing Microsoft® SQL Server® 2000 data
management software to build the database and SQL Server 2000 Analysis Services to
develop a BI product they called Business Analyzer. Business Analyzer saved considerable
time for financial advisors who previously had to jump between numerous software
applications to get the information they needed. In 2009, they upgraded to SQL Server
2008 Enterprise and made use of its Integration Services, which allowed them to add
transactional data to their data warehouse from over a dozen databases running on
different platforms.The company also made use of SQL Server 2008 Analysis Services and
Reporting Services to create a new BI tool, Reports Center, that provides more data
analysis and richer graphic tools. Reports Center responds to queries much faster, cutting
fact-processing time from 10 hours to 40 minutes. Increased efficiency meant more
people were using the new system. In the months following deployment, usage rose
40 percent as the user base swelled from 4,000 to 6,600. The upgrade also meant
significant cost savings for the company, not only because of increased efficiency but also
because of the data compression feature that compressed parts of the database by as
much as 80 percent.The size of the database fell from 2 terabytes to about 600 gigabytes.
The cost savings was important to Raymond James at a time when financial service firms
were looking to the U.S. government for bailout loans. In May 2009, Raymond James
decided that its position was secure enough to allow it to turn down the federal monies.
The company predicted that even if the economy worsened, it could sustain about
$300 million in losses and it would be able to stay afloat during these tough times.
B
V
usiness
ignette
S E V E N
7
Introduction to Structured Query Language (SQL)
In this chapter, you will learn:
쐍 The basic commands and functions of SQL
쐍 How to use SQL for data administration (to create tables, indexes, and views)
쐍 How to use SQL for data manipulation (to add, modify, delete, and retrieve data)
쐍 How to use SQL to query a database for useful information
In this chapter, you will learn the basics of Structured Query Language (SQL). SQL,
pronounced S-Q-L by some and “sequel” by others, is composed of commands that enable
users to create database and table structures, perform various types of data manipulation
and data administration, and query the database to extract useful information. All relational
DBMS software supports SQL, and many software vendors have developed extensions to
the basic SQL command set.
Because SQL’s vocabulary is simple, the language is relatively easy to learn. Its simplicity is
enhanced by the fact that much of its work takes place behind the scenes. For example, a
single command creates the complex table structures required to store and manipulate data
successfully. Furthermore, SQL is a nonprocedural language; that is, the user specifies what
must be done, but not how it is to be done. To issue SQL commands, end users and
programmers do not need to know the physical data storage format or the complex
activities that take place when a SQL command is executed.
Although quite useful and powerful, SQL is not meant to stand alone in the applications
arena. Data entry with SQL is possible but awkward, as are data corrections and additions.
SQL itself does not create menus, special report forms, overlays, pop-ups, or any of the
other utilities and screen devices that end users usually expect. Instead, those features are
available as vendor-supplied enhancements. SQL focuses on data definition (creating tables,
indexes, and views) and data manipulation (adding, modifying, deleting, and retrieving data);
we will cover these basic functions in this chapter. In spite of its limitations, SQL is a
powerful tool for extracting information and managing data.
P
review
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
7.1 INTRODUCTION TO SQL
Ideally, a database language allows you to create database and table structures, to perform basic data management
chores (add, delete, and modify), and to perform complex queries designed to transform the raw data into useful
information. Moreover, a database language must perform such basic functions with minimal user effort, and its
command structure and syntax must be easy to learn. Finally, it must be portable; that is, it must conform to some basic
standard so that an individual does not have to relearn the basics when moving from one RDBMS to another. SQL
meets those ideal database language requirements well.
SQL functions fit into two broad categories:
쐌
TABLE
7.1
It is a data definition language (DDL): SQL includes commands to create database objects such as tables,
indexes, and views, as well as commands to define access rights to those database objects. The data definition
commands you will learn in this chapter are listed in Table 7.1.
SQL Data Definition Commands
COMMAND OR OPTION
CREATE SCHEMA AUTHORIZATION
CREATE TABLE
NOT NULL
UNIQUE
PRIMARY KEY
FOREIGN KEY
DEFAULT
CHECK
CREATE INDEX
CREATE VIEW
ALTER TABLE
CREATE TABLE AS
DROP TABLE
DROP INDEX
DROP VIEW
쐌
TABLE
7.2
DESCRIPTION
Creates a database schema
Creates a new table in the user's database schema
Ensures that a column will not have null values
Ensures that a column will not have duplicate values
Defines a primary key for a table
Defines a foreign key for a table
Defines a default value for a column (when no value is given)
Validates data in an attribute
Creates an index for a table
Creates a dynamic subset of rows/columns from one or more
tables
Modifies a tables definition (adds, modifies, or deletes attributes
or constraints)
Creates a new table based on a query in the user's database
schema
Permanently deletes a table (and its data)
Permanently deletes an index
Permanently deletes a view
It is a data manipulation language (DML): SQL includes commands to insert, update, delete, and retrieve
data within the database tables. The data manipulation commands you will learn in this chapter are listed in
Table 7.2.
SQL Data Manipulation Commands
COMMAND OR OPTION
INSERT
SELECT
WHERE
GROUP BY
HAVING
ORDER BY
UPDATE
DESCRIPTION
Inserts row(s) into a table
Selects attributes from rows in one or more tables or views
Restricts the selection of rows based on a conditional expression
Groups the selected rows based on one or more attributes
Restricts the selection of grouped rows based on a condition
Orders the selected rows based on one or more attributes
Modifies an attribute’s values in one or more table’s rows
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TABLE
7.2
7
SQL Data Manipulation Commands (continued)
COMMAND OR OPTION
DELETE
COMMIT
ROLLBACK
COMPARISON OPERATORS
=, <, >, <=, >=, <>
LOGICAL OPERATORS
AND/OR/NOT
SPECIAL OPERATORS
BETWEEN
IS NULL
LIKE
IN
EXISTS
DISTINCT
AGGREGATE FUNCTIONS
COUNT
MIN
MAX
SUM
AVG
DESCRIPTION
Deletes one or more rows from a table
Permanently saves data changes
Restores data to their original values
Used in conditional expressions
Used in conditional expressions
Used in conditional expressions
Checks whether an attribute value is within a range
Checks whether an attribute value is null
Checks whether an attribute value matches a given string pattern
Checks whether an attribute value matches any value within a value list
Checks whether a subquery returns any rows
Limits values to unique values
Used with SELECT to return mathematical summaries on columns
Returns the number of rows with non-null values for a given column
Returns the minimum attribute value found in a given column
Returns the maximum attribute value found in a given column
Returns the sum of all values for a given column
Returns the average of all values for a given column
You will be happy to know that SQL is relatively easy to learn. Its basic command set has a vocabulary of fewer than
100 words. Better yet, SQL is a nonprocedural language: you merely command what is to be done; you don’t have to
worry about how it is to be done. The American National Standards Institute (ANSI) prescribes a standard SQL—the
current fully approved version is SQL-2003. The ANSI SQL standards are also accepted by the International
Organization for Standardization (ISO), a consortium composed of national standards bodies of more than
150 countries. Although adherence to the ANSI/ISO SQL standard is usually required in commercial and government
contract database specifications, many RDBMS vendors add their own special enhancements. Consequently, it is
seldom possible to move a SQL-based application from one RDBMS to another without making some changes.
However, even though there are several different SQL “dialects,” the differences among them are minor. Whether you
use Oracle, Microsoft SQL Server, MySQL, IBM’s DB2, Microsoft Access, or any other well-established RDBMS, a
software manual should be sufficient to get you up to speed if you know the material presented in this chapter.
At the heart of SQL is the query. In Chapter 1, Database Systems, you learned that a query is a spur-of-the-moment
question. Actually, in the SQL environment, the word query covers both questions and actions. Most SQL queries are
used to answer questions such as these: “What products currently held in inventory are priced over $100, and what
is the quantity on hand for each of those products?” “How many employees have been hired since January 1, 2008
by each of the company’s departments?” However, many SQL queries are used to perform actions such as adding or
deleting table rows or changing attribute values within tables. Still other SQL queries create new tables or indexes. In
short, for a DBMS, a query is simply a SQL statement that must be executed. But before you can use SQL to query
a database, you must define the database environment for SQL with its data definition commands.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
7.2 DATA DEFINITION COMMANDS
Before examining the SQL syntax for creating and defining tables and other elements, let’s first examine the simple
database model and the database tables that will form the basis for the many SQL examples you’ll explore in this
chapter.
7.2.1 The Database Model
A simple database composed of the following tables is used to illustrate the SQL commands in this chapter:
CUSTOMER, INVOICE, LINE, PRODUCT, and VENDOR. This database model is shown in Figure 7.1.
FIGURE
The database model
7.1
The database model in Figure 7.1 reflects the following business rules:
쐌
A customer may generate many invoices. Each invoice is generated by one customer.
쐌
An invoice contains one or more invoice lines. Each invoice line is associated with one invoice.
쐌
Each invoice line references one product. A product may be found in many invoice lines. (You can sell more
than one hammer to more than one customer.)
쐌
A vendor may supply many products. Some vendors do not (yet?) supply products. (For example, a vendor list
may include potential vendors.)
쐌
If a product is vendor-supplied, that product is supplied by only a single vendor.
쐌
Some products are not supplied by a vendor. (For example, some products may be produced in-house or
bought on the open market.)
As you can see in Figure 7.1, the database model contains many tables. However, to illustrate the initial set of data
definition commands, the focus of attention will be the PRODUCT and VENDOR tables. You will have the opportunity
to use the remaining tables later in this chapter and in the problem section.
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Online Content
The database model in Figure 7.1 is implemented in the Microsoft Access Ch07_SaleCo database located in
the Premium Website for this book. (This database contains a few additional tables that are not reflected in
Figure 7.1. These tables are used for discussion purposes only.) If you use MS Access, you can use the database
supplied online. However, it is strongly suggested that you create your own database structures so you can
practice the SQL commands illustrated in this chapter.
SQL script files for creating the tables and loading the data in Oracle and MS SQL Server are also located in the
Premium Website. How you connect to your database depends on how the software was installed on your
computer. Follow the instructions provided by your instructor or school.
So that you have a point of reference for understanding the effect of the SQL queries, the contents of the PRODUCT
and VENDOR tables are listed in Figure 7.2.
FIGURE
The VENDOR and PRODUCT tables
7.2
Table name: VENDOR
Database name: Ch07_SaleCo
Table name: PRODUCT
Note the following about these tables. (The features correspond to the business rules reflected in the ERD shown in
Figure 7.1.)
쐌
The VENDOR table contains vendors who are not referenced in the PRODUCT table. Database designers note
that possibility by saying that PRODUCT is optional to VENDOR; a vendor may exist without a reference to
a product. You examined such optional relationships in detail in Chapter 4, Entity Relationship (ER) Modeling.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
쐌
Existing V_CODE values in the PRODUCT table must (and do) have a match in the VENDOR table to ensure
referential integrity.
쐌
A few products are supplied factory-direct, a few are made in-house, and a few may have been bought in a
warehouse sale. In other words, a product is not necessarily supplied by a vendor. Therefore, VENDOR is
optional to PRODUCT.
A few of the conditions just described were made for the sake of illustrating specific SQL features. For example, null
V_CODE values were used in the PRODUCT table to illustrate (later) how you can track such nulls using SQL.
7.2.2 Creating the Database
Before you can use a new RDBMS, you must complete two tasks: first, create the database structure, and second,
create the tables that will hold the end-user data. To complete the first task, the RDBMS creates the physical files that
will hold the database. When you create a new database, the RDBMS automatically creates the data dictionary tables
in which to store the metadata and creates a default database administrator. Creating the physical files that will hold
the database means interacting with the operating system and the file systems supported by the operating system.
Therefore, creating the database structure is the one feature that tends to differ substantially from one RDBMS to
another. The good news is that it is relatively easy to create a database structure, regardless of which RDBMS you use.
If you use Microsoft Access, creating the database is simple: start Access, select File → New → Blank Database, specify
the folder in which you want to store the database, and then name the database. However, if you work in a database
environment typically used by larger organizations, you will probably use an enterprise RDBMS such as Oracle, SQL
Server, MySQL, or DB2. Given their security requirements and greater complexity, those database products require
a more elaborate database creation process. (See Appendix N, Creating a New Database using Oracle 11g, for an
illustration of specific instructions to create a database structure in Oracle.)
You will be relieved to discover that, with the exception of the database creation process, most RDBMS vendors use
SQL that deviates little from the ANSI standard SQL. For example, most RDBMSs require that each SQL command
ends with a semicolon. However, some SQL implementations do not use a semicolon. Important syntax differences
among implementations will be highlighted in the Note boxes.
If you are using an enterprise RDBMS, before you can start creating tables you must be authenticated by the RDBMS.
Authentication is the process through which the DBMS verifies that only registered users may access the database.
To be authenticated, you must log on to the RDBMS using a user ID and a password created by the database
administrator. In an enterprise RDBMS, every user ID is associated with a database schema.
7.2.3 The Database Schema
In the SQL environment, a schema is a group of database objects—such as tables and indexes—that are related to
each other. Usually, the schema belongs to a single user or application. A single database can hold multiple schemas
belonging to different users or applications. Think of a schema as a logical grouping of database objects, such as tables,
indexes, and views. Schemas are useful in that they group tables by owner (or function) and enforce a first level of
security by allowing each user to see only the tables that belong to that user.
ANSI SQL standards define a command to create a database schema:
CREATE SCHEMA AUTHORIZATION {creator};
Therefore, if the creator is JONES, use the command:
CREATE SCHEMA AUTHORIZATION JONES;
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Most enterprise RDBMSs support that command. However, the command is seldom used directly—that is, from the
command line. (When a user is created, the DBMS automatically assigns a schema to that user.) When the DBMS is
used, the CREATE SCHEMA AUTHORIZATION command must be issued by the user who owns the schema. That
is, if you log on as JONES, you can only use CREATE SCHEMA AUTHORIZATION JONES.
For most RDBMSs, the CREATE SCHEMA AUTHORIZATION is optional. That is why this chapter focuses on the
ANSI SQL commands required to create and manipulate tables.
7.2.4 Data Types
In the data dictionary in Table 7.3, note particularly the data types selected. Keep in mind that data-type selection is
usually dictated by the nature of the data and by the intended use. For example:
쐌
P_PRICE clearly requires some kind of numeric data type; defining it as a character field is not acceptable.
쐌
Just as clearly, a vendor name is an obvious candidate for a character data type. For example, VARCHAR2(35)
fits well because vendor names are “variable-length” character strings, and in this case, such strings may be up
to 35 characters long.
쐌
At first glance, it might seem logical to select a numeric data type for V_AREACODE because it contains only
digits. However, adding and subtracting area codes does not yield meaningful results. Therefore, selecting a
character data type is more appropriate. This is true for many common attributes found in business data models.
For example, even though zip codes contain all digits, they must be defined as character data because some zip
codes begin with the digit zero (0), and a numeric data type would cause the leading zero to be dropped.
쐌
U.S. state abbreviations are always two characters, so CHAR(2) is a logical choice.
쐌
Selecting P_INDATE to be a (Julian) DATE field rather than a character field is desirable because the Julian
dates allow you to make simple date comparisons and to perform date arithmetic. For instance, if you have
used DATE fields, you can determine how many days there are between them.
If you use DATE fields, you can also determine what the date will be in say, 60 days from a given P_INDATE by using
P_INDATE + 60. Or you can use the RDBMS’s system date—SYSDATE in Oracle, GETDATE() in MS SQL Server,
and Date() in Access—to determine the answer to questions such as, “What will be the date 60 days from today?” For
example, you might use SYSDATE + 60 (in Oracle), GETDATE() + 60 (in MS SQL Server), or Date() + 60 (in Access).
Date arithmetic capability is particularly useful in billing. Perhaps you want your system to start charging interest on
a customer balance 60 days after the invoice is generated. Such simple date arithmetic would be impossible if you used
a character data type.
Data-type selection sometimes requires professional judgment. For example, you must make a decision about the
V_CODE’s data type as follows:
쐌
If you want the computer to generate new vendor codes by adding 1 to the largest recorded vendor code, you
must classify V_CODE as a numeric attribute. (You cannot perform mathematical procedures on character
data.) The designation INTEGER will ensure that only the counting numbers (integers) can be used. Most SQL
implementations also permit the use of SMALLINT for integer values up to six digits.
쐌
If you do not want to perform mathematical procedures based on V_CODE, you should classify it as a character
attribute, even though it is composed entirely of numbers. Character data are “quicker” to process in queries.
Therefore, when there is no need to perform mathematical procedures on the attribute, store it as a character
attribute.
The first option is used to demonstrate the SQL procedures in this chapter.
V_CODE
V_NAME
V_CONTACT
V_AREACODE
V_PHONE
V_STATE
V_ORDER
VENDOR
INTEGER
CHAR(35)
CHAR(25)
CHAR(3)
CHAR(8)
CHAR(2)
CHAR(1)
CHAR(10)
VARCHAR(35)
DATE
SMALLINT
SMALLINT
NUMBER(8,2)
NUMBER(5,2)
INTEGER
Product code
Product description
Stocking date
Units available
Minimum units
Product price
Discount rate
Vendor code
Vendor code
Vendor name
Contact person
Area code
Phone number
State
Previous order
TYPE
CONTENTS
#####
Xxxxxxxxxxxxxx
Xxxxxxxxxxxxxx
999
999-9999
XX
X
XXXXXXXXXX
Xxxxxxxxxxxx
DD-MON-YYYY
####
####
####.##
0.##
###
FORMAT
1000-9999
NA
NA
NA
NA
NA
Y or N
NA
NA
NA
0-9999
0-9999
0.00-9999.00
0.00-0.20
100-999
RANGE*
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
REQUIRED
PK
FK
PK
OR
FK
PK
VENDOR
FK
REFERENCED
TABLE
T O
FK
= Foreign key
PK
= Primary key
CHAR
= Fixed character length data, 1 to 255 characters
VARCHAR = Variable character length data, 1 to 2,000 characters. VARCHAR is automatically converted to VARCHAR2 in Oracle.
NUMBER = Numeric data. NUMBER(9,2) is used to specify numbers with two decimal places and up to nine digits long, including the decimal places. Some RDBMSs
permit the use of a MONEY or a CURRENCY data type.
INT
= Integer values only
SMALLINT = Small integer values only
DATE formats vary. Commonly accepted formats are: ’DD-MON-YYYY’, ’DD-MON-YY’, ’MM/DD/YYYY’, and ’MM/DD/YY’
* Not all the ranges shown here will be illustrated in this chapter. However, you can use these constraints to practice writing your own constraints.
ATTRIBUTE
NAME
P_CODE
P_DESCRIPT
P_INDATE
P_QOH
P_MIN
P_PRICE
P_DISCOUNT
V_CODE
Data Dictionary for the CH07_SALECO Database
TABLE
NAME
PRODUCT
7.3
TABLE
I N T R O D U C T I O N
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
227
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7
When you define the attribute’s data type, you must pay close attention to the expected use of the attributes for sorting
and data-retrieval purposes. For example, in a real estate application, an attribute that represents the numbers of
bathrooms in a home (H_BATH_NUM) could be assigned the CHAR(3) data type because it is highly unlikely the
application will do any addition, multiplication, or division with the number of bathrooms. Based on the CHAR(3)
data-type definition, valid H_BATH_NUM values would be '2','1','2.5','10'. However, this data-type decision creates
potential problems. For example, if an application sorts the homes by number of bathrooms, a query would “see” the
value '10' as less than '2', which is clearly incorrect. So you must give some thought to the expected use of the data
in order to properly define the attribute data type.
The data dictionary in Table 7.3 contains only a few of the data types supported by SQL. For teaching purposes, the
selection of data types is limited to ensure that almost any RDBMS can be used to implement the examples. If your
RDBMS is fully compliant with ANSI SQL, it will support many more data types than the ones shown in Table 7.4.
And many RDBMSs support data types beyond the ones specified in ANSI SQL.
TABLE
7.4
DATA TYPE
Numeric
Character
Some Common SQL Data Types
FORMAT
NUMBER(L,D)
COMMENTS
The declaration NUMBER(7,2) indicates numbers that will be stored with
two decimal places and may be up to seven digits long, including the sign
and the decimal place. Examples: 12.32, -134.99.
INTEGER
May be abbreviated as INT. Integers are (whole) counting numbers, so they
cannot be used if you want to store numbers that require decimal places.
SMALLINT
Like INTEGER but limited to integer values up to six digits. If your integer
values are relatively small, use SMALLINT instead of INT.
DECIMAL(L,D)
Like the NUMBER specification, but the storage length is a minimum
specification. That is, greater lengths are acceptable, but smaller ones are not.
DECIMAL(9,2), DECIMAL(9), and DECIMAL are all acceptable.
Fixed-length character data for up to 255 characters. If you store strings that
are not as long as the CHAR parameter value, the remaining spaces are left
unused. Therefore, if you specify CHAR(25), strings such as Smith and
Katzenjammer are each stored as 25 characters. However, a U.S. area code
is always three digits long, so CHAR(3) would be appropriate if you wanted
to store such codes.
CHAR(L)
VARCHAR(L) or
VARCHAR2(L)
Date
DATE
Variable-length character data. The designation VARCHAR2(25) will let you
store characters up to 25 characters long. However, VARCHAR will not leave
unused spaces. Oracle automatically converts VARCHAR to VARCHAR2.
Stores dates in the Julian date format.
In addition to the data types shown in Table 7.4, SQL supports several other data types, including TIME, TIMESTAMP,
REAL, DOUBLE, FLOAT, and intervals such as INTERVAL DAY TO HOUR. Many RDBMSs have also expanded the
list to include other types of data, such as LOGICAL, CURRENCY, AutoNumber (Access), and sequence (Oracle).
However, because this chapter is designed to introduce the SQL basics, the discussion is limited to the data types
summarized in Table 7.4.
I N T R O D U C T I O N
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7.2.5 Creating Table Structures
Now you are ready to implement the PRODUCT and VENDOR table structures with the help of SQL, using the
CREATE TABLE syntax shown next.
CREATE TABLE tablename
column1
column2
PRIMARY KEY
FOREIGN KEY
CONSTRAINT
(
data type
data type
(column1
(column1
constraint ] );
[constraint] [,
[constraint] ] [,
[, column2]) ] [,
[, column2]) REFERENCES tablename] [,
Online Content
All the SQL commands you will see in this chapter are located in script files in the Premium Website for this
book. You can copy and paste the SQL commands into your SQL program. Script files are provided for Oracle
and SQL Server users.
To make the SQL code more readable, most SQL programmers use one line per column (attribute) definition. In
addition, spaces are used to line up the attribute characteristics and constraints. Finally, both table and attribute names
are fully capitalized. Those conventions are used in the following examples that create VENDOR and PRODUCT tables
and throughout the book.
Note
SQL SYNTAX
Syntax notation for SQL commands used in this book:
CAPITALS
Required SQL command keywords
italics
An end-user-provided parameter (generally required)
{a | b | ..}
A mandatory parameter; use one option from the list separated by |
[ѧѧ]
An optional parameter—anything inside square brackets is optional
Tablename
The name of a table
Column
The name of an attribute in a table
data type
A valid data-type definition
constraint
A valid constraint definition
condition
A valid conditional expression (evaluates to true or false)
columnlist
One or more column names or expressions separated by commas
tablelist
One or more table names separated by commas
conditionlist
One or more conditional expressions separated by logical operators
expression
A simple value (such as 76 or Married) or a formula (such as P_PRICE − 10)
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CREATE TABLE VENDOR (
V_CODE
INTEGER
V_NAME
VARCHAR(35)
V_CONTACT
VARCHAR(15)
V_AREACODE
CHAR(3)
V_PHONE
CHAR(8)
V_STATE
CHAR(2)
V_ORDER
CHAR(1)
PRIMARY KEY (V_CODE));
NOT
NOT
NOT
NOT
NOT
NOT
NOT
NULL UNIQUE,
NULL,
NULL,
NULL,
NULL,
NULL,
NULL,
Note
• Because the PRODUCT table contains a foreign key that references the VENDOR table, create the
VENDOR table first. (In fact, the M side of a relationship always references the 1 side. Therefore, in a
1:M relationship, you must always create the table for the 1 side first.)
• If your RDBMS does not support the VARCHAR2 and FCHAR format, use CHAR.
• Oracle accepts the VARCHAR data type and automatically converts it to VARCHAR2.
• If your RDBMS does not support SINT or SMALLINT, use INTEGER or INT. If INTEGER is not supported,
use NUMBER.
• If you use Access, you can use the NUMBER data type, but you cannot use the number delimiters at the
SQL level. For example, using NUMBER(8,2) to indicate numbers with up to eight characters and two
decimal places is fine in Oracle, but you cannot use it in Access—you must use NUMBER without the
delimiters.
• If your RDBMS does not support primary and foreign key designations or the UNIQUE specification,
delete them from the SQL code shown here.
• If you use the PRIMARY KEY designation in Oracle, you do not need the NOT NULL and UNIQUE
specifications.
• The ON UPDATE CASCADE clause is part of the ANSI standard, but it may not be supported by your
RDBMS. In that case, delete the ON UPDATE CASCADE clause.
CREATE TABLE PRODUCT (
P_CODE
VARCHAR(10)
P_DESCRIPT
VARCHAR(35)
P_INDATE
DATE
P_QOH
SMALLINT
P_MIN
SMALLINT
P_PRICE
NUMBER(8,2)
P_DISCOUNT
NUMBER(5,2)
V_CODE
INTEGER,
PRIMARY KEY (P_CODE),
FOREIGN KEY (V_CODE) REFERENCES VENDOR
NOT
NOT
NOT
NOT
NOT
NOT
NOT
NULL
NULL,
NULL,
NULL,
NULL,
NULL,
NULL,
UNIQUE,
ON UPDATE CASCADE);
As you examine the preceding SQL table-creating command sequences, note the following features:
쐌
The NOT NULL specifications for the attributes ensure that a data entry will be made. When it is crucial to have
the data available, the NOT NULL specification will not allow the end user to leave the attribute empty (with
no data entry at all). Because this specification is made at the table level and stored in the data dictionary,
application programs can use this information to create the data dictionary validation automatically.
쐌
The UNIQUE specification creates a unique index in the respective attribute. Use it to avoid having duplicated
values in a column.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
쐌
The primary key attributes contain both a NOT NULL and a UNIQUE specification. Those specifications
enforce the entity integrity requirements. If the NOT NULL and UNIQUE specifications are not supported, use
PRIMARY KEY without the specifications. (For example, if you designate the PK in MS Access, the NOT
NULL and UNIQUE specifications are automatically assumed and are not spelled out.)
쐌
The entire table definition is enclosed in parentheses. A comma is used to separate each table element
(attributes, primary key, and foreign key) definition.
Note
If you are working with a composite primary key, all of the primary key’s attributes are contained within the
parentheses and are separated with commas. For example, the LINE table in Figure 7.1 has a primary key that
consists of the two attributes INV_NUMBER and LINE_NUMBER. Therefore, you would define the primary key
by typing:
PRIMARY KEY (INV_NUMBER, LINE_NUMBER),
The order of the primary key components is important because the indexing starts with the first-mentioned
attribute, then proceeds with the next attribute, and so on. In this example, the line numbers would be ordered
within each of the invoice numbers:
INV_NUMBER
LINE_NUMBER
1001
1001
1002
1003
1003
1
2
1
1
2
쐌
The ON UPDATE CASCADE specification ensures that if you make a change in any VENDOR’s V_CODE,
that change is automatically applied to all foreign key references throughout the system (cascade) to ensure that
referential integrity is maintained. (Although the ON UPDATE CASCADE clause is part of the ANSI standard,
some RDBMSs, such as Oracle, do not support ON UPDATE CASCADE. If your RDBMS does not support
the clause, delete it from the code shown here.)
쐌
An RDBMS will automatically enforce referential integrity for foreign keys. That is, you cannot have an invalid
entry in the foreign key column; at the same time, you cannot delete a vendor row as long as a product row
references that vendor.
쐌
The command sequence ends with a semicolon. (Remember, your RDBMS may require that you omit the
semicolon.)
Note
NOTE ABOUT COLUMN NAMES
Do not use mathematical symbols such as +, −, and / in your column names; instead, use an underscore to
separate words, if necessary. For example, PER-NUM might generate an error message, but PER_NUM is
acceptable. Also, do not use reserved words. Reserved words are words used by SQL to perform specific functions.
For example, in some RDBMSs, the column name INITIAL will generate the message invalid column name.
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Note
NOTE TO ORACLE USERS
When you press the Enter key after typing each line, a line number is automatically generated as long as you do
not type a semicolon before pressing the Enter key. For example, Oracle's execution of the CREATE TABLE
command will look like this:
CREATE TABLE PRODUCT (
2
3
4
5
6
7
8
9
10
11
12
13
P_CODE
CONSTRAINT
P_DESCRIPT
P_INDATE
P_QOH
P_MIN
P_PRICE
P_DISCOUNT
V_CODE
CONSTRAINT
FOREIGN KEY
;
VARCHAR2(10)
PRODUCT_P_CODE_PK PRIMARY KEY,
VARCHAR2(35)
NOT NULL,
DATE
NOT NULL,
NUMBER
NOT NULL,
NUMBER
NOT NULL,
NUMBER(8,2)
NOT NULL,
NUMBER(5,2)
NOT NULL,
NUMBER,
PRODUCT_V_CODE_FK
V_CODE REFERENCES VENDOR)
In the preceding SQL command sequence, note the following:
• The attribute definition for P_CODE starts in line 2 and ends with a comma at the end of line 3.
• The CONSTRAINT clause (line 3) allows you to define and name a constraint in Oracle. You can name the
constraint to meet your own naming conventions. In this case, the constraint was named PRODUCT_P_
CODE_PK.
• Examples of constraints are NOT NULL, UNIQUE, PRIMARY KEY, FOREIGN KEY, and CHECK. For
additional details about constraints, see below.
• To define a PRIMARY KEY constraint, you could also use the following syntax: P_CODE VARCHAR2(10)
PRIMARY KEY,.
• In this case, Oracle would automatically name the constraint.
• Lines 11 and 12 define a FOREIGN KEY constraint name PRODUCT_V_CODE_FK for the attribute
V_CODE. The CONSTRAINT clause is generally used at the end of the CREATE TABLE command
sequence.
• If you do not name the constraints yourself, Oracle will automatically assign a name. Unfortunately, the
Oracle-assigned name makes sense only to Oracle, so you will have a difficult time deciphering it later. You
should assign a name that makes sense to human beings!
7.2.6 SQL Constraints
In Chapter 3, The Relational Database Model, you learned that adherence to rules on entity integrity and referential
integrity is crucial in a relational database environment. Fortunately, most SQL implementations support both integrity
rules. Entity integrity is enforced automatically when the primary key is specified in the CREATE TABLE command
sequence. For example, you can create the VENDOR table structure and set the stage for the enforcement of entity
integrity rules by using:
PRIMARY KEY (V_CODE)
In the PRODUCT table’s CREATE TABLE sequence, note that referential integrity has been enforced by specifying in
the PRODUCT table:
FOREIGN KEY (V_CODE) REFERENCES VENDOR ON UPDATE CASCADE
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
That foreign key constraint definition ensures that:
쐌
You cannot delete a vendor from the VENDOR table if at least one product row references that vendor. This
is the default behavior for the treatment of foreign keys.
쐌
On the other hand, if a change is made in an existing VENDOR table’s V_CODE, that change must be reflected
automatically in any PRODUCT table V_CODE reference (ON UPDATE CASCADE). That restriction makes
it impossible for a V_CODE value to exist in the PRODUCT table pointing to a nonexistent VENDOR table
V_CODE value. In other words, the ON UPDATE CASCADE specification ensures the preservation of
referential integrity. (Oracle does not support ON UPDATE CASCADE.)
In general, ANSI SQL permits the use of ON DELETE and ON UPDATE clauses to cover CASCADE, SET NULL, or
SET DEFAULT.
Online Content
For a more detailed discussion of the options for the ON DELETE and ON UPDATE clauses, see Appendix D,
Converting an ER Model into a Database Structure, Section D.2, General Rules Governing Relationships
Among Tables. Appendix D is in the Premium Website.
Note
NOTE ABOUT REFERENTIAL CONSTRAINT ACTIONS
The support for the referential constraints actions varies from product to product. For example:
• MS Access, SQL Server, and Oracle support ON DELETE CASCADE.
• MS Access and SQL Server support ON UPDATE CASCADE.
• Oracle does not support ON UPDATE CASCADE.
• Oracle supports SET NULL.
• MS Access and SQL Server do not support SET NULL.
• Refer to your product manuals for additional information on referential constraints.
While MS Access does not support ON DELETE CASCADE or ON UPDATE CASCADE at the SQL
command-line level, it does support them through the relationship window interface. In fact, whenever you try
to establish a relationship between two tables in Access, the relationship window interface will automatically
pop up.
Besides the PRIMARY KEY and FOREIGN KEY constraints, the ANSI SQL standard also defines the following
constraints:
쐌
The NOT NULL constraint ensures that a column does not accept nulls.
쐌
The UNIQUE constraint ensures that all values in a column are unique.
쐌
The DEFAULT constraint assigns a value to an attribute when a new row is added to a table. The end user may,
of course, enter a value other than the default value.
쐌
The CHECK constraint is used to validate data when an attribute value is entered. The CHECK constraint does
precisely what its name suggests: it checks to see that a specified condition exists. Examples of such constraints
include the following:
-
The minimum order value must be at least 10.
-
The date must be after April 15, 2010.
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If the CHECK constraint is met for the specified attribute (that is, the condition is true), the data are accepted for that
attribute. If the condition is found to be false, an error message is generated and the data are not accepted.
Note that the CREATE TABLE command lets you define constraints in two different places:
쐌
When you create the column definition (known as a column constraint).
쐌
When you use the CONSTRAINT keyword (known as a table constraint).
A column constraint applies to just one column; a table constraint may apply to many columns. Those constraints are
supported at varying levels of compliance by enterprise RDBMSs.
In this chapter, Oracle is used to illustrate SQL constraints. For example, note that the following SQL command
sequence uses the DEFAULT and CHECK constraints to define the table named CUSTOMER.
CREATE TABLE CUSTOMER (
CUS_CODE
NUMBER
CUS_LNAME
VARCHAR(15)
CUS_FNAME
VARCHAR(15)
CUS_INITIAL
CHAR(1),
CUS_AREACODE
CHAR(3)
PRIMARY KEY,
NOT NULL,
NOT NULL,
DEFAULT '615'
NOT NULL
CHECK(CUS_AREACODE IN ('615','713','931')),
CUS_PHONE
CHAR(8)
NOT NULL,
CUS_BALANCE
NUMBER(9,2)
DEFAULT 0.00,
CONSTRAINT CUS_UI1 UNIQUE (CUS_LNAME, CUS_FNAME));
In this case, the CUS_AREACODE attribute is assigned a default value of '615'. Therefore, if a new CUSTOMER table row
is added and the end user makes no entry for the area code, the '615' value will be recorded. Also note that the CHECK
condition restricts the values for the customer’s area code to 615, 713, and 931; any other values will be rejected.
It is important to note that the DEFAULT value applies only when new rows are added to a table and then only when
no value is entered for the customer’s area code. (The default value is not used when the table is modified.) In contrast,
the CHECK condition is validated whether a customer row is added or modified. However, while the CHECK
condition may include any valid expression, it applies only to the attributes in the table being checked. If you want to
check for conditions that include attributes in other tables, you must use triggers. (See Chapter 8, Advanced SQL.)
Finally, the last line of the CREATE TABLE command sequence creates a unique index constraint (named CUS_UI1)
on the customer’s last name and first name. The index will prevent the entry of two customers with the same last name
and first name. (This index merely illustrates the process. Clearly, it should be possible to have more than one person
named John Smith in the CUSTOMER table.)
Note
NOTE TO MS ACCESS USERS
MS Access does not accept the DEFAULT or CHECK constraints. However, MS Access will accept the
CONSTRAINT CUS_UI1 UNIQUE (CUS_LNAME, CUS_FNAME) line and create the unique index.
In the following SQL command to create the INVOICE table, the DEFAULT constraint assigns a default date to a new
invoice, and the CHECK constraint validates that the invoice date is greater than January 1, 2010.
CREATE TABLE INVOICE (
INV_NUMBER
NUMBER
PRIMARY KEY,
CUS_CODE
NUMBER
NOT NULL REFERENCES CUSTOMER(CUS_CODE),
INV_DATE
DATE
DEFAULT SYSDATE NOT NULL,
CONSTRAINT INV_CK1 CHECK (INV_DATE > TO_DATE('01-JAN-2010','DD-MON-YYYY')));
I N T R O D U C T I O N
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Q U E R Y
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( S Q L )
In this case, notice the following:
쐌
The CUS_CODE attribute definition contains REFERENCES CUSTOMER (CUS_CODE) to indicate that the
CUS_CODE is a foreign key. This is another way to define a foreign key.
쐌
The DEFAULT constraint uses the SYSDATE special function. This function always returns today’s date.
쐌
The invoice date (INV_DATE) attribute is automatically given today’s date (returned by SYSDATE) when a new
row is added and no value is given for the attribute.
쐌
A CHECK constraint is used to validate that the invoice date is greater than 'January 1, 2010'. When
comparing a date to a manually entered date in a CHECK clause, Oracle requires the use of the TO_DATE
function. The TO_DATE function takes two parameters: the literal date and the date format used.
The final SQL command sequence creates the LINE table. The LINE table has a composite primary key (INV_
NUMBER, LINE_NUMBER) and uses a UNIQUE constraint in INV_NUMBER and P_CODE to ensure that the same
product is not ordered twice in the same invoice.
CREATE TABLE LINE (
INV_NUMBER
NUMBER
NOT NULL,
LINE_NUMBER
NUMBER(2,0)
NOT NULL,
P_CODE
VARCHAR(10)
NOT NULL,
LINE_UNITS
NUMBER(9,2)
DEFAULT 0.00
NOT NULL,
LINE_PRICE
NUMBER(9,2)
DEFAULT 0.00
NOT NULL,
PRIMARY KEY (INV_NUMBER, LINE_NUMBER),
FOREIGN KEY (INV_NUMBER) REFERENCES INVOICE ON DELETE CASCADE,
FOREIGN KEY (P_CODE) REFERENCES PRODUCT(P_CODE),
CONSTRAINT LINE_UI1 UNIQUE(INV_NUMBER, P_CODE));
In the creation of the LINE table, note that a UNIQUE constraint is added to prevent the duplication of an invoice line.
A UNIQUE constraint is enforced through the creation of a unique index. Also note that the ON DELETE CASCADE
foreign key action enforces referential integrity. The use of ON DELETE CASCADE is recommended for weak entities
to ensure that the deletion of a row in the strong entity automatically triggers the deletion of the corresponding rows
in the dependent weak entity. In that case, the deletion of an INVOICE row will automatically delete all of the LINE
rows related to the invoice. In the following section, you will learn more about indexes and how to use SQL commands
to create them.
7.2.7 SQL Indexes
You learned in Chapter 3 that indexes can be used to improve the efficiency of searches and to avoid duplicate column
values. In the previous section, you saw how to declare unique indexes on selected attributes when the table is created.
In fact, when you declare a primary key, the DBMS automatically creates a unique index. Even with this feature, you
often need additional indexes. The ability to create indexes quickly and efficiently is important. Using the CREATE
INDEX command, SQL indexes can be created on the basis of any selected attribute. The syntax is:
CREATE [UNIQUE] INDEX indexname ON tablename(column1 [, column2])
For example, based on the attribute P_INDATE stored in the PRODUCT table, the following command creates an
index named P_INDATEX:
CREATE INDEX P_INDATEX ON PRODUCT(P_INDATE);
SQL does not let you write over an existing index without warning you first, thus preserving the index structure within
the data dictionary. Using the UNIQUE index qualifier, you can even create an index that prevents you from using a
value that has been used before. Such a feature is especially useful when the index attribute is a candidate key whose
values must not be duplicated:
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CREATE UNIQUE INDEX P_CODEX ON PRODUCT(P_CODE);
If you now try to enter a duplicate P_CODE value, SQL produces the error message “duplicate value in index.” Many
RDBMSs, including Access, automatically create a unique index on the PK attribute(s) when you declare the PK.
A common practice is to create an index on any field that is used as a search key, in comparison operations in a
conditional expression, or when you want to list rows in a specific order. For example, if you want to create a report
of all products by vendor, it would be useful to create an index on the V_CODE attribute in the PRODUCT table.
Remember that a vendor can supply many products. Therefore, you should not create a UNIQUE index in this case.
Better yet, to make the search as efficient as possible, using a composite index is recommended.
Unique composite indexes are often used to prevent data duplication. For example, consider the case illustrated in
Table 7.5, in which required employee test scores are stored. (An employee can take a test only once on a given date.)
Given the structure of Table 7.5, the PK is EMP_NUM + TEST_NUM. The third test entry for employee 111 meets
entity integrity requirements—the combination 111,3 is unique—yet the WEA test entry is clearly duplicated.
TABLE
7.5
A Duplicated Test Record
EMP_NUM
110
110
111
111
111
112
TEST_NUM
1
2
1
2
3
1
TEST_CODE
WEA
WEA
HAZ
WEA
WEA
CHEM
TEST_DATE
15-Jan-2010
12-Jan-2010
14-Dec-2009
18-Feb-2010
18-Feb-2010
17-Aug-2009
TEST_SCORE
93
87
91
95
95
91
Such duplication could have been avoided through the use of a unique composite index, using the attributes
EMP_NUM, TEST_CODE, and TEST_DATE:
CREATE UNIQUE INDEX EMP_TESTDEX ON TEST(EMP_NUM, TEST_CODE, TEST_DATE);
By default, all indexes produce results that are listed in ascending order, but you can create an index that yields output
in descending order. For example, if you routinely print a report that lists all products ordered by price from highest
to lowest, you could create an index named PROD_PRICEX by typing:
CREATE INDEX PROD_PRICEX ON PRODUCT(P_PRICE DESC);
To delete an index, use the DROP INDEX command:
DROP INDEX indexname
For example, if you want to eliminate the PROD_PRICEX index, type:
DROP INDEX PROD_PRICEX;
After creating the tables and some indexes, you are ready to start entering data. The following sections use two tables
(VENDOR and PRODUCT) to demonstrate most of the data manipulation commands.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
7.3 DATA MANIPULATION COMMANDS
In this section, you will learn how to use the basic SQL data manipulation commands INSERT, SELECT, COMMIT,
UPDATE, ROLLBACK, and DELETE.
7.3.1 Adding Table Rows
SQL requires the use of the INSERT command to enter data into a table. The INSERT command’s basic syntax looks
like this:
INSERT INTO tablename VALUES (value1, value2, ... , valuen)
Because the PRODUCT table uses its V_CODE to reference the VENDOR table’s V_CODE, an integrity violation will
occur if those VENDOR table V_CODE values don’t yet exist. Therefore, you need to enter the VENDOR rows before
the PRODUCT rows. Given the VENDOR table structure defined earlier and the sample VENDOR data shown in
Figure 7.2, you would enter the first two data rows as follows:
INSERT INTO VENDOR
VALUES (21225,'Bryson, Inc.','Smithson','615','223-3234','TN','Y');
INSERT INTO VENDOR
VALUES (21226,'Superloo, Inc.','Flushing','904','215-8995','FL','N');
and so on, until all of the VENDOR table records have been entered.
(To see the contents of the VENDOR table, use the SELECT * FROM VENDOR; command.)
The PRODUCT table rows would be entered in the same fashion, using the PRODUCT data shown in Figure 7.2. For
example, the first two data rows would be entered as follows, pressing the Enter key at the end of each line:
INSERT INTO PRODUCT
VALUES ('11QER/31','Power painter, 15 psi., 3-nozzle','03-Nov-09',8,5,109.99,0.00,25595);
INSERT INTO PRODUCT
VALUES ('13-Q2/P2','7.25-in. pwr. saw blade','13-Dec-09',32,15,14.99, 0.05, 21344);
(To see the contents of the PRODUCT table, use the SELECT * FROM PRODUCT; command.)
Note
Date entry is a function of the date format expected by the DBMS. For example, March 25, 2010 might be
shown as 25-Mar-2010 in Access and Oracle, or it might be displayed in other presentation formats in another
RDBMS. MS Access requires the use of # delimiters when performing any computations or comparisons based
on date attributes, as in P_INDATE >= #25-Mar-10#.
In the preceding data entry lines, observe that:
쐌
The row contents are entered between parentheses. Note that the first character after VALUES is a parenthesis
and that the last character in the command sequence is also a parenthesis.
쐌
Character (string) and date values must be entered between apostrophes (').
쐌
Numerical entries are not enclosed in apostrophes.
쐌
Attribute entries are separated by commas.
쐌
A value is required for each column in the table.
This version of the INSERT commands adds one table row at a time.
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Inserting Rows with Null Attributes
Thus far, you have entered rows in which all of the attribute values are specified. But what do you do if a product does
not have a vendor or if you don’t yet know the vendor code? In those cases, you would want to leave the vendor code
null. To enter a null, use the following syntax:
INSERT INTO PRODUCT
VALUES ('BRT-345','Titanium drill bit','18-Oct-09', 75, 10, 4.50, 0.06, NULL);
Incidentally, note that the NULL entry is accepted only because the V_CODE attribute is optional—the NOT NULL
declaration was not used in the CREATE TABLE statement for this attribute.
Inserting Rows with Optional Attributes
There might be occasions when more than one attribute is optional. Rather than declaring each attribute as NULL in
the INSERT command, you can indicate just the attributes that have required values. You do that by listing the attribute
names inside parentheses after the table name. For the purpose of this example, assume that the only required
attributes for the PRODUCT table are P_CODE and P_DESCRIPT:
INSERT INTO PRODUCT(P_CODE, P_DESCRIPT) VALUES ('BRT-345','Titanium drill bit');
7.3.2 Saving Table Changes
Any changes made to the table contents are not saved on disk until you close the database, close the program you are
using, or use the COMMIT command. If the database is open and a power outage or some other interruption occurs
before you issue the COMMIT command, your changes will be lost and only the original table contents will be retained.
The syntax for the COMMIT command is:
COMMIT [WORK]
The COMMIT command permanently saves all changes—such as rows added, attributes modified, and rows
deleted—made to any table in the database. Therefore, if you intend to make your changes to the PRODUCT table
permanent, it is a good idea to save those changes by using:
COMMIT;
Note
NOTE TO MS ACCESS USERS
MS Access doesn't support the COMMIT command because it automatically saves changes after the execution
of each SQL command.
However, the COMMIT command’s purpose is not just to save changes. In fact, the ultimate purpose of the COMMIT
and ROLLBACK commands (see Section 7.3.5) is to ensure database update integrity in transaction management.
(You will see how such issues are addressed in Chapter 10, Transaction Management and Concurrency Control.)
7.3.3 Listing Table Rows
The SELECT command is used to list the contents of a table. The syntax of the SELECT command is as follows:
SELECT columnlist
FROM
tablename
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
The columnlist represents one or more attributes, separated by commas. You could use the * (asterisk) as a wildcard
character to list all attributes. A wildcard character is a symbol that can be used as a general substitute for other
characters or commands. For example, to list all attributes and all rows of the PRODUCT table, use:
SELECT * FROM PRODUCT;
Figure 7.3 shows the output generated by that command. (Figure 7.3 shows all of the rows in the PRODUCT table
that serve as the basis for subsequent discussions. If you entered only the PRODUCT table’s first two records, as shown
in the preceding section, the output of the preceding SELECT command would show only the rows you entered. Don’t
worry about the difference between your SELECT output and the output shown in Figure 7.3. When you complete the
work in this section, you will have created and populated your VENDOR and PRODUCT tables with the correct rows
for use in future sections.)
FIGURE
The contents of the PRODUCT table
7.3
Note
Your listing may not be in the order shown in Figure 7.3. The listings shown in the figure are the result of
system-controlled primary-key-based index operations. You will learn later how to control the output so that it
conforms to the order you have specified.
Note
NOTE TO ORACLE USERS
Some SQL implementations (such as Oracle's) cut the attribute labels to fit the width of the column. However,
Oracle lets you set the width of the display column to show the complete attribute name. You can also change
the display format, regardless of how the data are stored in the table. For example, if you want to display dollar
symbols and commas in the P_PRICE output, you can declare:
COLUMN P_PRICE FORMAT $99,999.99
to change the output 12347.67 to $12,347.67.
In the same manner, to display only the first 12 characters of the P_DESCRIPT attribute, use:
COLUMN P_DESCRIPT FORMAT A12 TRUNCATE
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Although SQL commands can be grouped together on a single line, complex command sequences are best shown on
separate lines, with space between the SQL command and the command’s components. Using that formatting
convention makes it much easier to see the components of the SQL statements, making it easy to trace the SQL logic,
and if necessary, to make corrections. The number of spaces used in the indention is up to you. For example, note
the following format for a more complex statement:
SELECT
FROM
P_CODE, P_DESCRIPT, P_INDATE, P_QOH, P_MIN, P_PRICE, P_DISCOUNT, V_CODE
PRODUCT;
When you run a SELECT command on a table, the RDBMS returns a set of one or more rows that have the same
characteristics as a relational table. In addition, the SELECT command lists all rows from the table you specified in the
FROM clause. This is a very important characteristic of SQL commands. By default, most SQL data manipulation
commands operate over an entire table (or relation). That is why SQL commands are said to be set-oriented
commands. A SQL set-oriented command works over a set of rows. The set may include one or more columns and
zero or more rows from one or more tables.
7.3.4 Updating Table Rows
Use the UPDATE command to modify data in a table. The syntax for this command is:
UPDATE
SET
[WHERE
tablename
columnname = expression [, columnname = expression]
conditionlist ];
For example, if you want to change P_INDATE from December 13, 2009, to January 18, 2010, in the second row of
the PRODUCT table (see Figure 7.3), use the primary key (13-Q2/P2) to locate the correct (second) row. Therefore, type:
UPDATE
SET
WHERE
PRODUCT
P_INDATE = '18-JAN-2010'
P_CODE = '13-Q2/P2';
If more than one attribute is to be updated in the row, separate the corrections with commas:
UPDATE
SET
WHERE
PRODUCT
P_INDATE = '18-JAN-2010', P_PRICE = 17.99, P_MIN = 10
P_CODE = '13-Q2/P2';
What would have happened if the previous UPDATE command had not included the WHERE condition? The
P_INDATE, P_PRICE, and P_MIN values would have been changed in all rows of the PRODUCT table. Remember,
the UPDATE command is a set-oriented operator. Therefore, if you don’t specify a WHERE condition, the UPDATE
command will apply the changes to all rows in the specified table.
Confirm the correction(s) by using this SELECT command to check the PRODUCT table’s listing:
SELECT * FROM PRODUCT;
7.3.5 Restoring Table Contents
If you have not yet used the COMMIT command to store the changes permanently in the database, you can restore
the database to its previous condition with the ROLLBACK command. ROLLBACK undoes any changes since the
last COMMIT command and brings the data back to the values that existed before the changes were made. To restore
the data to their “prechange” condition, type:
ROLLBACK;
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
and then press the Enter key. Use the SELECT statement again to see that the ROLLBACK did, in fact, restore the
data to their original values.
COMMIT and ROLLBACK work only with data manipulation commands that are used to add, modify, or delete table
rows. For example, assume that you perform these actions:
1.
CREATE a table called SALES.
2.
INSERT 10 rows in the SALES table.
3.
UPDATE two rows in the SALES table.
4.
Execute the ROLLBACK command.
Will the SALES table be removed by the ROLLBACK command? No, the ROLLBACK command will undo only the
results of the INSERT and UPDATE commands. All data definition commands (CREATE TABLE) are automatically
committed to the data dictionary and cannot be rolled back. The COMMIT and ROLLBACK commands are examined
in greater detail in Chapter 10.
Note
NOTE TO MS ACCESS USERS
MS Access does not support the ROLLBACK command.
Some RDBMSs, such as Oracle, automatically COMMIT data changes when issuing data definition commands. For
example, if you had used the CREATE INDEX command after updating the two rows in the previous example, all
previous changes would have been committed automatically; doing a ROLLBACK afterward wouldn’t have undone
anything. Check your RDBMS manual to understand these subtle differences.
7.3.6 Deleting Table Rows
It is easy to delete a table row using the DELETE statement; the syntax is:
DELETE FROM
[WHERE
tablename
conditionlist ];
For example, if you want to delete from the PRODUCT table the product that you added earlier whose code (P_CODE)
is 'BRT-345', use:
DELETE FROM
WHERE
PRODUCT
P_CODE = 'BRT-345';
In that example, the primary key value lets SQL find the exact record to be deleted. However, deletions are not limited
to a primary key match; any attribute may be used. For example, in your PRODUCT table, you will see that there are
several products for which the P_MIN attribute is equal to 5. Use the following command to delete all rows from the
PRODUCT table for which the P_MIN is equal to 5:
DELETE FROM
WHERE
PRODUCT
P_MIN = 5;
Check the PRODUCT table’s contents again to verify that all products with P_MIN equal to 5 have been deleted.
Finally, remember that DELETE is a set-oriented command. And keep in mind that the WHERE condition is optional.
Therefore, if you do not specify a WHERE condition, all rows from the specified table will be deleted!
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7.3.7 Inserting Table Rows with a Select Subquery
You learned in Section 7.3.1 how to use the INSERT statement to add rows to a table. In that section, you added rows
one at a time. In this section, you will learn how to add multiple rows to a table, using another table as the source of
the data. The syntax for the INSERT statement is:
INSERT INTO tablename
SELECT columnlist
FROM tablename;
In that case, the INSERT statement uses a SELECT subquery. A subquery, also known as a nested query or an
inner query, is a query that is embedded (or nested) inside another query. The inner query is always executed first by
the RDBMS. Given the previous SQL statement, the INSERT portion represents the outer query, and the SELECT
portion represents the subquery. You can nest queries (place queries inside queries) many levels deep; in every case,
the output of the inner query is used as the input for the outer (higher-level) query. In Chapter 8 you will learn more
about the various types of subqueries.
The values returned by the SELECT subquery should match the attributes and data types of the table in the INSERT
statement. If the table into which you are inserting rows has one date attribute, one number attribute, and one
character attribute, the SELECT subquery should return one or more rows in which the first column has date values,
the second column has number values, and the third column has character values.
Online Content
Before you execute the commands in the following sections, you MUST do the following:
• If you are using Oracle, run the sqlintrodbinit.sql script file in the Premium Website to create all tables
and load the data in the database.
• If you are using Access, copy the original Ch07_SaleCo.mbd file from the Premium Website.
7.4 SELECT QUERIES
In this section, you will learn how to fine-tune the SELECT command by adding restrictions to the search criteria. SELECT,
coupled with appropriate search conditions, is an incredibly powerful tool that enables you to transform data into
information. For example, in the following sections, you will learn how to create queries that can be used to answer
questions such as these: “What products were supplied by a particular vendor?” “Which products are priced below $10?”
“How many products supplied by a given vendor were sold between January 5, 2010 and March 20, 2010?”
7.4.1 Selecting Rows with Conditional Restrictions
You can select partial table contents by placing restrictions on the rows to be included in the output. This is done by
using the WHERE clause to add conditional restrictions to the SELECT statement. The following syntax enables you
to specify which rows to select:
SELECT
FROM
[WHERE
columnlist
tablelist
conditionlist ];
The SELECT statement retrieves all rows that match the specified condition(s)—also known as the conditional
criteria—you specified in the WHERE clause. The conditionlist in the WHERE clause of the SELECT statement is
represented by one or more conditional expressions, separated by logical operators. The WHERE clause is optional.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
If no rows match the specified criteria in the WHERE clause, you see a blank screen or a message that tells you that
no rows were retrieved. For example, the query:
SELECT
FROM
WHERE
P_DESCRIPT, P_INDATE, P_PRICE, V_CODE
PRODUCT
V_CODE = 21344;
returns the description, date, and price of products with a vendor code of 21344, as shown in Figure 7.4.
FIGURE
MS Access users can use the Access QBE (query by example)
query generator. Although the Access QBE generates its
own “native” version of SQL, you can also elect to type
standard SQL in the Access SQL window, as shown at the
bottom of Figure 7.5. Figure 7.5 shows the Access QBE
screen, the SQL window’s QBE-generated SQL, and the
listing of the modified SQL.
7.4
Selected PRODUCT table
attributes for VENDOR
code 21344
FIGURE
The Microsoft Access QBE and its SQL
7.5
Query options
Microsoft Access-generated SQL
User-entered SQL
Numerous conditional restrictions can be placed on the selected table contents. For example, the comparison
operators shown in Table 7.6 can be used to restrict output.
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7
Note
NOTE TO MS ACCESS USERS
The MS Access QBE interface automatically designates the data source by using the table name as a prefix. You
will discover later that the table name prefix is used to avoid ambiguity when the same column name appears
in multiple tables. For example, both the VENDOR and the PRODUCT tables contain the V_CODE attribute.
Therefore, if both tables are used (as they would be in a join), the source of the V_CODE attribute must be
specified.
TABLE
7.6
Comparison Operators
SYMBOL
=
<
<=
>
>=
<> or !=
MEANING
Equal to
Less than
Less than or equal to
Greater than
Greater than or equal to
Not equal to
The following example uses the “not equal to” operator:
SELECT
FROM
WHERE
P_DESCRIPT, P_INDATE, P_PRICE, V_CODE
PRODUCT
V_CODE <> 21344;
The output, shown in Figure 7.6, lists all of the rows for which the vendor code is not 21344.
Note that, in Figure 7.6, rows with nulls in the V_CODE column (see Figure 7.3) are not included in the SELECT
command’s output.
FIGURE
7.6
Selected PRODUCT table
attributes for VENDOR codes
other than 21344
The command sequence:
SELECT
FROM
WHERE
P_DESCRIPT, P_QOH, P_MIN, P_PRICE
PRODUCT
P_PRICE <= 10;
yields the output shown in Figure 7.7.
Using Comparison Operators on Character
Attributes
FIGURE
7.7
Selected PRODUCT table
attributes with a P_PRICE
restriction
Because computers identify all characters by their (numeric)
American Standard Code for Information Interchange
(ASCII) codes, comparison operators may even be used to
place restrictions on character-based attributes. Therefore,
the command:
SELECT
FROM
WHERE
P_CODE, P_DESCRIPT, P_QOH, P_MIN,
P_PRICE
PRODUCT
P_CODE < '1558-QW1';
would be correct and would yield a list of all rows in which the
P_CODE is alphabetically less than 1558-QW1. (Because the
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
ASCII code value for the letter B is greater than the value of the letter A, it follows that A is less than B.) Therefore,
the output will be generated as shown in Figure 7.8.
String (character) comparisons are made from left to right.
This left-to-right comparison is especially useful when
7.8
attributes such as names are to be compared. For example,
the string “Ardmore” would be judged greater than the
string “Aarenson” but less than the string “Brown”; such
results may be used to generate alphabetical listings like
those found in a phone directory. If the characters 0−9 are
stored as strings, the same left-to-right string comparisons
can lead to apparent anomalies. For example, the ASCII
code for the character “5” is, as expected, greater than the
ASCII code for the character “4.” Yet the same “5” will also be judged greater than the string “44” because the first
character in the string “44” is less than the string “5.” For that reason, you may get some unexpected results from
comparisons when dates or other numbers are stored in character format. This also applies to date comparisons. For
example, the left-to-right ASCII character comparison would force the conclusion that the date “01/01/2010”
occurred before “12/31/2009.” Because the leftmost character “0” in “01/01/2010” is less than the leftmost
character “1” in “12/31/2009,” “01/01/2010” is less than “12/31/2009.” Naturally, if date strings are stored in
a yyyy/mm/dd format, the comparisons will yield appropriate results, but this is a nonstandard date presentation.
That’s why all current RDBMSs support “date” data types; you should use them. In addition, using “date” data types
gives you the benefit of date arithmetic.
FIGURE
Selected PRODUCT table
attributes: the ASCII code
effect
Using Comparison Operators on Dates
Date procedures are often more software-specific than other SQL procedures. For example, the query to list all of the
rows in which the inventory stock dates occur on or after January 20, 2010 will look like this:
SELECT
FROM
WHERE
P_DESCRIPT, P_QOH, P_MIN, P_PRICE, P_INDATE
PRODUCT
P_INDATE >= '20-Jan-2010';
(Remember that MS Access users must use the # delimiters for dates. For example, you would use #20-Jan-10# in the
above WHERE clause.) The date-restricted output is shown in Figure 7.9.
FIGURE
7.9
Selected PRODUCT table
attributes: date restriction
Using Computed
Aliases
Columns
and Column
Suppose that you want to determine the total value of each
of the products currently held in inventory. Logically, that
determination requires the multiplication of each product’s
quantity on hand by its current price. You can accomplish
this task with the following command:
SELECT
FROM
P_DESCRIPT, P_QOH, P_PRICE, P_QOH *
P_PRICE
PRODUCT;
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FIGURE
7.10
7
SELECT statement with a
computed column
Entering that SQL command in Access generates the output
shown in Figure 7.10.
SQL accepts any valid expressions (or formulas) in the
computed columns. Such formulas can contain any valid
mathematical operators and functions that are applied to
attributes in any of the tables specified in the FROM clause
of the SELECT statement. Note also that Access automatically adds an Expr label to all computed columns. (The first
computed column would be labeled Expr1; the second,
Expr2; and so on.) Oracle uses the actual formula text as the
label for the computed column.
To make the output more readable, the SQL standard
permits the use of aliases for any column in a SELECT
statement. An alias is an alternative name given to a
column or table in any SQL statement.
For example, you can rewrite the previous SQL statement as:
SELECT
FROM
P_DESCRIPT, P_QOH, P_PRICE, P_QOH * P_PRICE AS TOTVALUE
PRODUCT;
The output of that command is shown in Figure 7.11.
FIGURE
7.11
SELECT statement with a
computed column and an alias
You could also use a computed column, an alias, and date
arithmetic in a single query. For example, assume that you
want to get a list of out-of-warranty products that have been
stored more than 90 days. In that case, the P_INDATE is at
least 90 days less than the current (system) date. The MS
Access version of this query is:
SELECT
FROM
WHERE
P_CODE, P_INDATE, DATE() - 90 AS
CUTDATE
PRODUCT
P_INDATE <= DATE() - 90;
The Oracle version of the same query is shown here:
SELECT
FROM
WHERE
P_CODE, P_INDATE, SYSDATE - 90 AS
CUTDATE
PRODUCT
P_INDATE <= SYSDATE - 90;
Note that DATE() and SYSDATE are special functions that return the current date in MS Access and Oracle,
respectively. You can use the DATE() and SYSDATE functions anywhere a date literal is expected, such as in the value
list of an INSERT statement, in an UPDATE statement when changing the value of a date attribute, or in a SELECT
statement as shown here. Of course, the previous query output would change based on the current date.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
Suppose that a manager wants a list of all products, the dates they were received, and the warranty expiration date
(90 days from when the product was received). To generate that list, type:
SELECT
FROM
P_CODE, P_INDATE, P_INDATE + 90 AS EXPDATE
PRODUCT;
Note that you can use all arithmetic operators with date attributes as well as with numeric attributes.
7.4.2 Arithmetic Operators: The Rule of Precedence
As you saw in the previous example, you can use arithmetic operators with table attributes in a column list or in a
conditional expression. In fact, SQL commands are often used in conjunction with the arithmetic operators shown in
Table 7.7.
TABLE
The Arithmetic Operators
7.7
ARITHMETIC OPERATOR
+
*
/
^
DESCRIPTION
Add
Subtract
Multiply
Divide
Raise to the power of (some applications use ** instead of ^)
Do not confuse the multiplication symbol (*) with the wildcard symbol used by some SQL implementations, such as MS
Access; the latter is used only in string comparisons, while the former is used in conjunction with mathematical
procedures.
As you perform mathematical operations on attributes, remember the rules of precedence. As the name suggests, the
rules of precedence are the rules that establish the order in which computations are completed. For example, note
the order of the following computational sequence:
1.
Perform operations within parentheses.
2.
Perform power operations.
3.
Perform multiplications and divisions.
4.
Perform additions and subtractions.
The application of the rules of precedence will tell you that 8 + 2 * 5 = 8 + 10 = 18, but (8 + 2) * 5 = 10 * 5 = 50.
Similarly, 4 + 5^2 * 3 = 4 + 25 * 3 = 79, but (4 + 5)^2 * 3 = 81 * 3 = 243, while the operation expressed by
(4 + 5^2) * 3 yields the answer (4 + 25) * 3 = 29 * 3 = 87.
7.4.3 Logical Operators: AND, OR, and NOT
In the real world, a search of data normally involves multiple conditions. For example, when you are buying a new
house, you look for a certain area, a certain number of bedrooms, bathrooms, stories, and so on. In the same way,
SQL allows you to include multiple conditions in a query through the use of logical operators. The logical operators
are AND, OR, and NOT. For example, if you want a list of the table contents for either the V_CODE = 21344 or the
V_CODE = 24288, you can use the OR operator, as in the following command sequence:
SELECT
FROM
WHERE
P_DESCRIPT, P_INDATE, P_PRICE, V_CODE
PRODUCT
V_CODE = 21344 OR V_CODE = 24288;
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That command generates the six rows shown in Figure 7.12 that match the logical restriction.
FIGURE
7.12
Selected PRODUCT table
attributes: the logical OR
The logical AND has the same SQL syntax requirement.
The following command generates a list of all rows for which
P_PRICE is less than $50 and for which P_INDATE is a date
occurring after January 15, 2010:
SELECT
FROM
WHERE
AND
P_DESCRIPT,
P_INDATE,
V_CODE
PRODUCT
P_PRICE < 50
P_INDATE > '15-Jan-2010';
P_PRICE,
This command will produce the output shown in Figure 7.13.
FIGURE
7.13
Selected PRODUCT table
attributes: the logical AND
You can combine the logical OR with the logical AND
to place further restrictions on the output. For
example, suppose that you want a table listing for the
following conditions:
쐌
The P_INDATE is after January 15, 2010, and the
P_PRICE is less than $50.
쐌
Or the V_CODE is 24288.
The required listing can be produced by using:
SELECT
FROM
WHERE
OR
P_DESCRIPT, P_INDATE, P_PRICE, V_CODE
PRODUCT
(P_PRICE < 50 AND P_INDATE > '15-Jan-2010')
V_CODE = 24288;
Note the use of parentheses to combine logical restrictions. Where you place the parentheses depends on how you
want the logical restrictions to be executed. Conditions listed within parentheses are always executed first. The
preceding query yields the output shown in Figure 7.14.
FIGURE
7.14
Selected PRODUCT table
attributes: the logical AND
and OR
Note that the three rows with the V_CODE = 24288 are
included regardless of the P_INDATE and P_PRICE entries
for those rows.
The use of the logical operators OR and AND can become
quite complex when numerous restrictions are placed on the
query. In fact, a specialty field in mathematics known as
Boolean algebra is dedicated to the use of logical
operators.
The logical operator NOT is used to negate the result of a
conditional expression. That is, in SQL, all conditional
expressions evaluate to true or false. If an expression is true,
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
the row is selected; if an expression is false, the row is not selected. The NOT logical operator is typically used to find
the rows that do not match a certain condition. For example, if you want to see a listing of all rows for which the
vendor code is not 21344, use the command sequence:
SELECT
FROM
WHERE
*
PRODUCT
NOT (V_CODE = 21344);
Note that the condition is enclosed in parentheses; that practice is optional, but it is highly recommended for clarity.
The logical NOT can be combined with AND and OR.
Note
If your SQL version does not support the logical NOT, you can generate the required output by using the
condition:
WHERE V_CODE <> 21344
If your version of SQL does not support <>, use:
WHERE V_CODE != 21344
7.4.4 Special Operators
ANSI-standard SQL allows the use of special operators in conjunction with the WHERE clause. These special operators
include:
BETWEEN: Used to check whether an attribute value is within a range
IS NULL: Used to check whether an attribute value is null
LIKE: Used to check whether an attribute value matches a given string pattern
IN: Used to check whether an attribute value matches any value within a value list
EXISTS: Used to check whether a subquery returns any rows
The BETWEEN Special Operator
If you use software that implements a standard SQL, the operator BETWEEN may be used to check whether an
attribute value is within a range of values. For example, if you want to see a listing for all products whose prices are
between $50 and $100, use the following command sequence:
SELECT
FROM
WHERE
*
PRODUCT
P_PRICE BETWEEN 50.00 AND 100.00;
Note
NOTE TO ORACLE USERS
When using the BETWEEN special operator, always specify the lower range value first. If you list the higher range
value first, Oracle will return an empty result set.
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7
If your DBMS does not support BETWEEN, you can use:
SELECT
FROM
WHERE
*
PRODUCT
P_PRICE > 50.00 AND P_PRICE < 100.00;
The IS NULL Special Operator
Standard SQL allows the use of IS NULL to check for a null attribute value. For example, suppose that you want to
list all products that do not have a vendor assigned (V_CODE is null). Such a null entry could be found by using the
command sequence:
SELECT
FROM
WHERE
P_CODE, P_DESCRIPT, V_CODE
PRODUCT
V_CODE IS NULL;
Similarly, if you want to check a null date entry, the command sequence is:
SELECT
FROM
WHERE
P_CODE, P_DESCRIPT, P_INDATE
PRODUCT
P_INDATE IS NULL;
Note that SQL uses a special operator to test for nulls. Why? Couldn’t you just enter a condition such as ⬙V_CODE
= NULL⬙? No. Technically, NULL is not a “value” the way the number 0 (zero) or the blank space is, but instead a
NULL is a special property of an attribute that represents precisely the absence of any value.
The LIKE Special Operator
The LIKE special operator is used in conjunction with wildcards to find patterns within string attributes. Standard SQL
allows you to use the percent sign (%) and underscore (_) wildcard characters to make matches when the entire string
is not known:
쐌
% means any and all following or preceding characters are eligible. For example,
'J%' includes Johnson, Jones, Jernigan, July, and J-231Q.
'Jo%' includes Johnson and Jones.
'%n' includes Johnson and Jernigan.
쐌
_ means any one character may be substituted for the underscore. For example,
'_23-456-6789' includes 123-456-6789, 223-456-6789, and 323-456-6789.
'_23-_56-678_' includes 123-156-6781, 123-256-6782, and 823-956-6788.
'_o_es' includes Jones, Cones, Cokes, totes, and roles.
Note
Some RDBMSs, such as Microsoft Access, use the wildcard characters * and ? instead of % and _.
For example, the following query would find all VENDOR rows for contacts whose last names begin with Smith.
SELECT
FROM
WHERE
V_NAME, V_CONTACT, V_AREACODE, V_PHONE
VENDOR
V_CONTACT LIKE 'Smith%';
If you check the original VENDOR data in Figure 7.2 again, you’ll see that this SQL query yields three records: two
Smiths and one Smithson.
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Keep in mind that most SQL implementations yield case-sensitive searches. For example, Oracle will not yield a result
that includes Jones if you use the wildcard search delimiter 'jo%' in a search for last names. The reason is that Jones
begins with a capital J, and your wildcard search starts with a lowercase j. On the other hand, MS Access searches are
not case sensitive.
For example, suppose that you typed the following query in Oracle:
SELECT
FROM
WHERE
V_NAME, V_CONTACT, V_AREACODE, V_PHONE
VENDOR
V_CONTACT LIKE 'SMITH%';
No rows will be returned because character-based queries may be case sensitive. That is, an uppercase character has
a different ASCII code than a lowercase character, causing SMITH, Smith, and smith to be evaluated as different
(unequal) entries. Because the table contains no vendor whose last name begins with (uppercase) SMITH, the
(uppercase) 'SMITH%' used in the query cannot be matched. Matches can be made only when the query entry is written
exactly like the table entry.
Some RDBMSs, such as Microsoft Access, automatically make the necessary conversions to eliminate case sensitivity.
Others, such as Oracle, provide a special UPPER function to convert both table and query character entries to
uppercase. (The conversion is done in the computer’s memory only; the conversion has no effect on how the value
is actually stored in the table.) So if you want to avoid a no-match result based on case sensitivity, and if your RDBMS
allows the use of the UPPER function, you can generate the same results by using the query:
SELECT
FROM
WHERE
V_NAME, V_CONTACT, V_AREACODE, V_PHONE
VENDOR
UPPER(V_CONTACT) LIKE 'SMITH%';
The preceding query produces a list that includes all rows containing a last name that begins with Smith, regardless
of uppercase or lowercase letter combinations such as Smith, smith, and SMITH.
The logical operators may be used in conjunction with the special operators. For instance, the query:
SELECT
FROM
WHERE
V_NAME, V_CONTACT, V_AREACODE, V_PHONE
VENDOR
V_CONTACT NOT LIKE 'Smith%';
will yield an output of all vendors whose names do not start with Smith.
Suppose that you do not know whether a person’s name is spelled Johnson or Johnsen. The wildcard character _ lets
you find a match for either spelling. The proper search would be instituted by the query:
SELECT
FROM
WHERE
*
VENDOR
V_CONTACT LIKE 'Johns_n';
Thus, the wildcards allow you to make matches when only approximate spellings are known. Wildcard characters may
be used in combinations. For example, the wildcard search based on the string '_l%' can yield the strings Al, Alton,
Elgin, Blakeston, blank, bloated, and eligible.
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The IN Special Operator
Many queries that would require the use of the logical OR can be more easily handled with the help of the special
operator IN. For example, the query:
SELECT
FROM
WHERE
OR
*
PRODUCT
V_CODE = 21344
V_CODE = 24288;
can be handled more efficiently with:
SELECT
FROM
WHERE
*
PRODUCT
V_CODE IN (21344, 24288);
Note that the IN operator uses a value list. All of the values in the list must be of the same data type. Each of the values
in the value list is compared to the attribute—in this case, V_CODE. If the V_CODE value matches any of the values
in the list, the row is selected. In this example, the rows selected will be only those in which the V_CODE is either
21344 or 24288.
If the attribute used is of a character data type, the list values must be enclosed in single quotation marks. For instance,
if the V_CODE had been defined as CHAR(5) when the table was created, the preceding query would have read:
SELECT
FROM
WHERE
*
PRODUCT
V_CODE IN ('21344', '24288');
The IN operator is especially valuable when it is used in conjunction with subqueries. For example, suppose that you
want to list the V_CODE and V_NAME of only those vendors who provide products. In that case, you could use a
subquery within the IN operator to automatically generate the value list. The query would be:
SELECT
FROM
WHERE
V_CODE, V_NAME
VENDOR
V_CODE IN (SELECT V_CODE FROM PRODUCT);
The preceding query will be executed in two steps:
1.
The inner query or subquery will generate a list of V_CODE values from the PRODUCT tables. Those
V_CODE values represent the vendors who supply products.
2.
The IN operator will compare the values generated by the subquery to the V_CODE values in the VENDOR
table and will select only the rows with matching values—that is, the vendors who provide products.
The IN special operator will receive additional attention in Chapter 8, where you will learn more about subqueries.
The EXISTS Special Operator
The EXISTS special operator can be used whenever there is a requirement to execute a command based on the result
of another query. That is, if a subquery returns any rows, run the main query; otherwise, don’t. For example, the
following query will list all vendors, but only if there are products to order:
SELECT
FROM
WHERE
*
VENDOR
EXISTS (SELECT * FROM PRODUCT WHERE P_QOH <= P_MIN);
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The EXISTS special operator is used in the following example to list all vendors, but only if there are products with
the quantity on hand, less than double the minimum quantity:
SELECT
FROM
WHERE
*
VENDOR
EXISTS (SELECT * FROM PRODUCT WHERE P_QOH < P_MIN * 2);
The EXISTS special operator will receive additional attention in Chapter 8, where you will learn more about
subqueries.
7.5 ADDITIONAL DATA DEFINITION COMMANDS
In this section, you will learn how to change (alter) table structures by changing attribute characteristics and by adding
columns. Then you will learn how to do advanced data updates to the new columns. Finally, you will learn how to copy
tables or parts of tables and how to delete tables.
All changes in the table structure are made by using the ALTER TABLE command, followed by a keyword that
produces the specific change you want to make. Three options are available: ADD, MODIFY, and DROP. You use
ADD to add a column, MODIFY to change column characteristics, and DROP to delete a column from a table. Most
RDBMSs do not allow you to delete a column (unless the column does not contain any values) because such an action
might delete crucial data that are used by other tables. The basic syntax to add or modify columns is:
ALTER TABLE tablename
{ADD | MODIFY} ( columnname datatype [ {ADD | MODIFY} columnname datatype] ) ;
The ALTER TABLE command can also be used to add table constraints. In those cases, the syntax would be:
ALTER TABLE tablename
ADD constraint [ ADD constraint ] ;
where constraint refers to a constraint definition similar to those you learned in Section 7.2.6.
You could also use the ALTER TABLE command to remove a column or table constraint. The syntax would be as follows:
ALTER TABLE tablename
DROP{PRIMARY KEY | COLUMN columnname | CONSTRAINT constraintname };
Notice that when removing a constraint, you need to specify the name given to the constraint. That is one reason why
you should always name your constraints in your CREATE TABLE or ALTER TABLE statement.
7.5.1 Changing a Column’s Data Type
Using the ALTER syntax, the (integer) V_CODE in the PRODUCT table can be changed to a character V_CODE by using:
ALTER TABLE PRODUCT
MODIFY (V_CODE CHAR(5));
Some RDBMSs, such as Oracle, do not let you change data types unless the column to be changed is empty. For
example, if you want to change the V_CODE field from the current number definition to a character definition, the
above command will yield an error message, because the V_CODE column already contains data. The error message
is easily explained. Remember that the V_CODE in PRODUCT references the V_CODE in VENDOR. If you change
the V_CODE data type, the data types don’t match, and there is a referential integrity violation, which triggers the
error message. If the V_CODE column does not contain data, the preceding command sequence will produce the
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expected table structure alteration (if the foreign key reference was not specified during the creation of the
PRODUCT table).
7.5.2 Changing a Column’s Data Characteristics
If the column to be changed already contains data, you can make changes in the column’s characteristics if those
changes do not alter the data type. For example, if you want to increase the width of the P_PRICE column to nine
digits, use the command:
ALTER TABLE PRODUCT
MODIFY (P_PRICE DECIMAL(9,2));
If you now list the table contents, you can see that the column width of P_PRICE has increased by one digit.
Note
Some DBMSs impose limitations on when it's possible to change attribute characteristics. For example, Oracle
lets you increase (but not decrease) the size of a column. The reason for this restriction is that an attribute
modification will affect the integrity of the data in the database. In fact, some attribute changes can be done only
when there are no data in any rows for the affected attribute.
7.5.3 Adding a Column
You can alter an existing table by adding one or more columns. In the following example, you add the column named
P_SALECODE to the PRODUCT table. (This column will be used later to determine whether goods that have been
in inventory for a certain length of time should be placed on special sale.)
Suppose that you expect the P_SALECODE entries to be 1, 2, or 3. Because there will be no arithmetic performed
with the P_SALECODE, the P_SALECODE will be classified as a single-character attribute. Note the inclusion of all
required information in the following ALTER command:
ALTER TABLE PRODUCT
ADD (P_SALECODE CHAR(1));
Online Content
If you are using the MS Access databases provided in the Premium Website, you can track each of the updates
in the following sections. For example, look at the copies of the PRODUCT table in the Ch07_SaleCo database,
one named Product_2 and one named PRODUCT_3. Each of the two copies includes the new P_SALECODE
column. If you want to see the cumulative effect of all UPDATE commands, you can continue using the
PRODUCT table with the P_SALECODE modification and all of the changes you will make in the following
sections. (You might even want to use both options, first to examine the individual effects of the update queries
and then to examine the cumulative effects.)
When adding a column, be careful not to include the NOT NULL clause for the new column. Doing so will cause an
error message; if you add a new column to a table that already has rows, the existing rows will default to a value of
null for the new column. Therefore, it is not possible to add the NOT NULL clause for this new column. (You can, of
course, add the NOT NULL clause to the table structure after all of the data for the new column have been entered
and the column no longer contains nulls.)
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7.5.4 Dropping a Column
Occasionally, you might want to modify a table by deleting a column. Suppose that you want to delete the V_ORDER
attribute from the VENDOR table. To accomplish that, you would use the following command:
ALTER TABLE VENDOR
DROP COLUMN V_ORDER;
Again, some RDBMSs impose restrictions on attribute deletion. For example, you may not drop attributes that are
involved in foreign key relationships, nor may you delete an attribute of a table that contains only that one attribute.
7.5.5 Advanced Data Updates
To make data entries in an existing row’s columns, SQL allows the UPDATE command. The UPDATE command
updates only data in existing rows. For example, to enter the P_SALECODE value '2' in the fourth row, use the
UPDATE command together with the primary key P_CODE '1546-QQ2'. Enter the value by using the command
sequence:
UPDATE
SET
WHERE
PRODUCT
P_SALECODE = '2'
P_CODE = '1546-QQ2';
Subsequent data can be entered the same way, defining each entry location by its primary key (P_CODE) and its
column location (P_SALECODE). For example, if you want to enter the P_SALECODE value '1' for the P_CODE
values '2232/QWE' and '2232/QTY', you use:
UPDATE
SET
WHERE
PRODUCT
P_SALECODE = '1'
P_CODE IN ('2232/QWE', '2232/QTY');
If your RDBMS does not support IN, use the following command:
UPDATE
SET
WHERE
PRODUCT
P_SALECODE = '1'
P_CODE = '2232/QWE' OR P_CODE = '2232/QTY';
The results of your efforts can be checked by using:
SELECT
FROM
P_CODE, P_DESCRIPT, P_INDATE, P_PRICE, P_SALECODE
PRODUCT;
Although the UPDATE sequences just shown allow you to enter values into specified table cells, the process is very
cumbersome. Fortunately, if a relationship can be established between the entries and the existing columns, the
relationship can be used to assign values to their appropriate slots. For example, suppose that you want to place sales
codes based on the P_INDATE into the table, using the following schedule:
P_INDATE
P_SALECODE
before December 25, 2009
2
between January 16, 2010, and February 10, 2010 1
Using the PRODUCT table, the following two command sequences make the appropriate assignments:
UPDATE
SET
WHERE
PRODUCT
P_SALECODE = '2'
P_INDATE < '25-Dec-2009';
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UPDATE
SET
WHERE
7
PRODUCT
P_SALECODE = '1'
P_INDATE >= '16-Jan-2010' AND P_INDATE <='10-Feb-2010';
To check the results of those two command sequences, use:
SELECT
FROM
P_CODE, P_DESCRIPT, P_INDATE, P_PRICE, P_SALECODE
PRODUCT;
If you have made all of the updates shown in this section using Oracle, your PRODUCT table should look like Figure
7.15. Make sure that you issue a COMMIT statement to save these changes.
FIGURE
The cumulative effect of the multiple updates in the PRODUCT table (Oracle)
7.15
The arithmetic operators are particularly useful in data updates. For example, if the quantity on hand in your
PRODUCT table has dropped below the minimum desirable value, you’ll order more of the product. Suppose, for
example, you have ordered 20 units of product 2232/QWE. When the 20 units arrive, you’ll want to add them to
inventory, using:
UPDATE
SET
WHERE
PRODUCT
P_QOH = P_QOH + 20
P_CODE = '2232/QWE';
If you want to add 10 percent to the price for all products that have current prices below $50, you can use:
UPDATE
SET
WHERE
PRODUCT
P_PRICE = P_PRICE * 1.10
P_PRICE < 50.00;
If you are using Oracle, issue a ROLLBACK command to undo the changes made by the last two UPDATE statements.
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Note
If you fail to roll back the changes of the preceding UPDATE queries, the output of the subsequent queries will
not match the results shown in the figures. Therefore:
• If you are using Oracle, use the ROLLBACK command to restore the database to its previous state.
• If you are using Access, copy the original Ch07_SaleCo.mdb file from the Premium Website for this book.
7.5.6 Copying Parts of Tables
As you will discover in later chapters on database design, sometimes it is necessary to break up a table structure into
several component parts (or smaller tables). Fortunately, SQL allows you to copy the contents of selected table columns
so that the data need not be reentered manually into the newly created table(s). For example, if you want to copy
P_CODE, P_DESCRIPT, P_PRICE, and V_CODE from the PRODUCT table to a new table named PART, you create
the PART table structure first, as follows:
CREATE TABLE PART(
PART_CODE
CHAR(8)
PART_DESCRIPT
CHAR(35),
PART_PRICE
DECIMAL(8,2),
V_CODE
INTEGER,
PRIMARY KEY (PART_CODE));
NOT NULL
UNIQUE,
Note that the PART column names need not be identical to those of the original table and that the new table need not
have the same number of columns as the original table. In this case, the first column in the PART table is PART_CODE,
rather than the original P_CODE found in the PRODUCT table. And the PART table contains only four columns rather
than the seven columns found in the PRODUCT table. However, column characteristics must match; you cannot copy
a character-based attribute into a numeric structure and vice versa.
Next, you need to add the rows to the new PART table, using the PRODUCT table rows. To do that, you use the
INSERT command you learned in Section 7.3.7. The syntax is:
INSERT INTO
SELECT
FROM
target_tablename[(target_columnlist)]
source_columnlist
source_tablename;
Note that the target column list is required if the source column list doesn’t match all of the attribute names and
characteristics of the target table (including the order of the columns). Otherwise, you do not need to specify the target
column list. In this example, you must specify the target column list in the INSERT command below because the
column names of the target table are different:
INSERT INTO PART
SELECT
(PART_CODE, PART_DESCRIPT, PART_PRICE, V_CODE)
P_CODE, P_DESCRIPT, P_PRICE, V_CODE FROM PRODUCT;
The contents of the PART table can now be examined by using the query:
SELECT
* FROM PART;
to generate the new PART table’s contents, shown in Figure 7.16.
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FIGURE
PART table attributes copied
from the PRODUCT table
7.16
SQL also provides another way to rapidly create a new table
based on selected columns and rows of an existing table. In
this case, the new table will copy the attribute names, data
characteristics, and rows of the original table. The Oracle
version of the command is:
CREATE TABLE PART AS
SELECT
P_CODE AS PART_CODE, P_DESCRIPT
AS PART_DESCRIPT,
P_PRICE AS PART_PRICE, V_CODE
FROM
PRODUCT;
If the PART table already exists, Oracle will not let you
overwrite the existing table. To run this command, you must
first delete the existing PART table. (See Section 7.5.8.)
The MS Access version of this command is:
SELECT
FROM
P_CODE AS PART_CODE, P_DESCRIPT AS PART_DESCRIPT,
P_PRICE AS PART_PRICE, V_CODE INTO PART
PRODUCT;
If the PART table already exists, MS Access will ask if you want to delete the existing table and continue with the
creation of the new PART table.
The SQL command just shown creates a new PART table with PART_CODE, PART_DESCRIPT, PART_PRICE, and
V_CODE columns. In addition, all of the data rows (for the selected columns) will be copied automatically. However,
note that no entity integrity (primary key) or referential integrity (foreign key) rules are automatically applied to
the new table. In the next section, you will learn how to define the PK to enforce entity integrity and the FK to enforce
referential integrity.
7.5.7 Adding Primary and Foreign Key Designations
When you create a new table based on another table, the new table does not include integrity rules from the old table.
In particular, there is no primary key. To define the primary key for the new PART table, use the following command:
ALTER TABLE
ADD
PART
PRIMARY KEY (PART_CODE);
Aside from the fact that the integrity rules are not automatically transferred to a new table that derives its data from
one or more other tables, several other scenarios could leave you without entity and referential integrity. For example,
you might have forgotten to define the primary and foreign keys when you created the original tables. Or if you
imported tables from a different database, you might have discovered that the importing procedure did not transfer the
integrity rules. In any case, you can reestablish the integrity rules by using the ALTER command. For example, if the
PART table’s foreign key has not yet been designated, it can be designated by:
ALTER TABLE
ADD
PART
FOREIGN KEY (V_CODE) REFERENCES VENDOR;
Alternatively, if neither the PART table’s primary key nor its foreign key has been designated, you can incorporate both
changes at once, using:
ALTER TABLE
ADD
ADD
PART
PRIMARY KEY (PART_CODE)
FOREIGN KEY (V_CODE) REFERENCES VENDOR;
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Even composite primary keys and multiple foreign keys can be designated in a single SQL command. For example,
if you want to enforce the integrity rules for the LINE table shown in Figure 7.1, you can use:
ALTER TABLE
ADD
ADD
ADD
LINE
PRIMARY KEY (INV_NUMBER, LINE_NUMBER)
FOREIGN KEY (INV_NUMBER) REFERENCES INVOICE
FOREIGN KEY (PROD_CODE) REFERENCES PRODUCT;
7.5.8 Deleting a Table from the Database
A table can be deleted from the database using the DROP TABLE command. For example, you can delete the PART
table you just created with:
DROP TABLE PART;
You can drop a table only if that table is not the “one” side of any relationship. If you try to drop a table otherwise,
the RDBMS will generate an error message indicating that a foreign key integrity violation has occurred.
7.6 ADDITIONAL SELECT QUERY KEYWORDS
One of the most important advantages of SQL is its ability to produce complex free-form queries. The logical operators
that were introduced earlier to update table contents work just as well in the query environment. In addition, SQL
provides useful functions that count, find minimum and maximum values, calculate averages, and so on. Better yet,
SQL allows the user to limit queries to only those entries that have no duplicates or entries whose duplicates can be
grouped.
7.6.1 Ordering a Listing
The ORDER BY clause is especially useful when the listing order is important to you. The syntax is:
SELECT
FROM
[WHERE
[ORDER BY
columnlist
tablelist
conditionlist ]
columnlist [ASC | DESC] ] ;
Although you have the option of declaring the order type—ascending or descending—the default order is ascending.
For example, if you want the contents of the PRODUCT table listed by P_PRICE in ascending order, use:
SELECT
FROM
ORDER BY
P_CODE, P_DESCRIPT, P_INDATE, P_PRICE
PRODUCT
P_PRICE;
The output is shown in Figure 7.17. Note that ORDER BY yields an ascending price listing.
Comparing the listing in Figure 7.17 to the actual table contents shown earlier in Figure 7.2, you will see that in
Figure 7.17, the lowest-priced product is listed first, followed by the next lowest-priced product, and so on. However,
although ORDER BY produces a sorted output, the actual table contents are unaffected by the ORDER BY command.
To produce the list in descending order, you would enter:
SELECT
FROM
ORDER BY
P_CODE, P_DESCRIPT, P_INDATE, P_PRICE
PRODUCT
P_PRICE DESC;
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7.17
7
Selected PRODUCT table
attributes: ordered by
(ascending) P_PRICE
Ordered listings are used frequently. For example, suppose
that you want to create a phone directory. It would be helpful
if you could produce an ordered sequence (last name, first
name, initial) in three stages:
1.
ORDER BY last name.
2.
Within the last names, ORDER BY first name.
3.
Within the first and last names, ORDER BY middle
initial.
Such a multilevel ordered sequence is known as a
cascading order sequence, and it can be created easily by
listing several attributes, separated by commas, after the
ORDER BY clause.
The cascading order sequence is the basis for any telephone
directory. To illustrate a cascading order sequence, use the
following SQL command on the EMPLOYEE table:
SELECT
FROM
ORDER BY
EMP_LNAME, EMP_FNAME, EMP_INITIAL, EMP_AREACODE, EMP_PHONE
EMPLOYEE
EMP_LNAME, EMP_FNAME, EMP_INITIAL;
That command yields the results shown in Figure 7.18.
FIGURE
Telephone list query results
7.18
The ORDER BY clause is useful in many applications, especially because the DESC qualifier can be invoked. For
example, listing the most recent items first is a standard procedure. Typically, invoice due dates are listed in descending
order. Or if you want to examine budgets, it’s probably useful to list the largest budget line items first.
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You can use the ORDER BY clause in conjunction with other SQL commands, too. For example, note the use of
restrictions on date and price in the following command sequence:
SELECT
FROM
WHERE
AND
ORDER BY
P_DESCRIPT, V_CODE, P_INDATE, P_PRICE
PRODUCT
P_INDATE < '21-Jan-2010'
P_PRICE <= 50.00
V_CODE, P_PRICE DESC;
The output is shown in Figure 7.19. Note that within each V_CODE, the P_PRICE values are in descending order.
FIGURE
7.19
A query based on multiple
restrictions
7.6.2 Listing Unique Values
How many different vendors are currently represented in
the PRODUCT table? A simple listing (SELECT) is not very
useful if the table contains several thousand rows and you
have to sift through the vendor codes manually. Fortunately,
SQL’s DISTINCT clause produces a list of only those values
that are different from one another. For example, the
command:
SELECT
FROM
DISTINCT V_CODE
PRODUCT;
yields only the different (distinct) vendor codes (V_CODE) that are encountered in the PRODUCT table, as shown in
Figure 7.20. Note that the first output row shows the null. (By default, Access places the null V_CODE at the top of
the list, while Oracle places it at the bottom. The placement of nulls does not affect the list contents. In Oracle, you
could use ORDER BY V_CODE NULLS FIRST to place nulls at the top of the list.)
Note
If the ordering column has nulls, they are listed either first or last, depending on the RDBMS.
The ORDER BY clause must always be listed last in the SELECT command sequence.
FIGURE
7.20
A listing of distinct (different)
V_CODE values in the
PRODUCT table
7.6.3 Aggregate Functions
SQL can perform various mathematical summaries for you,
such as counting the number of rows that contain a specified
condition, finding the minimum or maximum values for
some specified attribute, summing the values in a specified
column, and averaging the values in a specified column.
Those aggregate functions are shown in Table 7.8.
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TABLE
7.8
FUNCTION
COUNT
MIN
MAX
SUM
AVG
Some Basic SQL Aggregate Functions
OUTPUT
The number of rows containing non-null values
The minimum attribute value encountered in a given column
The maximum attribute value encountered in a given column
The sum of all values for a given column
The arithmetic mean (average) for a specified column
To illustrate another standard SQL command format, most of the remaining input and output sequences are presented
using the Oracle RDBMS.
COUNT
The COUNT function is used to tally the number of non-null values of an attribute. COUNT can be used in conjunction
with the DISTINCT clause. For example, suppose that you want to find out how many different vendors are in the
PRODUCT table. The answer, generated by the first SQL code set shown in Figure 7.21, is 6. The answer indicates
that six different VENDOR codes are found in the PRODUCT table. (Note that the nulls are not counted as V_CODE
values.)
FIGURE
COUNT function output examples
7.21
The aggregate functions can be combined with the SQL commands explored earlier. For example, the second SQL
command set in Figure 7.21 supplies the answer to the question, “How many vendors referenced in the PRODUCT
table have supplied products with prices that are less than or equal to $10?” The answer is three, indicating that three
vendors referenced in the PRODUCT table have supplied products that meet the price specification.
The COUNT aggregate function uses one parameter within parentheses, generally a column name such as
COUNT(V_CODE) or COUNT(P_CODE). The parameter may also be an expression such as COUNT(DISTINCT
V_CODE) or COUNT(P_PRICE+10). Using that syntax, COUNT always returns the number of non-null values in the
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given column. (Whether the column values are computed or show stored table row values is immaterial.) In contrast,
the syntax COUNT(*) returns the number of total rows returned by the query, including the rows that contain nulls. In
the example in Figure 7.21, SELECT COUNT(P_CODE) FROM PRODUCT and SELECT COUNT(*) FROM
PRODUCT will yield the same answer because there are no null values in the P_CODE primary key column.
Note that the third SQL command set in Figure 7.21 uses the COUNT(*) command to answer the question, “How many
rows in the PRODUCT table have a P_PRICE value less than or equal to $10?” The answer, five, indicates that five
products have a listed price that meets the price specification. The COUNT(*) aggregate function is used to count rows in
a query result set. In contrast, the COUNT(column) aggregate function counts the number of non-null values in a given
column. For example, in Figure 7.20, the COUNT(*) function would return a value of 7 to indicate seven rows returned
by the query. The COUNT(V_CODE) function would return a value of 6 to indicate the six non-null vendor code values.
Note
NOTE TO MS ACCESS USERS
MS Access does not support the use of COUNT with the DISTINCT clause. If you want to use such queries in
MS Access, you must create subqueries with DISTINCT and NOT NULL clauses. For example, the equivalent
MS Access queries for the first two queries shown in Figure 7.21 are:
SELECT
FROM
COUNT(*)
(SELECT DISTINCT V_CODE FROM PRODUCT WHERE V_CODE IS NOT NULL)
and
SELECT COUNT(*)
FROM (SELECT DISTINCT(V_CODE)
FROM (SELECT V_CODE, P_PRICE FROM PRODUCT
WHERE V_CODE IS NOT NULL AND P_PRICE<10))
Those two queries can be found in the Premium Website in the Ch07_SaleCo (Access) database. MS Access
does add a trailer at the end of the query after you have executed it, but you can delete that trailer the next time
you use the query.
MAX and MIN
The MAX and MIN functions help you find answers to problems such as the:
쐌
Highest (maximum) price in the PRODUCT table.
쐌
Lowest (minimum) price in the PRODUCT table.
The highest price, $256.99, is supplied by the first SQL command set in Figure 7.22. The second SQL command set
shown in Figure 7.22 yields the minimum price of $4.99.
The third SQL command set in Figure 7.22 demonstrates that the numeric functions can be used in conjunction with
more complex queries. However, you must remember that the numeric functions yield only one value based on all
of the values found in the table: a single maximum value, a single minimum value, a single count, or a single average
value. It is easy to overlook this warning. For example, examine the question, “Which product has the highest price?”
Although that query seems simple enough, the SQL command sequence:
SELECT
FROM
WHERE
P_CODE, P_DESCRIPT, P_PRICE
PRODUCT
P_PRICE = MAX(P_PRICE);
does not yield the expected results. This is because the use of MAX(P_PRICE) to the right side of a comparison
operator is incorrect, thus producing an error message. The aggregate function MAX(columnname) can be used only
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7
in the column list of a SELECT statement. Also, in a comparison that uses an equality symbol, you can use only a single
value to the right of the equals sign.
To answer the question, therefore, you must compute the maximum price first, then compare it to each price returned
by the query. To do that, you need a nested query. In this case, the nested query is composed of two parts:
쐌
The inner query, which is executed first.
쐌
The outer query, which is executed last. (Remember that the outer query is always the first SQL command you
encounter—in this case, SELECT.)
Using the following command sequence as an example, note that the inner query first finds the maximum price value,
which is stored in memory. Because the outer query now has a value to which to compare each P_PRICE value, the
query executes properly.
SELECT
FROM
WHERE
P_CODE, P_DESCRIPT, P_PRICE
PRODUCT
P_PRICE = (SELECT MAX(P_PRICE) FROM PRODUCT);
The execution of that nested query yields the correct answer, shown below the third (nested) SQL command set in
Figure 7.22.
FIGURE
MAX and MIN output examples
7.22
The MAX and MIN aggregate functions can also be used with date columns. For example, to find out the product that
has the oldest date, you would use MIN(P_INDATE). In the same manner, to find out the most recent product, you
would use MAX(P_INDATE).
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
Note
You can use expressions anywhere a column name is expected. Suppose that you want to know what product
has the highest inventory value. To find the answer, you can write the following query:
SELECT
FROM
WHERE
*
PRODUCT
P_QOH*P_PRICE = (SELECT MAX(P_QOH*P_PRICE) FROM PRODUCT);
SUM
The SUM function computes the total sum for any specified attribute, using whatever condition(s) you have imposed.
For example, if you want to compute the total amount owed by your customers, you could use the following command:
SELECT
FROM
SUM(CUS_BALANCE) AS TOTBALANCE
CUSTOMER;
You could also compute the sum total of an expression. For example, if you want to find the total value of all items
carried in inventory, you could use:
SELECT
FROM
SUM(P_QOH * P_PRICE) AS TOTVALUE
PRODUCT;
because the total value is the sum of the product of the quantity on hand and the price for all items. (See Figure 7.23.)
AVG
FIGURE
The total value of all items in the PRODUCT table
7.23
The AVG function format is similar to those of MIN and MAX and is subject to the same operating restrictions. The
first SQL command set shown in Figure 7.24 shows how a simple average P_PRICE value can be generated to yield
the computed average price of 56.42125. The second SQL command set in Figure 7.24 produces five output lines
that describe products whose prices exceed the average product price. Note that the second query uses nested SQL
commands and the ORDER BY clause examined earlier.
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7
FIGURE
AVG function output examples
7.24
7.6.4 Grouping Data
Frequency distributions can be created quickly and easily using the GROUP BY clause within the SELECT statement.
The syntax is:
SELECT
FROM
[WHERE
[GROUP BY
[HAVING
[ORDER BY
columnlist
tablelist
conditionlist ]
columnlist ]
conditionlist ]
columnlist [ASC | DESC] ] ;
The GROUP BY clause is generally used when you have attribute columns combined with aggregate functions in the
SELECT statement. For example, to determine the minimum price for each sales code, use the first SQL command
set shown in Figure 7.25.
The second SQL command set in Figure 7.25 generates the average price within each sales code. Note that the
P_SALECODE nulls are included within the grouping.
The GROUP BY clause is valid only when used in conjunction with one of the SQL aggregate functions, such as
COUNT, MIN, MAX, AVG, and SUM. For example, as shown in the first command set in Figure 7.26, if you try to
group the output by using:
SELECT
FROM
GROUP
V_CODE, P_CODE, P_DESCRIPT, P_PRICE
PRODUCT
BY V_CODE;
you generate a “not a GROUP BY expression” error. However, if you write the preceding SQL command sequence
in conjunction with some aggregate function, the GROUP BY clause works properly. The second SQL command
sequence in Figure 7.26 properly answers the question, “How many products are supplied by each vendor?” because
it uses a COUNT aggregate function.
I N T R O D U C T I O N
FIGURE
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
GROUP BY clause output examples
7.25
FIGURE
Incorrect and correct use of the GROUP BY clause
7.26
Note that the last output line in Figure 7.26 shows a null for the V_CODE, indicating that two products were not
supplied by a vendor. Perhaps those products were produced in-house, or they might have been bought via a
nonvendor channel, or the person making the data entry might have merely forgotten to enter a vendor code.
(Remember that nulls can be the result of many things.)
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7
Note
When using the GROUP BY clause with a SELECT statement:
• The SELECTs columnlist must include a combination of column names and aggregate functions.
• The GROUP BY clauses columnlist must include all nonaggregate function columns specified in the
SELECTs columnlist. If required, you could also group by any aggregate function columns that appear in the
SELECTs columnlist.
• The GROUP BY clause columnlist can include any columns from the tables in the FROM clause of the
SELECT statement, even if they do not appear in the SELECTs columnlist.
The GROUP BY Feature's HAVING Clause
A particularly useful extension of the GROUP BY feature is the HAVING clause. The HAVING clause operates very
much like the WHERE clause in the SELECT statement. However, the WHERE clause applies to columns and
expressions for individual rows, while the HAVING clause is applied to the output of a GROUP BY operation. For
example, suppose that you want to generate a listing of the number of products in the inventory supplied by each
vendor. However, this time you want to limit the listing to products whose prices average below $10. The first part of
that requirement is satisfied with the help of the GROUP BY clause, as illustrated in the first SQL command set in
Figure 7.27. Note that the HAVING clause is used in conjunction with the GROUP BY clause in the second SQL
command set in Figure 7.27 to generate the desired result.
FIGURE
An application of the HAVING clause
7.27
If you use the WHERE clause instead of the HAVING clause—the second SQL command set in Figure 7.27 will
produce an error message.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
You can also combine multiple clauses and aggregate functions. For example, consider the following SQL statement:
SELECT
FROM
GROUP BY
HAVING
ORDER BY
V_CODE, SUM(P_QOH * P_PRICE) AS TOTCOST
PRODUCT
V_CODE
(SUM(P_QOH * P_PRICE) > 500)
SUM(P_QOH * P_PRICE) DESC;
This statement will do the following:
쐌
Aggregate the total cost of products grouped by V_CODE.
쐌
Select only the rows having totals that exceed $500.
쐌
List the results in descending order by the total cost.
Note the syntax used in the HAVING and ORDER BY clauses; in both cases, you must specify the column expression
(formula) used in the SELECT statement’s column list, rather than the column alias (TOTCOST). Some RDBMSs allow
you to replace the column expression with the column alias, while others do not.
7.7 VIRTUAL TABLES: CREATING A VIEW
As you learned earlier, the output of a relational operator such as SELECT is another relation (or table). Suppose that
at the end of every day, you would like to get a list of all products to reorder, that is, products with a quantity on hand
that is less than or equal to the minimum quantity. Instead of typing the same query at the end of every day, wouldn’t
it be better to permanently save that query in the database? That’s the function of a relational view. A view is a virtual
table based on a SELECT query. The query can contain columns, computed columns, aliases, and aggregate functions
from one or more tables. The tables on which the view is based are called base tables.
You can create a view by using the CREATE VIEW command:
CREATE VIEW viewname AS SELECT query
The CREATE VIEW statement is a data definition command that stores the subquery specification—the SELECT
statement used to generate the virtual table—in the data dictionary.
The first SQL command set in Figure 7.28 shows the syntax used to create a view named PRICEGT50. This view
contains only the designated three attributes (P_DESCRIPT, P_QOH, and P_PRICE) and only rows in which the price
is over $50. The second SQL command sequence in Figure 7.28 shows the rows that make up the view.
A relational view has several special characteristics:
쐌
You can use the name of a view anywhere a table name is expected in a SQL statement.
쐌
Views are dynamically updated. That is, the view is re-created on demand each time it is invoked. Therefore,
if new products are added (or deleted) to meet the criterion P_PRICE > 50.00, those new products will
automatically appear (or disappear) in the PRICEGT50 view the next time the view is invoked.
쐌
Views provide a level of security in the database because the view can restrict users to only specified columns
and specified rows in a table. For example, if you have a company with hundreds of employees in several
departments, you could give the secretary of each department a view of only certain attributes and for the
employees that belong only to that secretary’s department.
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C H A P T E R
FIGURE
7
Creating a virtual table with the CREATE VIEW command
7.28
Note
NOTE TO MS ACCESS USERS
The CREATE VIEW command is not directly supported in MS Access. To create a view in MS Access, you just
need to create a SQL query and then save it.
쐌
Views may also be used as the basis for reports. For example, if you need a report that shows a summary of
total product cost and quantity-on-hand statistics grouped by vendor, you could create a PROD_STATS
view as:
CREATE VIEW PROD_STATS AS
SELECT
V_CODE, SUM(P_QOH*P_PRICE) AS TOTCOST,
MAX(P_QOH) AS MAXQTY, MIN(P_QOH) AS MINQTY,
AVG(P_QOH) AS AVGQTY
FROM
PRODUCT
GROUP BY V_CODE;
In Chapter 8, you will learn more about views and, in particular, about updating data in base tables through views.
7.8 JOINING DATABASE TABLES
The ability to combine (join) tables on common attributes is perhaps the most important distinction between a relational
database and other databases. A join is performed when data are retrieved from more than one table at a time. (If
necessary, review the join definitions and examples in Chapter 3, The Relational Database Model.)
To join tables, you simply list the tables in the FROM clause of the SELECT statement. The DBMS will create the
Cartesian product of every table in the FROM clause. (Review Chapter 3 to revisit these terms, if necessary.) However,
to get the correct result—that is, a natural join—you must select only the rows in which the common attribute values
match. To do this, use the WHERE clause to indicate the common attributes used to link the tables (this WHERE clause
is sometimes referred to as the join condition).
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
The join condition is generally composed of an equality comparison between the foreign key and the primary key of
related tables. For example, suppose that you want to join the two tables VENDOR and PRODUCT. Because V_CODE
is the foreign key in the PRODUCT table and the primary key in the VENDOR table, the link is established on
V_CODE. (See Table 7.9.)
TABLE
7.9
TABLE
PRODUCT
VENDOR
Creating Links Through Foreign Keys
ATTRIBUTES TO BE SHOWN
P_DESCRIPT, P_PRICE
V_COMPANY, V_PHONE
LINKING ATTRIBUTE
V_CODE
V_CODE
When the same attribute name appears in more than one of the joined tables, the source table of the attributes listed
in the SELECT command sequence must be defined. To join the PRODUCT and VENDOR tables, you would use the
following, which produces the output shown in Figure 7.29:
SELECT
FROM
WHERE
FIGURE
P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONE
PRODUCT, VENDOR
PRODUCT.V_CODE = VENDOR.V_CODE;
The results of a join
7.29
Your output might be presented in a different order because the SQL command produces a listing in which the order
of the columns is not relevant. In fact, you are likely to get a different order of the same listing the next time you
execute the command. However, you can generate a more predictable list by using an ORDER BY clause:
SELECT
FROM
WHERE
ORDER BY
PRODUCT.P_DESCRIPT, PRODUCT.P_PRICE, VENDOR.V_NAME, VENDOR.V_CONTACT,
VENDOR.V_AREACODE, VENDOR.V_PHONE
PRODUCT, VENDOR
PRODUCT.V_CODE = VENDOR.V_CODE
PRODUCT.P_PRICE;
In that case, your listing will always be arranged from the lowest price to the highest price.
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7
Note
Table names were used as prefixes in the preceding SQL command sequence. For example, PRODUCT.P_
PRICE was used rather than P_PRICE. Most current-generation RDBMSs do not require table names to be used
as prefixes unless the same attribute name occurs in several of the tables being joined. In that case, V_CODE is
used as a foreign key in PRODUCT and as a primary key in VENDOR; therefore, you must use the table names
as prefixes in the WHERE clause. In other words, you can write the previous query as:
SELECT
FROM
P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONE
PRODUCT, VENDOR WHERE PRODUCT.V_CODE = VENDOR.V_CODE
ORDER BY P_PRICE;
Naturally, if an attribute name occurs in several places, its origin (table) must be specified. If you fail to
provide such a specification, SQL will generate an error message to indicate that you have been ambiguous
about the attribute’s origin.
The preceding SQL command sequence joins a row in the PRODUCT table with a row in the VENDOR table in which
the V_CODE values of these rows are the same, as indicated in the WHERE clause’s condition. Because any vendor
can deliver any number of ordered products, the PRODUCT table might contain multiple V_CODE entries for each
V_CODE entry in the VENDOR table. In other words, each V_CODE in VENDOR can be matched with many
V_CODE rows in PRODUCT.
If you do not specify the WHERE clause, the result will be the Cartesian product of PRODUCT and VENDOR. Because
the PRODUCT table contains 16 rows and the VENDOR table contains 11 rows, the Cartesian product will produce
a listing of (16 × 11) = 176 rows. (Each row in PRODUCT will be joined to each row in the VENDOR table.)
All of the SQL commands can be used on the joined tables. For example, the following command sequence is quite
acceptable in SQL and produces the output shown in Figure 7.30:
SELECT
FROM
WHERE
AND
FIGURE
P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONE
PRODUCT, VENDOR
PRODUCT.V_CODE = VENDOR.V_CODE
P_INDATE > '15-Jan-2010';
An ordered and limited listing after a join
7.30
When joining three or more tables, you need to specify a join condition for each pair of tables. The number of join
conditions will always be N-1, where N represents the number of tables listed in the FROM clause. For example, if you
have three tables, you must have two join conditions; if you have five tables, you must have four join conditions; and
so on.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
Remember, the join condition will match the foreign key of a table to the primary key of the related table. For example,
using Figure 7.1, if you want to list the customer last name, invoice number, invoice date, and product descriptions for
all invoices for customer 10014, you must type the following:
SELECT
FROM
WHERE
AND
AND
AND
ORDER BY
CUS_LNAME, INVOICE.INV_NUMBER, INV_DATE, P_DESCRIPT
CUSTOMER, INVOICE, LINE, PRODUCT
CUSTOMER.CUS_CODE = INVOICE.CUS_CODE
INVOICE.INV_NUMBER = LINE.INV_NUMBER
LINE.P_CODE = PRODUCT.P_CODE
CUSTOMER.CUS_CODE = 10014
INV_NUMBER;
Finally, be careful not to create circular join conditions. For example, if Table A is related to Table B, Table B is related
to Table C, and Table C is also related to Table A, create only two join conditions: join A with B and B with C. Do
not join C with A!
7.8.1 Joining Tables with an Alias
An alias may be used to identify the source table from which the data are taken. The aliases P and V are used to label
the PRODUCT and VENDOR tables in the next command sequence. Any legal table name may be used as an alias.
(Also notice that there are no table name prefixes because the attribute listing contains no duplicate names in the
SELECT statement.)
SELECT
FROM
WHERE
ORDER BY
P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONE
PRODUCT P, VENDOR V
P.V_CODE = V.V_CODE
P_PRICE;
7.8.2 Recursive Joins
An alias is especially useful when a table must be joined to itself in a recursive query. For example, suppose that you
are working with the EMP table shown in Figure 7.31.
FIGURE
7.31
The contents of the EMP table
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7
Using the data in the EMP table, you can generate a list of all employees with their managers’ names by joining the
EMP table to itself. In that case, you would also use aliases to differentiate the table from itself. The SQL command
sequence would look like this:
SELECT
FROM
WHERE
ORDER BY
E.EMP_MGR, M.EMP_LNAME, E.EMP_NUM, E.EMP_LNAME
EMP E, EMP M
E.EMP_MGR=M.EMP_NUM
E.EMP_MGR;
The output of the preceding command sequence is shown in Figure 7.32.
FIGURE
7.32
Using an alias to join a table
to itself
7.8.3 Outer Joins
Figure 7.29 showed the results of joining the PRODUCT
and VENDOR tables. If you examine the output, note that
14 product rows are listed. Compare the output to the
PRODUCT table in Figure 7.2, and note that two products
are missing. Why? The reason is that there are two products
with nulls in the V_CODE attribute. Because there is no
matching null “value” in the VENDOR table’s V_CODE
attribute, the products do not show up in the final output
based on the join. Also, note that in the VENDOR table in
Figure 7.2, several vendors have no matching V_CODE in
the PRODUCT table. To include those rows in the final join
output, you must use an outer join.
Note
In MS Access, add AS to the previous SQL command sequence, for example:
SELECT
FROM
WHERE
ORDER BY
E.EMP_MGR,M.EMP_LNAME,E.EMP_NUM,E.EMP_LNAME
EMP AS E, EMP AS M
E.EMP_MGR = M.EMP_NUM
E.EMP_MGR;
There are two types of outer joins: left and right. (See Chapter 3.) Given the contents of the PRODUCT and VENDOR
tables, the following left outer join will show all VENDOR rows and all matching PRODUCT rows:
SELECT
FROM
P_CODE, VENDOR.V_CODE, V_NAME
VENDOR LEFT JOIN PRODUCT
ON VENDOR.V_CODE = PRODUCT.V_CODE;
Figure 7.33 shows the output generated by the left outer join command in MS Access. Oracle yields the same result
but shows the output in a different order.
The right outer join will join both tables and show all product rows with all matching vendor rows. The SQL command
for the right outer join is:
SELECT
FROM
PRODUCT.P_CODE, VENDOR.V_CODE, V_NAME
VENDOR RIGHT JOIN PRODUCT
ON VENDOR.V_CODE = PRODUCT.V_CODE;
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
Figure 7.34 shows the output generated by the right outer join command sequence in MS Access. Again, Oracle yields
the same result but shows the output in a different order.
In Chapter 8, you will learn more about joins and how to use the latest ANSI SQL standard syntax.
FIGURE
7.33
The left outer
join results
FIGURE
7.34
The right outer
join results
Online Content
For a complete walk-through example of converting an ER model into a database structure and using SQL
commands to create tables, see Appendix D, Converting the ER Model into a Database Structure, in the
Premium Website for this book.
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S u m m a r y
◗
◗
◗
◗
◗
◗
The SQL commands can be divided into two overall categories: data definition language (DDL) commands and data
manipulation language (DML) commands.
The ANSI standard data types are supported by all RDBMS vendors in different ways. The basic data types are
NUMBER, INTEGER, CHAR, VARCHAR, and DATE.
The basic data definition commands allow you to create tables, indexes, and views. Many SQL constraints can be
used with columns. The commands are CREATE TABLE, CREATE INDEX, CREATE VIEW, ALTER TABLE,
DROP TABLE, DROP VIEW, and DROP INDEX.
DML commands allow you to add, modify, and delete rows from tables. The basic DML commands are SELECT,
INSERT, UPDATE, DELETE, COMMIT, and ROLLBACK.
The INSERT command is used to add new rows to tables. The UPDATE command is used to modify data values
in existing rows of a table. The DELETE command is used to delete rows from tables. The COMMIT and
ROLLBACK commands are used to permanently save or roll back changes made to the rows. Once you COMMIT
the changes, you cannot undo them with a ROLLBACK command.
The SELECT statement is the main data retrieval command in SQL. A SELECT statement has the following syntax:
SELECT
FROM
[WHERE
[GROUP BY
[HAVING
[ORDER BY
◗
◗
◗
◗
◗
◗
columnlist
tablelist
conditionlist ]
columnlist ]
conditionlist ]
columnlist [ASC | DESC] ] ;
The column list represents one or more column names separated by commas. The column list may also include
computed columns, aliases, and aggregate functions. A computed column is represented by an expression or
formula (for example, P_PRICE * P_QOH). The FROM clause contains a list of table names or view names.
The WHERE clause can be used with the SELECT, UPDATE, and DELETE statements to restrict the rows affected
by the DDL command. The condition list represents one or more conditional expressions separated by logical
operators (AND, OR, and NOT). The conditional expression can contain any comparison operators (=, >, <, >=,
<=, and <>) as well as special operators (BETWEEN, IS NULL, LIKE, IN, and EXISTS).
Aggregate functions (COUNT, MIN, MAX, and AVG) are special functions that perform arithmetic computations
over a set of rows. The aggregate functions are usually used in conjunction with the GROUP BY clause to group
the output of aggregate computations by one or more attributes. The HAVING clause is used to restrict the output
of the GROUP BY clause by selecting only the aggregate rows that match a given condition.
The ORDER BY clause is used to sort the output of a SELECT statement. The ORDER BY clause can sort by one
or more columns and can use either ascending or descending order.
You can join the output of multiple tables with the SELECT statement. The join operation is performed every time
you specify two or more tables in the FROM clause and use a join condition in the WHERE clause to match the
foreign key of one table to the primary key of the related table. If you do not specify a join condition, the DBMS
will automatically perform a Cartesian product of the tables you specify in the FROM clause.
The natural join uses the join condition to match only rows with equal values in the specified columns. You could also
do a right outer join and left outer join to select the rows that have no matching values in the other related table.
I N T R O D U C T I O N
K e y
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
T e r m s
alias, 246
DISTINCT, 261
NOT, 248
ALTER TABLE, 253
DROP INDEX, 236
OR, 247
AND, 248
DROP TABLE, 259
ORDER BY, 259
authentication, 225
EXISTS, 249
recursive query, 273
AVG, 265
GROUP BY, 266
reserved words, 231
base tables, 269
HAVING, 268
ROLLBACK, 240
BETWEEN, 249
IN, 249
rules of precedence, 247
Boolean algebra, 248
inner query, 242
schema, 225
cascading order sequence, 260
INSERT, 237
SELECT, 238
COMMIT, 238
IS NULL, 249
subquery, 242
COUNT, 262
LIKE, 249
SUM, 265
CREATE INDEX, 235
MAX, 263
UPDATE, 240
CREATE TABLE, 229
MIN, 263
view, 269
CREATE VIEW, 269
nested query, 242
wildcard character, 239
DELETE, 241
Online Content
Answers to selected Review Questions and Problems for this chapter are contained in the Premium Website for
this book.
R e v i e w
Q u e s t i o n s
1. In a SELECT query, what is the difference between a WHERE clause and a HAVING clause?
2. Explain why the following command would create an error and what changes could be made to fix the error.
SELECT V_CODE, SUM(P_QOH) FROM PRODUCT;
3. What type of integrity is enforced when a primary key is declared?
4. Explain why it might be more appropriate to declare an attribute that contains only digits as a character data type
instead of a numeric data type.
5. What is the difference between a column constraint and a table constraint?
6. What are “referential constraint actions”?
7. Rewrite the following WHERE clause without the use of the IN special operator.
WHERE V_STATE IN ('TN', 'FL', 'GA')
8. Explain the difference between an ORDER BY clause and a GROUP BY clause.
9. Explain why the two following commands produce different results:
SELECT DISTINCT COUNT (V_CODE) FROM PRODUCT;
SELECT COUNT (DISTINCT V_CODE) FROM PRODUCT;
10. What is the difference between the COUNT aggregate function and the SUM aggregate function?
11. Explain why it would be preferable to use a DATE data type to store date data instead of a character data type.
12. What is the difference between an inner join and an outer join?
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Online Content
Problems 1−25 are based on the Ch07_ConstructCo database located in the Premium Website. This database
is stored in Microsoft Access format. If you use another DBMS such as Oracle, SQL Server, MySQL, or DB2, use
its import utilities to import the Access database contents. The Premium Website provides Oracle and SQL
script files.
P r o b l e m s
The Ch07_ConstructCo database stores data for a consulting company that tracks all charges to projects. The
charges are based on the hours each employee works on each project. The structure and contents of the
Ch07_ConstructCo database are shown in Figure P7.1.
FIGURE
The Ch07_ConstructCo database
P7.1
Relational diagram
Table name: EMPLOYEE
Table name: JOB
Table name: ASSIGNMENT
Table name: PROJECT
Database name: Ch07_ConstructCo
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
Note that the ASSIGNMENT table in Figure P7.1 stores the JOB_CHG_HOUR values as an attribute (ASSIGN_
CHG_HR) to maintain historical accuracy of the data. The JOB_CHG_HOUR values are likely to change over time.
In fact, a JOB_CHG_HOUR change will be reflected in the ASSIGNMENT table. And, naturally, the employee primary
job assignment might change, so the ASSIGN_JOB is also stored. Because those attributes are required to maintain
the historical accuracy of the data, they are not redundant.
Given the structure and contents of the Ch07_ConstructCo database shown in Figure P7.1, use SQL commands to
answer Problems 1–25.
1. Write the SQL code that will create the table structure for a table named EMP_1. This table is a subset of the
EMPLOYEE table. The basic EMP_1 table structure is summarized in the following table. (Note that the
JOB_CODE is the FK to JOB.)
ATTRIBUTE (FIELD)
NAME
EMP_NUM
EMP_LNAME
EMP_FNAME
EMP_INITIAL
EMP_HIREDATE
JOB_CODE
DATA
DECLARATION
CHAR(3)
VARCHAR(15)
VARCHAR(15)
CHAR(1)
DATE
CHAR(3)
2. Having created the table structure in Problem 1, write the SQL code to enter the first two rows for the table
shown in Figure P7.2.
FIGURE
The contents of the EMP_1 table
P7.2
3. Assuming that the data shown in the EMP_1 table have been entered, write the SQL code that will list all
attributes for a job code of 502.
4. Write the SQL code that will save the changes made to the EMP_1 table.
5. Write the SQL code to change the job code to 501 for the person whose employee number (EMP_NUM) is 107.
After you have completed the task, examine the results, and then reset the job code to its original value.
6. Write the SQL code to delete the row for the person named William Smithfield, who was hired on June 22,
2004, and whose job code classification is 500. (Hint: Use logical operators to include all of the information
given in this problem.)
7. Write the SQL code that will restore the data to its original status; that is, the table should contain the data that
existed before you made the changes in Problems 5 and 6.
279
280
C H A P T E R
7
8. Write the SQL code to create a copy of EMP_1, naming the copy EMP_2. Then write the SQL code that will
add the attributes EMP_PCT and PROJ_NUM to its structure. The EMP_PCT is the bonus percentage to be paid
to each employee. The new attribute characteristics are:
EMP_PCTNUMBER(4,2)
PROJ_NUMCHAR(3)
(Note: If your SQL implementation allows it, you may use DECIMAL(4,2) rather than NUMBER(4,2).)
9. Write the SQL code to change the EMP_PCT value to 3.85 for the person whose employee number (EMP_NUM)
is 103. Next, write the SQL command sequences to change the EMP_PCT values as shown in Figure P7.9.
FIGURE
The EMP_2 table after the modifications
P7.9
10. Using a single command sequence, write the SQL code that will change the project number (PROJ_NUM) to 18
for all employees whose job classification (JOB_CODE) is 500.
11. Using a single command sequence, write the SQL code that will change the project number (PROJ_NUM) to 25
for all employees whose job classification (JOB_CODE) is 502 or higher. When you finish Problems 10 and 11,
the EMP_2 table will contain the data shown in Figure P7.11. (You may assume that the table has been saved
again at this point.)
FIGURE
The EMP_2 table contents after the modifications
P7.11
12. Write the SQL code that will change the PROJ_NUM to 14 for those employees who were hired before January
1, 1994 and whose job code is at least 501. (You may assume that the table will be restored to its condition
preceding this question.)
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
13. Write the two SQL command sequences required to:
a. Create a temporary table named TEMP_1 whose structure is composed of the EMP_2 attributes EMP_NUM
and EMP_PCT.
b. Copy the matching EMP_2 values into the TEMP_1 table.
14. Write the SQL command that will delete the newly created TEMP_1 table from the database.
15. Write the SQL code required to list all employees whose last names start with Smith. In other words, the rows
for both Smith and Smithfield should be included in the listing. Assume case sensitivity.
16. Using the EMPLOYEE, JOB, and PROJECT tables in the Ch07_ConstructCo database (see Figure P7.1), write
the SQL code that will produce the results shown in Figure P7.16.
FIGURE
The query results for Question 16
P7.16
17. Write the SQL code that will produce a virtual table named REP_1. The virtual table should contain the same
information that was shown in Problem 16.
18. Write the SQL code to find the average bonus percentage in the EMP_2 table you created in Problem 8.
19. Write the SQL code that will produce a listing for the data in the EMP_2 table in ascending order by the bonus
percentage.
20. Write the SQL code that will list only the distinct project numbers found in the EMP_2 table.
21. Write the SQL code to calculate the ASSIGN_CHARGE values in the ASSIGNMENT table in the Ch07_
ConstructCo database. (See Figure P7.1.) Note that ASSIGN_CHARGE is a derived attribute that is calculated
by multiplying ASSIGN_CHG_HR by ASSIGN_HOURS.
22. Using the data in the ASSIGNMENT table, write the SQL code that will yield the total number of hours worked
for each employee and the total charges stemming from those hours worked. The results of running that query
are shown in Figure P7.22.
FIGURE
Total hours and charges by employee
P7.22
23. Write a query to produce the total number of hours and charges for each of the projects represented in the
ASSIGNMENT table. The output is shown in Figure P7.23.
281
282
C H A P T E R
FIGURE
P7.23
FIGURE
P7.24
7
Total hour and charges by
project
Total hours and charges, all
employees
24. Write the SQL code to generate the total hours worked
and the total charges made by all employees. The
results are shown in Figure P7.24. (Hint: This is a
nested query. If you use Microsoft Access, you can
generate the result by using the query output shown in
Figure P7.22 as the basis for the query that will
produce the output shown in Figure P7.24.)
25. Write the SQL code to generate the total hours worked
and the total charges made to all projects. The results
should be the same as those shown in Figure P7.24.
(Hint: This is a nested query. If you use Microsoft
Access, you can generate the result by using the query
output shown in Figure P7.23 as the basis for this
query.)
The structure and contents of the Ch07_SaleCo
database are shown in Figure P7.26. Use this database
to answer the following problems. Save each query as
QXX, where XX is the problem number.
Online Content
Problems 26−43 are based on the Ch07_SaleCo database located in the Premium Website. This database is
stored in Microsoft Access format. If you use another DBMS such as Oracle, SQL Server, MySQL, or DB2, use
its import utilities to import the Access database contents. The Premium Website provides Oracle and SQL
script files.
26. Write a query to count the number of invoices.
27. Write a query to count the number of customers with a customer balance over $500.
28. Generate a listing of all purchases made by the customers, using the output shown in Figure P7.28 as your guide.
(Hint: Use the ORDER BY clause to order the resulting rows shown in Figure P7.28.)
I N T R O D U C T I O N
FIGURE
T O
S T R U C T U R E D
Q U E R Y
The Ch07_SaleCo database
P7.26
Relational diagram
Table name: CUSTOMER
Table name: INVOICE
FIGURE
P7.28
Table name: LINE
List of customer purchases
Table name: VENDOR
Table name: PRODUCT
L A N G U A G E
( S Q L )
283
284
C H A P T E R
7
29. Using the output shown in Figure P7.29 as your guide, generate a list of customer purchases, including the
subtotals for each of the invoice line numbers. (Hint: Modify the query format used to produce the list of customer
purchases in Problem 28, delete the INV_DATE column, and add the derived (computed) attribute LINE_UNITS
* LINE_PRICE to calculate the subtotals.)
FIGURE
Summary of customer purchases with subtotals
P7.29
30. Modify the query used in Problem 29 to produce the summary shown in Figure P7.30.
FIGURE
P7.30
FIGURE
P7.31
Customer purchase
summary
31. Modify the query in Problem 30 to include the number of individual
product purchases made by each customer. (In other words, if the
customer’s invoice is based on three products, one per LINE_
NUMBER, you count three product purchases. Note that in the original
invoice data, customer 10011 generated three invoices, which contained a total of six lines, each representing a product purchase.) Your
output values must match those shown in Figure P7.31.
32. Use a query to compute the average purchase amount per product
made by each customer. (Hint: Use the results of Problem 31 as the
basis for this query.) Your output values must match
those shown in Figure P7.32. Note that the average
Customer total purchase
purchase amount is equal to the total purchases divided
amounts and number of
purchases
by the number of purchases.
33. Create a query to produce the total purchase per
invoice, generating the results shown in Figure P7.33.
The invoice total is the sum of the product purchases in
the LINE that corresponds to the INVOICE.
I N T R O D U C T I O N
FIGURE
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
Average purchase amount by customer
P7.32
34. Use a query to show the invoices and invoice totals as shown in Figure P7.34. (Hint: Group by the CUS_CODE.)
FIGURE
Invoice totals
FIGURE
P7.33
P7.34
Invoice totals by
customer
35. Write a query to produce the number of invoices and the total purchase amounts by customer, using the output
shown in Figure P7.35 as your guide. (Compare this summary to the results shown in Problem 34.)
36. Using the query results in Problem 35 as your basis, write a query to generate the total number of invoices, the
invoice total for all of the invoices, the smallest invoice amount, the largest invoice amount, and the average of
all of the invoices. (Hint: Check the figure output in Problem 35.) Your output must match Figure P7.36.
FIGURE
P7.35
Number of invoices and total
purchase amounts by customer
FIGURE
P7.36
Number of invoices, invoice
totals, minimum, maximum,
and average sales
285
286
C H A P T E R
7
37. List the balance characteristics of the customers who have made purchases during the current invoice cycle—that
is, for the customers who appear in the INVOICE table. The results of this query are shown in Figure P7.37.
FIGURE
P7.37
Balances for
customers who
made purchases
FIGURE
P7.38
Balance summary of customers
who made purchases
38. Using the results of the query created in Problem 37, provide a summary of the customer balance characteristics
as shown in Figure P7.38.
39. Create a query to find the customer balance characteristics for all customers, including the total of the outstanding
balances. The results of this query are shown in Figure P7.39.
40. Find the listing of customers who did not make purchases during the invoicing period. Your output must match
the output shown in Figure P7.40.
FIGURE
P7.39
Balance summary for all
customers
FIGURE
P7.40
Balances of
customers who did
not make purchases
41. Find the customer balance summary for all customers who have not made purchases during the current invoicing
period. The results are shown in Figure P7.41.
FIGURE
P7.41
Summary of customer balances for customers who did not
make purchases
42. Create a query to produce the summary of the value of products currently in inventory. Note that the value of
each product is produced by the multiplication of the units currently in inventory and the unit price. Use the
ORDER BY clause to match the order shown in Figure P7.42.
43. Using the results of the query created in Problem 42, find the total value of the product inventory. The results
are shown in Figure P7.43.
I N T R O D U C T I O N
FIGURE
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
Value of products currently in inventory
P7.42
FIGURE
P7.43
Total value of all
products in
inventory
C a s e s
TinyVideo is a small movie rental company with a single store. TinyVideo needs a database system to track the rental
of movies to its members. TinyVideo can own several copies (VIDEO) of each movie (MOVIE). For example, the store
may have 10 copies of the movie “Twist in the Wind.” “Twist in the Wind” would be one MOVIE, and each copy would
be a VIDEO. A rental transaction (RENTAL) involves one or more videos being rented to a member (MEMBERSHIP).
A video can be rented many times over its lifetime; therefore, there is a M:N relationship between RENTAL
and VIDEO. DETAILRENTAL is the bridge table to resolve this relationship. The complete ERD is provided in
Figure P7.44.
44. Write the SQL code to create the table structures for the entities shown in Figure P7.44. The structures should
contain the attributes specified in the ERD. Use data types that are appropriate for the data that will need to be
stored in each attribute. Enforce primary key and foreign key constraints as indicated by the ERD.
287
288
C H A P T E R
FIGURE
7
The Ch07_MovieCo ERD
P7.44
45. The following tables provide a very small portion of the data that will be kept in the database. This data needs
to be inserted into the database for testing purposes. Write the INSERT commands necessary to place the
following data in the tables that were created in Problem 1.
MEMBERSHIP
MEM_ MEM_
NUM
FNAME
102
Tami
103
Curt
104
Jamal
105
Iva
106
Miranda
107
Rosario
108
Mattie
109
Clint
110
Lewis
111
Stacy
112
Luis
113
Minnie
MEM_
LNAME
Dawson
Knight
Melendez
Mcclain
Parks
Elliott
Guy
Ochoa
Rosales
Mann
Trujillo
Gonzales
MEM_STREET
MEM_CITY
2632 Takli Circle
4025 Cornell Court
788 East 145th Avenue
6045 Musket Ball Circle
4469 Maxwell Place
7578 Danner Avenue
4390 Evergreen Street
1711 Elm Street
4524 Southwind Circle
2789 East Cook Avenue
7267 Melvin Avenue
6430 Vasili Drive
Norene
Flatgap
Quebeck
Summit
Germantown
Columbia
Lily
Greeneville
Counce
Murfreesboro
Heiskell
Williston
MEM_
STATE
TN
KY
TN
KY
TN
TN
KY
TN
TN
TN
TN
TN
MEM_ZIP
37136
41219
38579
42783
38183
38402
40740
37745
38326
37132
37754
38076
MEM_
BALANCE
11
6
0
15
0
5
0
10
0
8
3
0
I N T R O D U C T I O N
RENTAL
RENT_NUM
1001
1002
1003
1004
1005
1006
1007
1008
1009
DETAILRENTAL
RENT_
VID_NUM
NUM
1001
34342
1001
61353
1002
59237
1003
54325
1003
61369
1003
61388
1004
44392
1004
34367
1004
34341
1005
34342
1005
44397
1006
34366
1006
61367
1007
34368
1008
34369
1009
54324
1001
34366
RENT_DATE
01-MAR-09
01-MAR-09
02-MAR-09
02-MAR-09
02-MAR-09
02-MAR-09
02-MAR-09
03-MAR-09
03-MAR-09
DETAIL_FEE
2
2
3.5
3.5
2
0
3.5
3.5
2
2
3.5
3.5
2
3.5
3.5
3.5
3.5
DETAIL_
DUEDATE
04-MAR-09
04-MAR-09
04-MAR-09
04-MAR-09
06-MAR-09
06-MAR-09
05-MAR-09
05-MAR-09
07-MAR-09
07-MAR-09
05-MAR-09
05-MAR-09
07-MAR-09
05-MAR-09
05-MAR-09
05-MAR-09
04-MAR-09
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
MEM_NUM
103
105
102
110
111
107
104
105
111
DETAIL_
RETURNDATE
02-MAR-09
03-MAR-09
04-MAR-09
09-MAR-09
09-MAR-09
09-MAR-09
07-MAR-09
07-MAR-09
07-MAR-09
05-MAR-09
05-MAR-09
04-MAR-09
05-MAR-09
02-MAR-09
DETAIL_
DAILYLATEFEE
1
1
3
3
1
1
3
3
1
1
3
3
1
3
3
3
3
( S Q L )
289
290
C H A P T E R
7
VIDEO
VID_NUM
54321
54324
54325
34341
34342
34366
34367
34368
34369
44392
44397
59237
61388
61353
61354
61367
61369
VID_INDATE
18-JUN-08
18-JUN-08
18-JUN-08
22-JAN-07
22-JAN-07
02-MAR-09
02-MAR-09
02-MAR-09
02-MAR-09
21-OCT-08
21-OCT-08
14-FEB-09
25-JAN-07
28-JAN-06
28-JAN-06
30-JUL-08
30-JUL-08
MOVIE_NUM
1234
1234
1234
1235
1235
1236
1236
1236
1236
1237
1237
1237
1239
1245
1245
1246
1246
MOVIE
MOVIE_NUM
1234
1235
1236
1237
1238
1239
1245
1246
PRICE
PRICE_CODE
1
2
3
4
MOVIE_NAME
The Cesar Family
Christmas
Smokey Mountain
Wildlife
Richard Goodhope
Beatnik Fever
Constant Companion
Where Hope Dies
Time to Burn
What He Doesn't Know
PRICE_DESCRIPTION
Standard
New Release
Discount
Weekly Special
MOVIE_YEAR
MOVIE_
COST
MOVIE_GENRE
PRICE_CODE
2007
39.95
FAMILY
2
2004
59.95
ACTION
1
2008
2007
2008
1998
2005
2006
59.95
29.95
89.95
25.49
45.49
58.29
DRAMA
COMEDY
DRAMA
DRAMA
ACTION
COMEDY
2
2
2
3
1
1
PRICE_RENTFEE
2
3.5
1.5
1
PRICE_DAILYLATEFEE
1
3
1
.5
For Questions 46–77, use the tables that were created in Problem 44 and the data that was loaded into those
tables in Problem 45.
46. Write the SQL command to save the rows inserted in Problem 45.
47. Write the SQL command to change the movie year for movie number 1245 to 2006.
48. Write the SQL command to change the price code for all Action movies to price code 3.
49. Write a single SQL command to increase all price rental fee values by $0.50.
50. Write the SQL command to save the changes made to the PRICE and MOVIE tables in Problems 45–49.
I N T R O D U C T I O N
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
51. Write a query to display the movie title, movie year, and movie genre for all movies. (The results are shown in
Figure P7.51.)
52. Write a query to display the movie year, movie title, and movie cost sorted by movie year in descending order.
(The results are shown in Figure P7.52.)
FIGURE
All movies
FIGURE
Movies by year
P7.52
P7.51
53. Write a query to display the movie title, movie year, and movie genre for all movies sorted by movie genre in
ascending order, then sorted by movie year in descending order within genre. (The results are shown in Figure
P7.53.)
54. Write a query to display the movie number, movie title, and price code for all movies with a title that starts with
the letter “R.” (The results are shown in Figure P7.54.)
FIGURE
P7.53
Movies with
multicolumn sort
FIGURE
P7.54
Movies starting
with R
55. Write a query to display the movie title, movie year, and movie cost for all movies that contain the word “hope”
anywhere in the title. Sort the results in ascending order by title. (The results are shown in figure P7.55.)
56. Write a query to display the movie title, movie year, and movie genre for all action movies. (The results are shown
in Figure P7.56.)
FIGURE
P7.55
Movies with
“Hope” in the title
FIGURE
Action movies
P7.56
57. Write a query to display the movie number, movie title, and movie cost for all movies with a cost greater than $40.
(The results are shown in Figure P7.57.)
291
292
C H A P T E R
7
58. Write a query to display the movie number, movie title, movie cost, and movie genre for all movies that are either
action or comedy movies and that have a cost that is less than $50. Sort the results in ascending order by genre.
(The results are shown in Figure P7.58.)
FIGURE
P7.57
Movies costing
less than $40
FIGURE
P7.58
Action or comedy
movies less than $50
59. Write a query to display the movie number, and movie description for all movies where the movie description is
a combination of the movie title, movie year, and movie genre with the movie year enclosed in parentheses. (The
results are shown in Figure P7.59.)
60. Write a query to display the movie genre and the number of movies in each genre. (The results are shown in
Figure P7.60.)
FIGURE
P7.59
Movies with concatenated
descriptions
FIGURE
Number of movies in genre
P7.60
61. Write a query to display the average cost of all of the movies. (The results are shown in Figure P7.61.)
62. Write a query to display the movie genre and average cost of movies in each genre. (The results are shown in
Figure P7.62.)
FIGURE
P7.61
Average movie cost
FIGURE
Average cost by genre
P7.62
63. Write a query to display the movie title, movie genre, price description, and price rental fee for all movies with
a price code. (The results are shown in Figure P7.63.)
64. Write a query to display the movie genre and average price rental fee for movies in each genre that have a price.
(The results are shown in Figure P7.64.)
I N T R O D U C T I O N
FIGURE
Rental fees for movies
P7.63
T O
S T R U C T U R E D
FIGURE
Q U E R Y
L A N G U A G E
( S Q L )
Average rental fee by genre
P7.64
65. Write a query to display the movie title, movie year, and the movie cost divided by the price rental fee for each
movie that has a price to determine the number of rentals it will take to break even on the purchase of the movie.
(The results are shown in Figure P7.65.)
66. Write a query to display the movie title and movie year for all movies that have a price code. (The results are
shown in Figure P7.66.)
FIGURE
Breakeven rentals
P7.65
FIGURE
Movies with a price
P7.66
67. Write a query to display the movie title, movie year, and movie cost for all movies that have a cost between
$44.99 and $49.99. (The results are shown in Figure P7.67.)
68. Write a query to display the movie title, movie year, price description, and price rental fee for all movies that are
in the genres family, comedy, or drama. (The results are shown in Figure P7.68.)
FIGURE
Movies costs within a range
P7.67
FIGURE
P7.68
Movies within specific genres
293
294
C H A P T E R
7
69. Write a query to display the movie number, movie title, and movie year for all movies that do not have a video.
(The results are shown in Figure P7.69.)
70. Write a query to display the membership number, first name, last name, and balance of the memberships that
have a rental. (The results are shown in Figure P7.70.)
FIGURE
Movies without videos
P7.69
FIGURE
P7.70
Balances of memberships
with rentals
71. Write a query to display the minimum balance, maximum balance, and average balance for memberships that
have a rental. (The results are shown in Figure P7.71.)
72. Write a query to display the membership name (concatenate the first name and last name with a space between
them into a single column), membership address (concatenate the street, city, state, and zip codes into a single
column with spaces. (The results are shown in Figure P7.72.)
FIGURE
P7.71
Minimum, maximum, and
average balances
FIGURE
P7.72
Concatenated
membership data
73. Write a query to display the rental number, rental date, video number, movie title, due date, and return date for
all videos that were returned after the due date. Sort the results by rental number and movie title. (The results are
shown in Figure P7.73.)
I N T R O D U C T I O N
FIGURE
T O
S T R U C T U R E D
Q U E R Y
L A N G U A G E
( S Q L )
Late video returns
P7.73
74. Write a query to display the rental number, rental date, video number, movie title, due date, return date, detail
fee, and number of days past the due date that the video was returned for each video that was returned after the
due date. Sort the results by rental number and movie title. (The results are shown in Figure P7.74.)
FIGURE
Number of days late
P7.74
75. Write a query to display the rental number, rental date, movie title, and detail fee for each movie that was returned
on or before the due date. (The results are shown in Figure P7.75.)
FIGURE
P7.75
Actual rental fees charged
295
296
C H A P T E R
7
76. Write a query to display the membership number, last name, first name, and total rental fees earned from that
membership. (The results are shown in Figure P7.76.) The total rental fee is the sum of all of the detail fees
(without the late fees) from all movies that the membership has rented.
FIGURE
Total rental fees paid by membership
P7.76
77. Write a query to display the movie number, movie genre, average movie cost of movies in that genre, movie cost
of that individual movie, and the percentage difference between the average movie cost and the individual movie
cost. (The results are shown in Figure P7.77.) (Note: The percentage difference is calculated as the cost of the
individual movie minus the average cost of movies in that genre, divided by the average cost of movies in that
genre multiplied by 100. For example, if the average cost of movies in the “family” genre is $25, if a given family
movie cost $26, then the calculation would be ((26 – 25) / 25 * 100), which would work out to be 4.00%. This
indicates that this movie costs 4% more than the average family movie.)
FIGURE
P7.77
Movie differences from genre average
In this chapter, you will learn:
쐍 About the relational set operators UNION, UNION ALL, INTERSECT, and MINUS
쐍 How to use the advanced SQL JOIN operator syntax
쐍 About the different types of subqueries and correlated queries
쐍 How to use SQL functions to manipulate dates, strings, and other data
쐍 How to create and use updatable views
쐍 How to create and use triggers and stored procedures
쐍 How to create embedded SQL
In Chapter 7, Introduction to Structured Query Language (SQL), you learned the basic SQL
data definition and data manipulation commands used to create and manipulate relational
data. In this chapter, you build on what you learned in Chapter 7 and learn how to use more
P
review
advanced SQL features.
In this chapter, you will learn about the SQL relational set operators (UNION, INTERSECT,
and MINUS) and how those operators are used to merge the results of multiple queries.
Joins are at the heart of SQL, so you must learn how to use the SQL JOIN statement to
extract information from multiple tables. In the previous chapter, you learned how cascading
queries inside other queries can be useful in certain circumstances. In this chapter, you will
also learn about the different styles of subqueries that can be implemented in a SELECT
statement. Finally, you will learn more of SQL’s many functions to extract information from
data, including manipulation of dates and strings and computations based on stored or even
derived data.
In the real world, business procedures require the execution of clearly defined actions when
a specific event occurs, such as the addition of a new invoice or a student’s enrollment in
a class. Such procedures can be applied within the DBMS through the use of triggers and
stored procedures. In addition, SQL facilitates the application of business procedures when
it is embedded in a programming language such as Visual Basic .NET, C#, or COBOL.
8
E I G H T
Advanced SQL
298
C H A P T E R
8
Online Content
Although most of the examples used in this chapter are shown in Oracle, you could also use MS SQL Server. The
Premium Website for this book provides you with the ADVSQLDBINIT.SQL script file (Oracle and MS SQL
versions) to create the tables and load the data used in this chapter. There you will also find additional SQL script
files to demonstrate each of the commands shown in this chapter.
8.1 RELATIONAL SET OPERATORS
In Chapter 3, The Relational Database Model, you learned about the eight general relational operators. In this section,
you will learn how to use three SQL commands (UNION, INTERSECT, and MINUS) to implement the union,
intersection, and difference relational operators.
In previous chapters, you learned that SQL data manipulation commands are set-oriented; that is, they operate over
entire sets of rows and columns (tables) at once. Using sets, you can combine two or more sets to create new sets (or
relations). That’s precisely what the UNION, INTERSECT, and MINUS statements do. In relational database terms, you
can use the words “sets,” “relations,” and “tables” interchangeably because they all provide a conceptual view of the
data set as it is presented to the relational database user.
Note
The SQL standard defines the operations that all DBMSs must perform on data, but it leaves the implementation
details to the DBMS vendors. Therefore, some advanced SQL features might not work on all DBMS
implementations. Also, some DBMS vendors might implement additional features not found in the SQL standard.
UNION, INTERSECT, and MINUS are the names of the SQL statements implemented in Oracle. The SQL
standard uses the keyword EXCEPT to refer to the difference (MINUS) relational operator. Other RDBMS vendors
might use a different command name or might not implement a given command at all. To learn more about the
ANSI/ISO SQL standards, check the ANSI Web site (www.ansi.org) to find out how to obtain the latest standard
documents in electronic form. As of this writing, the most recent fully approved standard is SQL-2003. The
SQL-2003 standard made revisions and additions to the previous standard; most notable is support for XML data.
The SQL-2006 standard extended support for XML and multimedia data. The SQL-2008 standard added INSTEAD
OF triggers and the TRUNCATE statement.
UNION, INTERSECT, and MINUS work properly only if relations are union-compatible, which means that the
number of attributes must be the same and their corresponding data types must be alike. In practice, some RDBMS
vendors require the data types to be “compatible” but not necessarily “exactly the same.” For example, compatible data
types are VARCHAR (35) and CHAR (15). In that case, both attributes store character (string) values; the only
difference is the string size. Another example of compatible data types is NUMBER and SMALLINT. Both data types
are used to store numeric values.
Note
Some DBMS products might require union-compatible tables to have identical data types.
A D V A N C E D
S Q L
Online Content
The Premium Website for this book provides SQL script files (Oracle and MS SQL Server) to demonstrate the
UNION, INTERSECT, and MINUS commands. It also provides the Ch08_SaleCo MS Access database containing
supported set operator alternative queries.
8.1.1 UNION
Suppose SaleCo has bought another company. SaleCo’s management wants to make sure that the acquired company’s
customer list is properly merged with SaleCo’s customer list. Because it is quite possible that some customers have
purchased goods from both companies, the two lists might contain common customers. SaleCo’s management wants to
make sure that customer records are not duplicated when the two customer lists are merged. The UNION query is a perfect
tool for generating a combined listing of customers—one that excludes duplicate records.
The UNION statement combines rows from two or more queries without including duplicate rows. The syntax of the
UNION statement is:
query UNION query
In other words, the UNION statement combines the output of two SELECT queries. (Remember that the SELECT
statements must be union-compatible. That is, they must return the same number of attributes and similar data types.)
To demonstrate the use of the UNION statement in SQL, let’s use the CUSTOMER and CUSTOMER_2 tables in the
Ch08_SaleCo database. To show the combined CUSTOMER and CUSTOMER_2 records without the duplicates, the
UNION query is written as follows:
SELECT
FROM
UNION
SELECT
FROM
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER_2;
Figure 8.1 shows the contents of the CUSTOMER and CUSTOMER_2 tables and the result of the UNION query.
Although MS Access is used to show the results here, similar results can be obtained with Oracle.
Note the following in Figure 8.1:
쐌
The CUSTOMER table contains 10 rows, while the CUSTOMER_2 table contains 7 rows.
쐌
Customers Dunne and Olowski are included in the CUSTOMER table as well as in the CUSTOMER_2 table.
쐌
The UNION query yields 15 records because the duplicate records of customers Dunne and Olowski are not
included. In short, the UNION query yields a unique set of records.
The UNION statement can be used to unite more than just two queries. For example, assume that you have four
union-compatible queries named T1, T2, T3, and T4. With the UNION statement, you can combine the output of all
four queries into a single result set. The SQL statement will be similar to this:
SELECT
UNION
SELECT
UNION
SELECT
UNION
SELECT
column-list FROM T1
column-list FROM T2
column-list FROM T3
column-list FROM T4;
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300
C H A P T E R
8
FIGURE
UNION query results
8.1
Database name: CH08_SaleCo
Table name: CUSTOMER
Query name: qryUNION-of-CUSTOMER-and-CUSTOMER_2
Table name: CUSTOMER_2
Note
The SQL-2003 standard calls for the elimination of duplicate rows when the UNION SQL statement is used.
However, some DBMS vendors might not adhere to that standard. Check your DBMS manual to see if the
UNION statement is supported and if so, how it is supported.
8.1.2 UNION ALL
If SaleCo’s management wants to know how many customers are on both the CUSTOMER and CUSTOMER_2 lists,
a UNION ALL query can be used to produce a relation that retains the duplicate rows. Therefore, the following query
will keep all rows from both queries (including the duplicate rows) and return 17 rows.
SELECT
FROM
UNION ALL
SELECT
FROM
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER_2;
Running the preceding UNION ALL query produces the result shown in Figure 8.2.
Like the UNION statement, the UNION ALL statement can be used to unite more than just two queries.
8.1.3 INTERSECT
If SaleCo’s management wants to know which customer records are duplicated in the CUSTOMER and
CUSTOMER_2 tables, the INTERSECT statement can be used to combine rows from two queries, returning only the
rows that appear in both sets. The syntax for the INTERSECT statement is:
query INTERSECT query
A D V A N C E D
FIGURE
S Q L
UNION ALL query results
8.2
Database name: CH08_SaleCo
Table name: CUSTOMER
Query name: qryUNION-ALL-of-CUSTOMER-and-CUSTOMER_2
Table name: CUSTOMER_2
To generate the list of duplicate customer records, you can use:
SELECT
FROM
INTERSECT
SELECT
FROM
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER_2;
The INTERSECT statement can be used to generate additional useful customer information. For example, the
following query returns the customer codes for all customers who are located in area code 615 and who have made
purchases. (If a customer has made a purchase, there must be an invoice record for that customer.)
SELECT
INTERSECT
SELECT
CUS_CODE FROM CUSTOMER WHERE CUS_AREACODE = '615'
DISTINCT CUS_CODE FROM INVOICE;
Figure 8.3 shows both sets of SQL statements and their output.
8.1.4 MINUS
The MINUS statement in SQL combines rows from two queries and returns only the rows that appear in the first set
but not in the second. The syntax for the MINUS statement is:
query MINUS query
For example, if the SaleCo managers want to know what customers in the CUSTOMER table are not found in the
CUSTOMER_2 table, they can use:
SELECT
FROM
MINUS
SELECT
FROM
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER_2;
301
302
C H A P T E R
FIGURE
8
INTERSECT query results
8.3
Note
MS Access does not support the INTERSECT query, nor does it support other complex queries you will explore
in this chapter. At least in some cases, Access might be able to give you the desired results if you use an
alternative query format or procedure. For example, although Access does not support SQL triggers and stored
procedures, you can use Visual Basic code to perform similar actions. However, the objective here is to show
you how some important standard SQL features may be used.
If the managers want to know what customers in the CUSTOMER_2 table are not found in the CUSTOMER table,
they merely switch the table designations:
SELECT
FROM
MINUS
SELECT
FROM
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER_2
CUS_LNAME, CUS_FNAME, CUS_INITIAL, CUS_AREACODE, CUS_PHONE
CUSTOMER;
You can extract much useful information by combining MINUS with various clauses such as WHERE. For example, the
following query returns the customer codes for all customers located in area code 615 minus the ones who have made
purchases, leaving the customers in area code 615 who have not made purchases.
SELECT
MINUS
SELECT
CUS_CODE FROM CUSTOMER WHERE CUS_AREACODE = '615'
DISTINCT CUS_CODE FROM INVOICE;
Figure 8.4 shows the preceding three SQL statements and their output.
A D V A N C E D
FIGURE
S Q L
MINUS query results
8.4
Note
Some DBMS products do not support the INTERSECT or MINUS statements, while others might implement the
difference relational operator in SQL as EXCEPT. Consult your DBMS manual to see if the statements illustrated
here are supported by your DBMS.
8.1.5 Syntax Alternatives
If your DBMS doesn’t support the INTERSECT or MINUS statements, you can use the IN and NOT IN subqueries to
obtain similar results. For example, the following query will produce the same results as the INTERSECT query shown
in Section 8.1.3:
SELECT
WHERE
CUS_CODE FROM CUSTOMER
CUS_AREACODE = '615' AND
CUS_CODE IN (SELECT DISTINCT CUS_CODE FROM INVOICE);
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304
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8
Figure 8.5 shows the use of the INTERSECT alternative.
FIGURE
INTERSECT alternative
8.5
Database name: CH08_SaleCo
Table name: CUSTOMER
Table name: INVOICE
Query name: qry-INTERSECT-Alternative
Note
MS Access will generate an input request for the CUS_AREACODE if you use apostrophes around the area code.
(If you supply the 615 area code, the query will execute properly.) You can eliminate that problem by using standard
double quotation marks, writing the WHERE clause in the second line of the preceding SQL statement as:
WHERE CUS_AREACODE = “615” AND
MS Access will also accept single quotation marks.
Using the same alternative to the MINUS statement, you can generate the output for the third MINUS query shown
in Section 8.1.4 by using:
SELECT
WHERE
CUS_CODE FROM CUSTOMER
CUS_AREACODE = '615' AND
CUS_CODE NOT IN (SELECT DISTINCT CUS_CODE FROM INVOICE);
The results of that query are shown in Figure 8.6. Note that the query output includes only the customers in area code
615 who have not made any purchases and, therefore, have not generated invoices.
A D V A N C E D
FIGURE
S Q L
MINUS alternative
8.6
Database name: CH08_SaleCo
Table name: CUSTOMER
Table name: INVOICE
Query name: qry-MINUS-Alternative
8.2 SQL JOIN OPERATORS
The relational join operation merges rows from two tables and returns the rows with one of the following conditions:
쐌
Have common values in common columns (natural join).
쐌
Meet a given join condition (equality or inequality).
쐌
Have common values in common columns or have no matching values (outer join).
In Chapter 7, you learned how to use the SELECT statement in conjunction with the WHERE clause to join two or
more tables. For example, you can join the PRODUCT and VENDOR tables through their common V_CODE by
writing:
SELECT
FROM
WHERE
P_CODE, P_DESCRIPT, P_PRICE, V_NAME
PRODUCT, VENDOR
PRODUCT.V_CODE = VENDOR.V_CODE;
The preceding SQL join syntax is sometimes referred to as an “old-style” join. Note that the FROM clause contains
the tables being joined and that the WHERE clause contains the condition(s) used to join the tables.
Note the following points about the preceding query:
쐌
The FROM clause indicates which tables are to be joined. If three or more tables are included, the join
operation takes place two tables at a time, from left to right. For example, if you are joining tables T1, T2, and
T3, the first join is table T1 with T2; the results of that join are then joined to table T3.
쐌
The join condition in the WHERE clause tells the SELECT statement which rows will be returned. In this case,
the SELECT statement returns all rows for which the V_CODE values in the PRODUCT and VENDOR tables
are equal.
쐌
The number of join conditions is always equal to the number of tables being joined minus one. For example,
if you join three tables (T1, T2, and T3), you will have two join conditions (j1 and j2). All join conditions are
connected through an AND logical operator. The first join condition (j1) defines the join criteria for T1 and T2.
The second join condition (j2) defines the join criteria for the output of the first join and T3.
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쐌
8
Generally, the join condition will be an equality comparison of the primary key in one table and the related
foreign key in the second table.
Join operations can be classified as inner joins and outer joins. The inner join is the traditional join in which only rows
that meet a given criteria are selected. The join criteria can be an equality condition (also called a natural join or an
equijoin) or an inequality condition (also called a theta join). An outer join returns not only the matching rows but also
the rows with unmatched attribute values for one table or both tables to be joined. The SQL standard also introduces
a special type of join, called a cross join, that returns the same result as the Cartesian product of two sets or tables.
In this section, you will learn various ways to express join operations that meet the ANSI SQL standard. These are
outlined in Table 8.1. It is useful to remember that not all DBMS vendors provide the same level of SQL support and
that some do not support the join styles shown in this section. Oracle 11g is used to demonstrate the use of the
following queries. Refer to your DBMS manual if you are using a different DBMS.
TABLE
8.1
SQL Join Expression Styles
JOIN
CLASSIFICATION
CROSS
JOIN
TYPE
CROSS
JOIN
INNER
Old-Style
JOIN
OUTER
SQL SYNTAX EXAMPLE
DESCRIPTION
SELECT *
FROM T1, T2
SELECT *
FROM T1 CROSS JOIN T2
SELECT *
FROM T1, T2
WHERE T1.C1=T2.C1
Returns the Cartesian product of T1 and
T2 (old style).
Returns the Cartesian product of T1
and T2.
Returns only the rows that meet the join
condition in the WHERE clause (old
style). Only rows with matching values
are selected.
Returns only the rows with matching
values in the matching columns. The
matching columns must have the same
names and similar data types.
Returns only the rows with matching
values in the columns indicated in the
USING clause.
Returns only the rows that meet the join
condition indicated in the ON clause.
NATURAL
JOIN
SELECT *
FROM T1 NATURAL JOIN T2
JOIN
USING
SELECT *
FROM T1 JOIN T2 USING (C1)
JOIN
ON
SELECT *
FROM T1 JOIN T2
ON T1.C1=T2.C1
SELECT *
FROM T1 LEFT OUTER JOIN T2
ON T1.C1=T2.C1
SELECT *
FROM T1 RIGHT OUTER JOIN T2
ON T1.C1=T2.C1
SELECT *
FROM T1 FULL OUTER JOIN T2
ON T1.C1=T2.C1
LEFT
JOIN
RIGHT
JOIN
FULL
JOIN
Returns rows with matching values and
includes all rows from the left table (T1)
with unmatched values.
Returns rows with matching values and
includes all rows from the right table
(T2) with unmatched values.
Returns rows with matching values and
includes all rows from both tables (T1
and T2) with unmatched values.
8.2.1 Cross Join
A cross join performs a relational product (also known as the Cartesian product) of two tables. The cross join
syntax is:
SELECT column-list FROM table1 CROSS JOIN table2
For example,
SELECT * FROM INVOICE CROSS JOIN LINE;
A D V A N C E D
S Q L
performs a cross join of the INVOICE and LINE tables. That CROSS JOIN query generates 144 rows. (There were
8 invoice rows and 18 line rows, yielding 8 × 18 = 144 rows.)
You can also perform a cross join that yields only specified attributes. For example, you can specify:
SELECT
FROM
INVOICE.INV_NUMBER, CUS_CODE, INV_DATE, P_CODE
INVOICE CROSS JOIN LINE;
The results generated through that SQL statement can also be generated by using the following syntax:
SELECT
FROM
INVOICE.INV_NUMBER, CUS_CODE, INV_DATE, P_CODE
INVOICE, LINE;
8.2.2 Natural Join
Recall from Chapter 3 that a natural join returns all rows with matching values in the matching columns and eliminates
duplicate columns. That style of query is used when the tables share one or more common attributes with common
names. The natural join syntax is:
SELECT column-list FROM table1 NATURAL JOIN table2
The natural join will perform the following tasks:
쐌
Determine the common attribute(s) by looking for attributes with identical names and compatible data types.
쐌
Select only the rows with common values in the common attribute(s).
쐌
If there are no common attributes, return the relational product of the two tables.
The following example performs a natural join of the CUSTOMER and INVOICE tables and returns only selected
attributes:
SELECT
FROM
CUS_CODE, CUS_LNAME, INV_NUMBER, INV_DATE
CUSTOMER NATURAL JOIN INVOICE;
The SQL code and its results are shown at the top of Figure 8.7.
You are not limited to two tables when performing a natural join. For example, you can perform a natural join of the
INVOICE, LINE, and PRODUCT tables and project only selected attributes by writing:
SELECT
FROM
INV_NUMBER, P_CODE, P_DESCRIPT, LINE_UNITS, LINE_PRICE
INVOICE NATURAL JOIN LINE NATURAL JOIN PRODUCT;
The SQL code and its results are shown at the bottom of Figure 8.7.
One important difference between the natural join and the “old-style” join syntax is that the natural join does not
require the use of a table qualifier for the common attributes. In the first natural join example, you projected
CUS_CODE. However, the projection did not require any table qualifier, even though the CUS_CODE attribute
appeared in both CUSTOMER and INVOICE tables. The same can be said of the INV_NUMBER attribute in the
second natural join example.
8.2.3 Join USING Clause
A second way to express a join is through the USING keyword. That query returns only the rows with matching values
in the column indicated in the USING clause—and that column must exist in both tables. The syntax is:
SELECT column-list FROM table1 JOIN table2 USING (common-column)
307
308
C H A P T E R
FIGURE
8
NATURAL JOIN results
8.7
To see the JOIN USING query in action, let’s perform a join of the INVOICE and LINE tables by writing:
SELECT
FROM
INV_NUMBER, P_CODE, P_DESCRIPT, LINE_UNITS, LINE_PRICE
INVOICE JOIN LINE USING (INV_NUMBER) JOIN PRODUCT USING (P_CODE);
The SQL statement produces the results shown in Figure 8.8.
As was the case with the NATURAL JOIN command, the JOIN USING operand does not require table qualifiers. As
a matter of fact, Oracle will return an error if you specify the table name in the USING clause.
8.2.4 JOIN ON Clause
The previous two join styles used common attribute names in the joining tables. Another way to express a join when
the tables have no common attribute names is to use the JOIN ON operand. That query will return only the rows that
meet the indicated join condition. The join condition will typically include an equality comparison expression of two
A D V A N C E D
FIGURE
S Q L
JOIN USING results
8.8
columns. (The columns may or may not share the same name but, obviously, must have comparable data types.) The
syntax is:
SELECT column-list FROM table1 JOIN table2 ON join-condition
The following example performs a join of the INVOICE and LINE tables, using the ON clause. The result is shown in
Figure 8.9.
SELECT
FROM
INVOICE.INV_NUMBER, PRODUCT.P_CODE, P_DESCRIPT, LINE_UNITS, LINE_PRICE
INVOICE JOIN LINE ON INVOICE.INV_NUMBER = LINE.INV_NUMBER
JOIN PRODUCT ON LINE.P_CODE = PRODUCT.P_CODE;
Note that unlike the NATURAL JOIN and the JOIN USING operands, the JOIN ON clause requires a table qualifier for
the common attributes. If you do not specify the table qualifier, you will get a “column ambiguously defined” error message.
Keep in mind that the JOIN ON syntax lets you perform a join even when the tables do not share a common
attribute name. For example, to generate a list of all employees with the managers’ names, you can use the following
(recursive) query:
SELECT
FROM
ORDER BY
E.EMP_MGR, M.EMP_LNAME, E.EMP_NUM, E.EMP_LNAME
EMP E JOIN EMP M ON E.EMP_MGR = M.EMP_NUM
E.EMP_MGR;
8.2.5 Outer Joins
An outer join returns not only the rows matching the join condition (that is, rows with matching values in the common
columns) but also the rows with unmatched values. The ANSI standard defines three types of outer joins: left, right,
and full. The left and right designations reflect the order in which the tables are processed by the DBMS. Remember
that join operations take place two tables at a time. The first table named in the FROM clause will be the left side, and
309
310
C H A P T E R
FIGURE
8
JOIN ON results
8.9
the second table named will be the right side. If three or more tables are being joined, the result of joining the first two
tables becomes the left side, and the third table becomes the right side.
The left outer join returns not only the rows matching the join condition (that is, rows with matching values in the
common column) but also the rows in the left side table with unmatched values in the right side table. The syntax is:
SELECT
FROM
column-list
table1 LEFT [OUTER] JOIN table2 ON join-condition
For example, the following query lists the product code, vendor code, and vendor name for all products and includes
those vendors with no matching products:
SELECT
FROM
P_CODE, VENDOR.V_CODE, V_NAME
VENDOR LEFT JOIN PRODUCT ON VENDOR.V_CODE = PRODUCT.V_CODE;
The preceding SQL code and its results are shown in Figure 8.10.
The right outer join returns not only the rows matching the join condition (that is, rows with matching values in the
common column) but also the rows in the right side table with unmatched values in the left side table. The syntax is:
SELECT
FROM
column-list
table1 RIGHT [OUTER] JOIN table2 ON join-condition
For example, the following query lists the product code, vendor code, and vendor name for all products and also
includes those products that do not have a matching vendor code:
SELECT
FROM
P_CODE, VENDOR.V_CODE, V_NAME
VENDOR RIGHT JOIN PRODUCT ON VENDOR.V_CODE = PRODUCT.V_CODE;
The SQL code and its output are shown in Figure 8.11.
A D V A N C E D
FIGURE
LEFT JOIN results
8.10
FIGURE
8.11
RIGHT JOIN results
S Q L
311
312
C H A P T E R
8
The full outer join returns not only the rows matching the join condition (that is, rows with matching values in the
common column) but also all of the rows with unmatched values in either side table. The syntax is:
SELECT
FROM
column-list
table1 FULL [OUTER] JOIN table2 ON join-condition
For example, the following query lists the product code, vendor code, and vendor name for all products and includes
all product rows (products without matching vendors) as well as all vendor rows (vendors without matching products):
SELECT
FROM
P_CODE, VENDOR.V_CODE, V_NAME
VENDOR FULL JOIN PRODUCT ON VENDOR.V_CODE = PRODUCT.V_CODE;
The SQL code and its results are shown in Figure 8.12.
FIGURE
FULL JOIN results
8.12
8.3 SUBQUERIES AND CORRELATED QUERIES
The use of joins in a relational database allows you to get information from two or more tables. For example, the
following query allows you to get the customer’s data with their respective invoices by joining the CUSTOMER and
INVOICE tables.
SELECT
FROM
WHERE
INV_NUMBER, INVOICE.CUS_CODE, CUS_LNAME, CUS_FNAME
CUSTOMER, INVOICE
CUSTOMER.CUS_CODE = INVOICE.CUS_CODE;
In the previous query, the data from both tables (CUSTOMER and INVOICE) are processed at once, matching rows
with shared CUS_CODE values.
A D V A N C E D
S Q L
However, it is often necessary to process data based on other processed data. Suppose, for example, that you want
to generate a list of vendors who provide products. (Recall that not all vendors in the VENDOR table have provided
products—some of them are only potential vendors.) In Chapter 7, you learned that you could generate such a list by
writing the following query:
SELECT
WHERE
V_CODE, V_NAME FROM VENDOR
V_CODE NOT IN (SELECT V_CODE FROM PRODUCT);
Similarly, to generate a list of all products with a price greater than or equal to the average product price, you can write
the following query:
SELECT
WHERE
P_CODE, P_PRICE FROM PRODUCT
P_PRICE >= (SELECT AVG(P_PRICE) FROM PRODUCT);
In both of those cases, you needed to get information that was not previously known:
쐌
What vendors provide products?
쐌
What is the average price of all products?
In both cases, you used a subquery to generate the required information that could then be used as input for the
originating query.
You learned how to use subqueries in Chapter 7; let’s review their basic characteristics:
쐌
A subquery is a query (SELECT statement) inside a query.
쐌
A subquery is normally expressed inside parentheses.
쐌
The first query in the SQL statement is known as the outer query.
쐌
The query inside the SQL statement is known as the inner query.
쐌
The inner query is executed first.
쐌
The output of an inner query is used as the input for the outer query.
쐌
The entire SQL statement is sometimes referred to as a nested query.
In this section, you learn more about the practical use of subqueries. You already know that a subquery is based on the
use of the SELECT statement to return one or more values to another query. But subqueries have a wide range of uses.
For example, you can use a subquery within a SQL data manipulation language (DML) statement (such as INSERT,
UPDATE, or DELETE) where a value or a list of values (such as multiple vendor codes or a table) is expected.
Table 8.2 uses simple examples to summarize the use of SELECT subqueries in DML statements.
TABLE
8.2
SELECT Subquery Examples
SELECT SUBQUERY EXAMPLES
INSERT INTO PRODUCT
SELECT * FROM P;
UPDATE PRODUCT
SET
P_PRICE = (SELECT AVG(P_PRICE)
FROM PRODUCT)
WHERE V_CODE IN (SELECT V_CODE
FROM VENDOR
WHERE V_AREACODE = '615')
DELETE FROM PRODUCT
WHERE V_CODE IN (SELECT V_CODE
FROM VENDOR
WHERE V_AREACODE = '615')
EXPLANATION
Inserts all rows from Table P into the PRODUCT table.
Both tables must have the same attributes. The subquery returns all rows from Table P.
Updates the product price to the average product price,
but only for the products that are provided by vendors
who have an area code equal to 615. The first subquery
returns the average price; the second subquery returns
the list of vendors with an area code equal to 615.
Deletes the PRODUCT table rows that are provided by
vendors with area code equal to 615. The subquery
returns the list of vendors codes with an area code
equal to 615.
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Using the examples shown in Table 8.2, note that the subquery is always at the right side of a comparison or assigning
expression. Also, a subquery can return one value or multiple values. To be precise, the subquery can return:
쐌
One single value (one column and one row). This subquery is used anywhere a single value is expected, as
in the right side of a comparison expression (such as in the preceding UPDATE example when you assign the
average price to the product’s price). Obviously, when you assign a value to an attribute, that value is a single
value, not a list of values. Therefore, the subquery must return only one value (one column, one row). If the
query returns multiple values, the DBMS will generate an error.
쐌
A list of values (one column and multiple rows). This type of subquery is used anywhere a list of values is
expected, such as when using the IN clause (that is, when comparing the vendor code to a list of vendors).
Again, in this case, there is only one column of data with multiple value instances. This type of subquery is used
frequently in combination with the IN operator in a WHERE conditional expression.
쐌
A virtual table (multicolumn, multirow set of values). This type of subquery can be used anywhere a table
is expected, such as when using the FROM clause. You will see this type of query later in this chapter.
It’s important to note that a subquery can return no values at all; it is a NULL. In such cases, the output of the outer
query might result in an error or a null empty set, depending where the subquery is used (in a comparison, an
expression, or a table set).
In the following sections, you will learn how to write subqueries within the SELECT statement to retrieve data from
the database.
8.3.1 WHERE Subqueries
The most common type of subquery uses an inner SELECT subquery on the right side of a WHERE comparison
expression. For example, to find all products with a price greater than or equal to the average product price, you write
the following query:
SELECT
WHERE
P_CODE, P_PRICE FROM PRODUCT
P_PRICE >= (SELECT AVG(P_PRICE) FROM PRODUCT);
The output of the preceding query is shown in Figure 8.13. Note that this type of query, when used in a >, <, =, >=,
or <= conditional expression, requires a subquery that returns only one single value (one column, one row). The value
generated by the subquery must be of a “comparable” data type; if the attribute to the left of the comparison symbol
is a character type, the subquery must return a character string. Also, if the query returns more than a single value,
the DBMS will generate an error.
Subqueries can also be used in combination with joins. For example, the following query lists all of the customers who
ordered the product “Claw hammer”:
SELECT
FROM
WHERE
DISTINCT CUS_CODE, CUS_LNAME, CUS_FNAME
CUSTOMER JOIN INVOICE USING (CUS_CODE)
JOIN LINE USING (INV_NUMBER)
JOIN PRODUCT USING (P_CODE)
P_CODE = (SELECT P_CODE FROM PRODUCT WHERE P_DESCRIPT = ‘Claw hammer’);
A D V A N C E D
S Q L
The result of that query is also shown in Figure 8.13.
FIGURE
WHERE subquery example
8.13
In the preceding example, the inner query finds the P_CODE for the product “Claw hammer.” The P_CODE is then
used to restrict the selected rows to only those where the P_CODE in the LINE table matches the P_CODE for “Claw
hammer.” Note that the previous query could have been written this way:
SELECT
FROM
WHERE
DISTINCT CUS_CODE, CUS_LNAME, CUS_FNAME
CUSTOMER JOIN INVOICE USING (CUS_CODE)
JOIN LINE USING (INV_NUMBER)
JOIN PRODUCT USING (P_CODE)
P_DESCRIPT = ‘Claw hammer’;
But what happens if the original query encounters the “Claw hammer” string in more than one product description? You
get an error message. To compare one value to a list of values, you must use an IN operand, as shown in the next section.
8.3.2 IN Subqueries
What would you do if you wanted to find all customers who purchased a “hammer” or any kind of saw or saw blade?
Note that the product table has two different types of hammers: “Claw hammer” and “Sledge hammer.” Also note that
there are multiple occurrences of products that contain “saw” in their product descriptions. There are saw blades,
jigsaws, and so on. In such cases, you need to compare the P_CODE not to one product code (single value) but to a
list of product code values. When you want to compare a single attribute to a list of values, you use the IN operator.
When the P_CODE values are not known beforehand, but they can be derived using a query, you must use an IN
subquery. The following example lists all customers who have purchased hammers, saws, or saw blades.
SELECT
FROM
WHERE
DISTINCT CUS_CODE, CUS_LNAME, CUS_FNAME
CUSTOMER JOIN INVOICE USING (CUS_CODE)
JOIN LINE USING (INV_NUMBER)
JOIN PRODUCT USING (P_CODE)
P_CODE IN ( SELECT
P_CODE FROM PRODUCT
WHERE
P_DESCRIPT LIKE '%hammer%'
OR
P_DESCRIPT LIKE '%saw%');
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The result of that query is shown in Figure 8.14.
FIGURE
IN subquery example
8.14
8.3.3 HAVING Subqueries
Just as you can use subqueries with the WHERE clause, you can use a subquery with a HAVING clause. Remember
that the HAVING clause is used to restrict the output of a GROUP BY query by applying a conditional criteria to the
grouped rows. For example, to list all products with the total quantity sold greater than the average quantity sold, you
would write the following query:
SELECT
FROM
GROUP BY
HAVING
P_CODE, SUM(LINE_UNITS)
LINE
P_CODE
SUM(LINE_UNITS) > (SELECT AVG(LINE_UNITS) FROM LINE);
The result of that query is shown in Figure 8.15.
FIGURE
8.15
HAVING subquery example
A D V A N C E D
S Q L
8.3.4 Multirow Subquery Operators: ANY and ALL
So far, you have learned that you must use an IN subquery when you need to compare a value to a list of values. But
the IN subquery uses an equality operator; that is, it selects only those rows that match (are equal to) at least one of
the values in the list. What happens if you need to do an inequality comparison (> or <) of one value to a list of values?
For example, suppose that you want to know which products have a product cost that is greater than all individual
product costs for products provided by vendors from Florida.
SELECT
FROM
WHERE
P_CODE, P_QOH * P_PRICE
PRODUCT
P_QOH * P_PRICE > ALL (SELECT P_QOH * P_PRICE
FROM PRODUCT
WHERE V_CODE IN (SELECT V_CODE
FROM VENDOR
WHERE V_STATE = ‘FL’));
The result of that query is shown in Figure 8.16.
FIGURE
Multirow subquery operator example
8.16
It’s important to note the following points about the query and its output in Figure 8.16:
쐌
The query is a typical example of a nested query.
쐌
The query has one outer SELECT statement with a SELECT subquery (call it sqA) containing a second SELECT
subquery (call it sqB).
쐌
The last SELECT subquery (sqB) is executed first and returns a list of all vendors from Florida.
쐌
The first SELECT subquery (sqA) uses the output of the SELECT subquery (sqB). The sqA subquery returns the
list of product costs for all products provided by vendors from Florida.
쐌
The use of the ALL operator allows you to compare a single value (P_QOH * P_PRICE) with a list of values
returned by the first subquery (sqA) using a comparison operator other than equals.
쐌
For a row to appear in the result set, it has to meet the criterion P_QOH * P_PRICE > ALL, of the individual
values returned by the subquery sqA. The values returned by sqA are a list of product costs. In fact, “greater
than ALL” is equivalent to “greater than the highest product cost of the list.” In the same way, a condition of
“less than ALL” is equivalent to “less than the lowest product cost of the list.”
Another powerful operator is the ANY multirow operator (the near cousin of the ALL multirow operator). The ANY
operator allows you to compare a single value to a list of values, selecting only the rows for which the inventory cost
is greater than any value of the list or less than any value of the list. You could use the equal to ANY operator, which
would be the equivalent of the IN operator.
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8.3.5 FROM Subqueries
So far you have seen how the SELECT statement uses subqueries within WHERE, HAVING, and IN statements and
how the ANY and ALL operators are used for multirow subqueries. In all of those cases, the subquery was part of a
conditional expression and it always appeared at the right side of the expression. In this section, you will learn how
to use subqueries in the FROM clause.
As you already know, the FROM clause specifies the table(s) from which the data will be drawn. Because the output
of a SELECT statement is another table (or more precisely a “virtual” table), you could use a SELECT subquery in the
FROM clause. For example, assume that you want to know all customers who have purchased products 13-Q2/P2
and 23109-HB. All product purchases are stored in the LINE table. It is easy to find out who purchased any given
product by searching the P_CODE attribute in the LINE table. But in this case, you want to know all customers who
purchased both products, not just one. You could write the following query:
SELECT
FROM
WHERE
DISTINCT CUSTOMER.CUS_CODE, CUSTOMER.CUS_LNAME
CUSTOMER,
(SELECT INVOICE.CUS_CODE FROM INVOICE NATURAL JOIN LINE
WHERE P_CODE = '13-Q2/P2') CP1,
(SELECT INVOICE.CUS_CODE FROM INVOICE NATURAL JOIN LINE
WHERE P_CODE = '23109-HB') CP2
CUSTOMER.CUS_CODE = CP1.CUS_CODE AND CP1.CUS_CODE = CP2.CUS_CODE;
The result of that query is shown in Figure 8.17.
FIGURE
FROM subquery example
8.17
Note in Figure 8.17 that the first subquery returns all customers who purchased product 13-Q2/P2, while the second
subquery returns all customers who purchased product 23109-HB. So in this FROM subquery, you are joining the
CUSTOMER table with two virtual tables. The join condition selects only the rows with matching CUS_CODE values
in each table (base or virtual).
In the previous chapter, you learned that a view is also a virtual table; therefore, you can use a view name anywhere a table
is expected. So in this example, you could create two views: one listing all customers who purchased product 13-Q2/P2
and another listing all customers who purchased product 23109-HB. Doing so, you would write the query as:
CREATE VIEW CP1 AS
SELECT
INVOICE.CUS_CODE FROM INVOICE NATURAL JOIN LINE
WHERE
P_CODE = '13-Q2/P2';
A D V A N C E D
S Q L
CREATE VIEW CP2 AS
SELECT
INVOICE.CUS_CODE FROM INVOICE NATURAL JOIN LINE
WHERE
P_CODE = '23109-HB';
SELECT
DISTINCT CUS_CODE, CUS_LNAME
FROM
CUSTOMER NATURAL JOIN CP1 NATURAL JOIN CP2;
You might speculate that the preceding query could also be written using the following syntax:
SELECT
FROM
WHERE
CUS_CODE, CUS_LNAME
CUSTOMER NATURAL JOIN INVOICE NATURAL JOIN LINE
P_CODE = '13-Q2/P2' AND P_CODE = '23109-HB';
But if you examine that query carefully, you will note that a P_CODE cannot be equal to two different values at the
same time. Therefore, the query will not return any rows.
8.3.6 Attribute List Subqueries
The SELECT statement uses the attribute list to indicate what columns to project in the resulting set. Those columns
can be attributes of base tables, computed attributes, or the result of an aggregate function. The attribute list can also
include a subquery expression, also known as an inline subquery. A subquery in the attribute list must return one single
value; otherwise, an error code is raised. For example, a simple inline query can be used to list the difference between
each product’s price and the average product price:
SELECT
FROM
P_CODE, P_PRICE, (SELECT AVG(P_PRICE) FROM PRODUCT) AS AVGPRICE,
P_PRICE – (SELECT AVG(P_PRICE) FROM PRODUCT) AS DIFF
PRODUCT;
Figure 8.18 shows the result of that query.
FIGURE
8.18
Inline subquery example
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In Figure 8.18, note that the inline query output returns one single value (the average product’s price) and that the
value is the same in every row. Note also that the query used the full expression instead of the column aliases when
computing the difference. In fact, if you try to use the alias in the difference expression, you will get an error message.
The column alias cannot be used in computations in the attribute list when the alias is defined in the same attribute
list. That DBMS requirement is the result of the way the DBMS parses and executes queries.
Another example will help you understand the use of attribute list subqueries and column aliases. For example, suppose
that you want to know the product code, the total sales by product, and the contribution by employee of each product’s
sales. To get the sales by-product, you need to use only the LINE table. To compute the contribution by employee, you
need to know the number of employees (from the EMPLOYEE table). As you study the tables’ structures, you can see
that the LINE and EMPLOYEE tables do not share a common attribute. In fact, you don’t need a common attribute.
You only need to know the total number of employees, not the total employees related to each product. So to answer
the query, you would write the following code:
SELECT
FROM
GROUP BY
P_CODE, SUM(LINE_UNITS * LINE_PRICE) AS SALES,
(SELECT COUNT(*) FROM EMPLOYEE) AS ECOUNT,
SUM(LINE_UNITS * LINE_PRICE)/(SELECT COUNT(*) FROM EMPLOYEE) AS CONTRIB
LINE
P_CODE;
The result of that query is shown in Figure 8.19.
FIGURE
Another example of an inline subquery
8.19
As you can see in Figure 8.19, the number of employees remains the same for each row in the result set. The use of
that type of subquery is limited to certain instances where you need to include data from other tables that are not
directly related to a main table or tables in the query. The value will remain the same for each row, like a constant in
a programming language. (You will learn another use of inline subqueries in Section 8.3.7, Correlated Subqueries.)
Note that you cannot use an alias in the attribute list to write the expression that computes the contribution per
employee.
A D V A N C E D
S Q L
Another way to write the same query by using column aliases requires the use of a subquery in the FROM clause, as follows:
SELECT
FROM
P_CODE, SALES, ECOUNT, SALES/ECOUNT AS CONTRIB
(SELECT P_CODE, SUM(LINE_UNITS * LINE_PRICE) AS SALES,
(SELECT COUNT(*) FROM EMPLOYEE) AS ECOUNT
FROM
LINE
GROUP BY P_CODE);
In that case, you are actually using two subqueries. The subquery in the FROM clause executes first and returns a virtual
table with three columns: P_CODE, SALES, and ECOUNT. The FROM subquery contains an inline subquery that
returns the number of employees as ECOUNT. Because the outer query receives the output of the inner query, you
can now refer to the columns in the outer subquery by using the column aliases.
8.3.7 Correlated Subqueries
Until now, all subqueries you have learned execute independently. That is, each subquery in a command sequence
executes in a serial fashion, one after another. The inner subquery executes first; its output is used by the outer query,
which then executes until the last outer query executes (the first SQL statement in the code).
In contrast, a correlated subquery is a subquery that executes once for each row in the outer query. That process
is similar to the typical nested loop in a programming language. For example:
FOR X = 1 TO 2
FOR Y = 1 TO 3
PRINT “X = “X, “Y = “Y
END
END
will yield the output:
X=1
Y=1
X=1
Y=2
X=1
Y=3
X=2
Y=1
X=2
Y=2
X=2
Y=3
Note that the outer loop X = 1 TO 2 begins the process by setting X = 1 and then the inner loop Y = 1 TO 3 is completed
for each X outer loop value. The relational DBMS uses the same sequence to produce correlated subquery results:
1.
It initiates the outer query.
2.
For each row of the outer query result set, it executes the inner query by passing the outer row to the inner query.
That process is the opposite of that of the subqueries as you have already seen. The query is called a correlated subquery
because the inner query is related to the outer query by the fact that the inner query references a column of the outer
subquery.
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To see the correlated subquery in action, suppose that you want to know all product sales in which the units sold value
is greater than the average units sold value for that product (as opposed to the average for all products). In that case,
the following procedure must be completed:
1.
Compute the average-units-sold value for a product.
2.
Compare the average computed in Step 1 to the units sold in each sale row and then select only the rows in
which the number of units sold is greater.
The following correlated query completes the preceding two-step process:
SELECT
FROM
WHERE
INV_NUMBER, P_CODE, LINE_UNITS
LINE LS
LS.LINE_UNITS > (SELECT
AVG ( LINE_UNITS)
FROM
LINE LA
WHERE
LA.P_CODE = LS.P_CODE);
The first example in Figure 8.20 shows the result of that query.
FIGURE
Correlated subquery examples
8.20
In the top query and its result in Figure 8.20, note that the LINE table is used more than once, so you must use table
aliases. In that case, the inner query computes the average units sold of the product that matches the P_CODE of the
outer query P_CODE. That is, the inner query runs once, using the first product code found in the (outer) LINE table,
and returns the average sale for that product. When the number of units sold in that (outer) LINE row is greater than
the average computed, the row is added to the output. Then the inner query runs again, this time using the second
product code found in the (outer) LINE table. The process repeats until the inner query has run for all rows in the (outer)
LINE table. In that case, the inner query will be repeated as many times as there are rows in the outer query.
A D V A N C E D
S Q L
To verify the results and to provide an example of how you can combine subqueries, you can add a correlated inline
subquery to the previous query. That correlated inline subquery will show the average units sold column for each
product. (See the second query and its results in Figure 8.20.) As you can see, the new query contains a correlated
inline subquery that computes the average units sold for each product. You not only get an answer, but you can also
verify that the answer is correct.
Correlated subqueries can also be used with the EXISTS special operator. For example, suppose that you want to know
all customers who have placed an order lately. In that case, you could use a correlated subquery like the first one shown
in Figure 8.21:
SELECT
FROM
WHERE
FIGURE
CUS_CODE, CUS_LNAME, CUS_FNAME
CUSTOMER
EXISTS
(SELECT
CUS_CODE FROM INVOICE
WHERE
INVOICE.CUS_CODE = CUSTOMER.CUS_CODE);
EXISTS correlated subquery examples
8.21
The second example of an EXISTS correlated subquery in Figure 8.21 will help you understand how to use correlated
queries. For example, suppose that you want to know what vendors you must contact to start ordering products that
are approaching the minimum quantity-on-hand value. In particular, you want to know the vendor code and name of
vendors for products having a quantity on hand that is less than double the minimum quantity. The query that answers
that question is as follows:
SELECT
FROM
WHERE
V_CODE, V_NAME
VENDOR
EXISTS
(SELECT
FROM
WHERE
AND
*
PRODUCT
P_QOH < P_MIN * 2
VENDOR.V_CODE = PRODUCT.V_CODE);
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In the second query in Figure 8.21, note that:
1.
The inner correlated subquery runs using the first vendor.
2.
If any products match the condition (quantity on hand is less than double the minimum quantity), the vendor
code and name are listed in the output.
3.
The correlated subquery runs using the second vendor, and the process repeats itself until all vendors are used.
8.4 SQL FUNCTIONS
The data in databases are the basis of critical business information. Generating information from data often requires
many data manipulations. Sometimes such data manipulation involves the decomposition of data elements. For
example, an employee’s date of birth can be subdivided into a day, a month, and a year. A product manufacturing code
(for example, SE-05-2-09-1234-1-3/12/04-19:26:48) can be designed to record the manufacturing region, plant,
shift, production line, employee number, date, and time. For years, conventional programming languages have had
special functions that enabled programmers to perform data transformations like those data decompositions. If you
know a modern programming language, it’s very likely that the SQL functions in this section will look familiar.
SQL functions are very useful tools. You’ll need to use functions when you want to list all employees ordered by year
of birth or when your marketing department wants you to generate a list of all customers ordered by zip code and the
first three digits of their telephone numbers. In both of those cases, you’ll need to use data elements that are not
present as such in the database; instead, you’ll need a SQL function that can be derived from an existing attribute.
Functions always use a numerical, date, or string value. The value may be part of the command itself (a constant or
literal) or it may be an attribute located in a table. Therefore, a function may appear anywhere in an SQL statement
where a value or an attribute can be used.
There are many types of SQL functions, such as arithmetic, trigonometric, string, date, and time functions. This section
will not explain all of those types of functions in detail, but it will give you a brief overview of the most useful ones.
Note
Although the main DBMS vendors support the SQL functions covered here, the syntax or degree of support will
probably differ. In fact, DBMS vendors invariably add their own functions to products to lure new customers.
The functions covered in this section represent just a small portion of functions supported by your DBMS. Read
your DBMS SQL reference manual for a complete list of available functions.
8.4.1 Date and Time Functions
All SQL-standard DBMSs support date and time functions. All date functions take one parameter (of a date or
character data type) and return a value (character, numeric, or date type). Unfortunately, date/time data types are
implemented differently by different DBMS vendors. The problem occurs because the ANSI SQL standard defines date
data types, but it does not say how those data types are to be stored. Instead, it lets the vendor deal with that issue.
Because date/time functions differ from vendor to vendor, this section will cover basic date/time functions for
MS Access/SQL Server and for Oracle. Table 8.3 shows a list of selected MS Access/SQL Server date/time functions.
A D V A N C E D
TABLE
8.3
S Q L
Selected MS Access/SQL Server Date/Time Functions
FUNCTION
YEAR
Returns a four-digit year
Syntax:
YEAR(date_value)
MONTH
Returns a two-digit month code
Syntax:
MONTH(date_value)
DAY
Returns the number of the day
Syntax:
DAY(date_value)
DATE() − MS Access
GETDATE() − SQL Server
Returns today’s date
DATEADD − SQL Server
Adds a number of selected time
periods to a date
Syntax:
DATEADD(datepart,
number, date)
DATEDIFF − SQL Server
Subtracts two dates
Syntax:
DATEDIFF(datepart, startdate,
enddate)
EXAMPLE(S)
Lists all employees born in 1966:
SELECT EMP_LNAME, EMP_FNAME, EMP_DOB,
YEAR(EMP_DOB) AS YEAR
FROM EMPLOYEE
WHERE YEAR(EMP_DOB) = 1966;
Lists all employees born in November:
SELECT EMP_LNAME, EMP_FNAME, EMP_DOB,
MONTH(EMP_DOB) AS MONTH
FROM EMPLOYEE
WHERE MONTH(EMP_DOB) = 11;
Lists all employees born on the 14th day of the month:
SELECT EMP_LNAME, EMP_FNAME, EMP_DOB,
DAY(EMP_DOB) AS DAY
FROM EMPLOYEE
WHERE DAY(EMP_DOB) = 14;
Lists how many days are left until Christmas:
SELECT #25-Dec-2010# − DATE();
Note two features:
• There is no FROM clause, which is acceptable in MS Access.
• The Christmas date is enclosed in # signs because you are doing date
arithmetic.
In MS SQL Server:
Use GETDATE() to get the current system date. To compute the difference
between dates, use the DATEDIFF function (see below).
Adds a number of dateparts to a given date. Dateparts can be minutes, hours,
days, weeks, months, quarters, or years. For example:
SELECT DATEADD(day,90, P_INDATE) AS DueDate
FROM PRODUCT;
The preceding example adds 90 days to P_INDATE.
In MS Access use:
SELECT P_INDATE+90 AS DueDate
FROM PRODUCT;
Returns the difference between two dates expressed in a selected datepart. For
example:
SELECT DATEDIFF(day, P_INDATE, GETDATE()) AS DaysAgo
FROM PRODUCT;
In MS Access use:
SELECT DATE() - P_INDATE AS DaysAgo
FROM PRODUCT;
Table 8.4 shows the equivalent date/time functions used in Oracle. Note that Oracle uses the same function
(TO_CHAR) to extract the various parts of a date. Also, another function (TO_DATE) is used to convert character
strings to a valid Oracle date format that can be used in date arithmetic.
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TABLE
8.4
8
Selected Oracle Date/Time Functions
FUNCTION
TO_CHAR
Returns a character string or a
formatted string from a date
value
Syntax:
TO_CHAR(date_value, fmt)
fmt = format used; can be:
MONTH: name of month
MON: three-letter month name
MM: two-digit month name
D: number for day of week
DD: number day of month
DAY: name of day of week
YYYY: four-digit year value
YY: two-digit year value
TO_DATE
Returns a date value using a
character string and a date format mask; also used to translate
a date between formats
Syntax:
TO_DATE(char_value, fmt)
fmt = format used; can be:
MONTH: name of month
MON: three-letter month name
MM: two-digit month name
D: number for day of week
DD: number day of month
DAY: name of day of week
YYYY: four-digit year value
YY: two-digit year value
SYSDATE
Returns today’s date
ADD_MONTHS
Adds a number of months to a
date; useful for adding months
or years to a date
Syntax:
ADD_MONTHS(date_value, n)
n = number of months
EXAMPLE(S)
Lists all employees born in 1982:
SELECT EMP_LNAME, EMP_FNAME, EMP_DOB,
TO_CHAR(EMP_DOB, 'YYYY') AS YEAR
FROM EMPLOYEE
WHERE TO_CHAR(EMP_DOB, 'YYYY') = '1982';
Lists all employees born in November:
SELECT EMP_LNAME, EMP_FNAME, EMP_DOB,
TO_CHAR(EMP_DOB, 'MM') AS MONTH
FROM EMPLOYEE
WHERE TO_CHAR(EMP_DOB, 'MM') = '11';
Lists all employees born on the 14th day of the month:
SELECT EMP_LNAME, EMP_FNAME, EMP_DOB,
TO_CHAR(EMP_DOB, 'DD') AS DAY
FROM EMPLOYEE
WHERE TO_CHAR(EMP_DOB, 'DD') = '14';
Lists the approximate age of the employees on the company’s tenth anniversary
date (11/25/2010):
SELECT EMP_LNAME, EMP_FNAME,
EMP_DOB, '11/25/2010' AS ANIV_DATE,
(TO_DATE('11/25/2000','MM/DD/YYYY') - EMP_DOB)/365 AS YEARS
FROM EMPLOYEE
ORDER BY YEARS;
Note the following:
• '11/25/2010' is a text string, not a date.
• The TO_DATE function translates the text string to a valid Oracle date used
in date arithmetic.
How many days between Thanksgiving and Christmas 2008?
SELECT TO_DATE('2010/12/25','YYYY/MM/DD') −
TO_DATE('NOVEMBER 27, 2010','MONTH DD, YYYY')
FROM DUAL;
Note the following:
• The TO_DATE function translates the text string to a valid Oracle date used
in date arithmetic.
• DUAL is Oracle’s pseudo-table used only for cases where a table is not
really needed.
Lists how many days are left until Christmas:
SELECT TO_DATE('25-Dec-2010','DD-MON-YYYY') SYSDATE
FROM DUAL;
Notice two things:
• DUAL is Oracle’s pseudo-table used only for cases where a table is not
really needed.
• The Christmas date is enclosed in a TO_DATE function to translate the
date to a valid date format.
Lists all products with their expiration date (two years from the purchase date):
SELECT
P_CODE, P_INDATE, ADD_MONTHS(P_INDATE,24)
FROM
PRODUCT
ORDER BY ADD_MONTHS(P_INDATE,24);
A D V A N C E D
TABLE
8.4
S Q L
Selected Oracle Date/Time Functions (continued)
FUNCTION
LAST_DAY
Returns the date of the last day
of the month given in a date
Syntax:
LAST_DAY(date_value)
EXAMPLE(S)
Lists all employees who were hired within the last seven days of a month:
SELECT EMP_LNAME, EMP_FNAME, EMP_HIRE_DATE
FROM EMPLOYEE
WHERE EMP_HIRE_DATE >=LAST_DAY(EMP_HIRE_DATE)-7;
8.4.2 Numeric Functions
Numeric functions can be grouped in many different ways, such as algebraic, trigonometric, and logarithmic. In this
section, you will learn two very useful functions. Do not confuse the SQL aggregate functions you saw in the previous
chapter with the numeric functions in this section. The first group operates over a set of values (multiple rows—hence,
the name aggregate functions), while the numeric functions covered here operate over a single row. Numeric
functions take one numeric parameter and return one value. Table 8.5 shows a selected group of numeric functions
available.
TABLE
8.5
Selected Numeric Functions
FUNCTION
ABS
Returns the absolute value of a number
Syntax:
ABS(numeric_value)
ROUND
Rounds a value to a specified precision
(number of digits)
Syntax:
ROUND(numeric_value, p)
p = precision
CEIL/CEILING/FLOOR
Returns the smallest integer greater than or
equal to a number or returns the largest
integer equal to or less than a number,
respectively
Syntax:
CEIL(numeric_value) − Oracle
CEILING(numeric_value) − SQL Server
FLOOR(numeric_value)
EXAMPLE(S)
In Oracle use:
SELECT 1.95, -1.93, ABS(1.95), ABS(-1.93)
FROM DUAL;
In MS Access/SQL Server use:
SELECT 1.95, -1.93, ABS(1.95), ABS(-1.93);
Lists the product prices rounded to one and zero decimal places:
SELECT P_CODE, P_PRICE,
ROUND(P_PRICE,1) AS PRICE1,
ROUND(P_PRICE,0) AS PRICE0
FROM PRODUCT;
Lists the product price, smallest integer greater than or equal to the
product price, and the largest integer equal to or less than the
product price.
In Oracle use:
SELECT P_PRICE, CEIL(P_PRICE), FLOOR(P_PRICE)
FROM PRODUCT;
In SQL Server use:
SELECT P_PRICE, CEILING(P_PRICE), FLOOR(P_PRICE)
FROM PRODUCT;
MS Access does not support these functions.
8.4.3 String Functions
String manipulations are among the most-used functions in programming. If you have ever created a report using any
programming language, you know the importance of properly concatenating strings of characters, printing names in
uppercase, or knowing the length of a given attribute. Table 8.6 shows a subset of useful string manipulation functions.
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TABLE
8.6
8
Selected String Functions
FUNCTION
Concatenation
|| − Oracle
+ − MS Access/SQL Server
Concatenates data from two different
character columns and returns a
single column
Syntax:
strg_value || strg_value
strg_value + strg_value
UPPER/LOWER
Returns a string in all capital or all
lowercase letters
Syntax:
UPPER(strg_value)
LOWER(strg_value)
SUBSTRING
Returns a substring or part of a given
string parameter
Syntax:
SUBSTR(strg_value, p, l) − Oracle
SUBSTRING(strg_value,p,l) −
SQL Server
p = start position
l = length of characters
LENGTH
Returns the number of characters in
a string value
Syntax:
LENGTH(strg_value) − Oracle
LEN(strg_value) − SQL Server
EXAMPLE(S)
Lists all employee names (concatenated).
In Oracle use:
SELECT EMP_LNAME || ', ' || EMP_FNAME AS NAME
FROM EMPLOYEE;
In MS Access / SQL Server use:
SELECT EMP_LNAME + ', ' + EMP_FNAME AS NAME
FROM EMPLOYEE;
Lists all employee names in all capital letters (concatenated).
In Oracle use:
SELECT UPPER(EMP_LNAME) || ', ' || UPPER(EMP_FNAME) AS NAME
FROM EMPLOYEE;
In SQL Server use:
SELECT UPPER(EMP_LNAME) + ', ' + UPPER(EMP_FNAME) AS NAME
FROM EMPLOYEE;
Lists all employee names in all lowercase letters (concatenated).
In Oracle use:
SELECT LOWER(EMP_LNAME) || ', ' || LOWER(EMP_FNAME) AS NAME
FROM EMPLOYEE;
In SQL Server use:
SELECT LOWER(EMP_LNAME) + ', ' + LOWER(EMP_FNAME) AS NAME
FROM EMPLOYEE;
Not supported by MS Access.
Lists the first three characters of all employee phone numbers.
In Oracle use:
SELECT EMP_PHONE, SUBSTR(EMP_PHONE,1,3) AS PREFIX
FROM EMPLOYEE;
In SQL Server use:
SELECT EMP_PHONE, SUBSTRING(EMP_PHONE,1,3) AS PREFIX
FROM EMPLOYEE;
Not supported by MS Access.
Lists all employee last names and the length of their names; ordered
descended by last name length.
In Oracle use:
SELECT EMP_LNAME, LENGTH(EMP_LNAME) AS NAMESIZE
FROM EMPLOYEE;
In MS Access / SQL Server use:
SELECT EMP_LNAME, LEN(EMP_LNAME) AS NAMESIZE
FROM EMPLOYEE;
8.4.4 Conversion Functions
Conversion functions allow you to take a value of a given data type and convert it to the equivalent value in another
data type. In Section 8.4.1, you learned about two of the basic Oracle SQL conversion functions: TO_CHAR and
TO_DATE. Note that the TO_CHAR function takes a date value and returns a character string representing a day, a
month, or a year. In the same way, the TO_DATE function takes a character string representing a date and returns
an actual date in Oracle format. SQL Server uses the CAST and CONVERT functions to convert one data type to
another. A summary of the selected functions is shown in Table 8.7.
A D V A N C E D
TABLE
8.7
S Q L
Selected Conversion Functions
FUNCTION
Numeric to Character:
TO_CHAR − Oracle
CAST − SQL Server
CONVERT − SQL Server
Returns a character string from a
numeric value.
Syntax:
Oracle: TO_CHAR(numeric_value,
fmt)
SQL Server:
CAST (numeric AS varchar(length))
CONVERT(varchar(length), numeric)
Date to Character:
TO_CHAR − Oracle
CAST − SQL Server
CONVERT − SQL Server
Returns a character string or a formatted character string from a date value
Syntax:
Oracle: TO_CHAR(date_value, fmt)
SQL Server:
CAST (date AS varchar(length))
CONVERT(varchar(length), date)
String to Number:
TO_NUMBER
Returns a formatted number from a
character string, using a given format
Syntax:
Oracle:
TO_NUMBER(char_value, fmt)
fmt = format used; can be:
9 = displays a digit
0 = displays a leading zero
, = displays the comma
. = displays the decimal point
$ = displays the dollar sign
B = leading blank
S = leading sign
MI = trailing minus sign
EXAMPLE(S)
Lists all product prices, quantity on hand, percent discount, and total
inventory cost using formatted values.
In Oracle use:
SELECT P_CODE,
TO_CHAR(P_PRICE,'999.99') AS PRICE,
TO_CHAR(P_QOH,'9,999.99') AS QUANTITY,
TO_CHAR(P_DISCOUNT,'0.99') AS DISC,
TO_CHAR(P_PRICE*P_QOH,'99,999.99')
AS TOTAL_COST
FROM PRODUCT;
In SQL Server use:
SELECT P_CODE, CAST(P_PRICE AS VARCHAR(8)) AS PRICE,
CONVERT(VARCHAR(4),P_QOH) AS QUANTITY,
CAST(P_DISCOUNT AS VARCHAR(4)) AS DISC,
CAST(P_PRICE*P_QOH AS VARCHAR(10)) AS TOTAL_COST
FROM PRODUCT;
Not supported in MS Access.
Lists all employee dates of birth, using different date formats.
In Oracle use:
SELECT EMP_LNAME, EMP_DOB,
TO_CHAR(EMP_DOB, ’DAY, MONTH DD, YYYY’)
AS ’DATEOFBIRTH’
FROM EMPLOYEE;
SELECT EMP_LNAME, EMP_DOB,
TO_CHAR(EMP_DOB, ’YYYY/MM/DD’)
AS ’DATEOFBIRTH’
FROM EMPLOYEE;
In SQL Server use:
SELECT EMP_LNAME, EMP_DOB,
CONVERT(varchar(11),EMP_DOB) AS “DATE OF BIRTH”
FROM EMPLOYEE;
SELECT EMP_LNAME, EMP_DOB,
CAST(EMP_DOB as varchar(11)) AS “DATE OF BIRTH”
FROM EMPLOYEE;
Not supported in MS Access.
Converts text strings to numeric values when importing data to a table
from another source in text format; for example, the query shown below
uses the TO_NUMBER function to convert text formatted to Oracle
default numeric values using the format masks given.
In Oracle use:
SELECT TO_NUMBER('-123.99', 'S999.99'),
TO_NUMBER('99.78-','B999.99MI')
FROM DUAL;
In SQL Server use:
SELECT CAST('-123.99' AS NUMERIC(8,2)),
CAST('-99.78' AS NUMERIC(8,2))
The SQL Server CAST function does not support the trailing sign on the
character string.
Not supported in MS Access.
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TABLE
8.7
8
Selected Conversion Functions (continued)
FUNCTION
CASE − SQL Server
DECODE − Oracle
Compares an attribute or expression
with a series of values and returns an
associated value or a default value if
no match is found
Syntax:
Oracle:
DECODE(e, x, y, d)
e = attribute or expression
x = value with which to compare e
y = value to return in e = x
d = default value to return if e is not
equal to x
SQL Server:
CASE When condition
THEN value1 ELSE value2 END
EXAMPLE(S)
The following example returns the sales tax rate for specified states:
• Compares V_STATE to 'CA'; if the values match, it returns .08.
• Compares V_STATE to 'FL'; if the values match, it returns .05.
• Compares V_STATE to 'TN'; if the values match, it returns .085.
If there is no match, it returns 0.00 (the default value).
SELECT V_CODE, V_STATE,
DECODE(V_STATE,'CA',.08,'FL',.05, 'TN',.085, 0.00)
AS TAX
FROM VENDOR;
In SQL Server use:
SELECT V_CODE, V_STATE,
CASE WHEN V_STATE = 'CA' THEN .08
WHEN V_STATE = 'FL' THEN .05
WHEN V_STATE = 'TN' THEN .085
ELSE 0.00 END AS TAX
FROM VENDOR
Not supported in MS Access.
8.5 ORACLE SEQUENCES
If you use MS Access, you might be familiar with the AutoNumber data type, which you can use to define a column
in your table that will be automatically populated with unique numeric values. In fact, if you create a table in MS Access
and forget to define a primary key, MS Access will offer to create a primary key column; if you accept, you will notice
that MS Access creates a column named ID with an AutoNumber data type. After you define a column as an
AutoNumber type, every time you insert a row in the table, MS Access will automatically add a value to that column,
starting with 1 and increasing the value by 1 in every new row you add. Also, you cannot include that column in your
INSERT statements—Access will not let you edit that value at all. MS SQL Server uses the Identity column property
to serve a similar purpose. In MS SQL Server a table can have at most one column defined as an Identity column. This
column behaves similarly to an MS Access column with the AutoNumber data type.
Oracle does not support the AutoNumber data type or the Identity column property. Instead, you can use a “sequence”
to assign values to a column on a table. But an Oracle sequence is very different from the Access AutoNumber data
type and deserves close scrutiny:
쐌
Oracle sequences are an independent object in the database. (Sequences are not a data type.)
쐌
Oracle sequences have a name and can be used anywhere a value is expected.
쐌
Oracle sequences are not tied to a table or a column.
쐌
Oracle sequences generate a numeric value that can be assigned to any column in any table.
쐌
The table attribute to which you assigned a value based on a sequence can be edited and modified.
쐌
An Oracle sequence can be created and deleted anytime.
The basic syntax to create a sequence in Oracle is:
CREATE SEQUENCE name [START WITH n] [INCREMENT BY n] [CACHE | NOCACHE]
where:
쐌
name is the name of the sequence.
쐌
n is an integer value that can be positive or negative.
쐌
START WITH specifies the initial sequence value. (The default value is 1.)
A D V A N C E D
S Q L
쐌
INCREMENT BY determines the value by which the sequence is incremented. (The default increment value
is 1. The sequence increment can be positive or negative to enable you to create ascending or descending
sequences.)
쐌
The CACHE or NOCACHE clause indicates whether Oracle will preallocate sequence numbers in memory.
(Oracle preallocates 20 values by default.)
For example, you could create a sequence to automatically assign values to the customer code each time a new
customer is added and create another sequence to automatically assign values to the invoice number each time a new
invoice is added. The SQL code to accomplish those tasks is:
CREATE SEQUENCE CUS_CODE_SEQ START WITH 20010 NOCACHE;
CREATE SEQUENCE INV_NUMBER_SEQ START WITH 4010 NOCACHE;
You can check all of the sequences you have created by using the following SQL command, illustrated in Figure 8.22:
SELECT * FROM USER_SEQUENCES;
FIGURE
Oracle sequence
8.22
To use sequences during data entry, you must use two special pseudo-columns: NEXTVAL and CURRVAL. NEXTVAL
retrieves the next available value from a sequence, and CURRVAL retrieves the current value of a sequence. For
example, you can use the following code to enter a new customer:
INSERT INTO CUSTOMER
VALUES (CUS_CODE_SEQ.NEXTVAL, ‘Connery’, ‘Sean’, NULL, ‘615’, ‘898-2008’, 0.00);
The preceding SQL statement adds a new customer to the CUSTOMER table and assigns the value 20010 to the
CUS_CODE attribute. Let’s examine some important sequence characteristics:
쐌
CUS_CODE_SEQ.NEXTVAL retrieves the next available value from the sequence.
쐌
Each time you use NEXTVAL, the sequence is incremented.
쐌
Once a sequence value is used (through NEXTVAL), it cannot be used again. If, for some reason, your SQL
statement rolls back, the sequence value does not roll back. If you issue another SQL statement (with another
NEXTVAL), the next available sequence value will be returned to the user—it will look as though the sequence
skips a number.
쐌
You can issue an INSERT statement without using the sequence.
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CURRVAL retrieves the current value of a sequence—that is, the last sequence number used, which was generated with
a NEXTVAL. You cannot use CURRVAL unless a NEXTVAL was issued previously in the same session. The main use
for CURRVAL is to enter rows in dependent tables. For example, the INVOICE and LINE tables are related in a
one-to-many relationship through the INV_NUMBER attribute. You can use the INV_NUMBER_SEQ sequence to
automatically generate invoice numbers. Then, using CURRVAL, you can get the latest INV_NUMBER used and assign
it to the related INV_NUMBER foreign key attribute in the LINE table. For example:
INSERT INTO INVOICE VALUES (INV_NUMBER_SEQ.NEXTVAL, 20010, SYSDATE);
INSERT INTO LINE
VALUES (INV_NUMBER_SEQ.CURRVAL, 1,'13-Q2/P2', 1, 14.99);
INSERT INTO LINE
VALUES (INV_NUMBER_SEQ.CURRVAL, 2,'23109-HB', 1, 9.95);
COMMIT;
The results are shown in Figure 8.23.
FIGURE
Oracle sequence examples
8.23
In the example shown in Figure 8.23, INV_NUMBER_SEQ.NEXTVAL retrieves the next available sequence number
(4010) and assigns it to the INV_NUMBER column in the INVOICE table. Also note the use of the SYSDATE attribute
to automatically insert the current date in the INV_DATE attribute. Next, the following two INSERT statements add the
A D V A N C E D
S Q L
products being sold to the LINE table. In this case, INV_NUMBER_SEQ.CURRVAL refers to the last-used INV_
NUMBER_SEQ sequence number (4010). In this way, the relationship between INVOICE and LINE is established
automatically. The COMMIT statement at the end of the command sequence makes the changes permanent. Of
course, you can also issue a ROLLBACK statement, in which case the rows you inserted in INVOICE and LINE tables
would be rolled back (but remember that the sequence number would not). Once you use a sequence number (with
NEXTVAL), there is no way to reuse it! This “no-reuse” characteristic is designed to guarantee that the sequence will
always generate unique values.
Remember these points when you think about sequences:
쐌
The use of sequences is optional. You can enter the values manually.
쐌
A sequence is not associated with a table. As in the examples in Figure 8.23, two distinct sequences were
created (one for customer code values and one for invoice number values), but you could have created just one
sequence and used it to generate unique values for both tables.
Note
The SQL-2003 standard defined the use of Identity columns and sequence objects. However, some DBMS
vendors might not adhere to the standard. Check your DBMS documentation.
Finally, you can drop a sequence from a database with a DROP SEQUENCE command. For example, to drop the
sequences created earlier, you would type:
DROP SEQUENCE CUS_CODE_SEQ;
DROP SEQUENCE INV_NUMBER_SEQ;
Dropping a sequence does not delete the values you assigned to table attributes (CUS_CODE and INV_NUMBER); it
deletes only the sequence object from the database. The values you assigned to the table columns (CUS_CODE and
INV_NUMBER) remain in the database.
Because the CUSTOMER and INVOICE tables are used in the following examples, you’ll want to keep the original data
set. Therefore, you can delete the customer, invoice, and line rows you just added by using the following commands:
DELETE FROM INVOICE WHERE INV_NUMBER = 4010;
DELETE FROM CUSTOMER WHERE CUS_CODE = 20010;
COMMIT;
Those commands delete the recently added invoice and all of the invoice line rows associated with the invoice (the LINE
table’s INV_NUMBER foreign key was defined with the ON DELETE CASCADE option) and the recently added
customer. The COMMIT statement saves all changes to permanent storage.
Note
At this point, you’ll need to re-create the CUS_CODE_SEQ and INV_NUMBER_SEQ sequences, as they will be
used again later in the chapter. Enter:
CREATE SEQUENCE CUS_CODE_SEQ START WITH 20010 NOCACHE;
CREATE SEQUENCE INV_NUMBER_SEQ START WITH 4010 NOCACHE;
8.6 UPDATABLE VIEWS
In Chapter 7, you learned how to create a view and why and how views are used. You will now take a look at how
views can be made to serve common data management tasks executed by database administrators.
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One of the most common operations in production database environments is using batch update routines to update
a master table attribute (field) with transaction data. As the name implies, a batch update routine pools multiple
transactions into a single batch to update a master table field in a single operation. For example, a batch update
routine is commonly used to update a product’s quantity on hand based on summary sales transactions. Such routines
are typically run as overnight batch jobs to update the quantity on hand of products in inventory. The sales transactions
performed, for example, by traveling salespeople were entered during periods when the system was offline.
Online Content
For MS Access users, the PRODMASTER and PRODSALES tables are located in the Ch08_UV database, which
is located in the Premium Website for this book.
Online Content
For Oracle users, all SQL commands you see in this section are located in the Premium Website for this book.
After you locate the script files (uv-01.sql through uv-04.sql), you can copy and paste the command sequences
into your SQL*Plus program.
To demonstrate a batch update routine, let’s begin by defining the master product table (PRODMASTER) and the product
monthly sales totals table (PRODSALES) shown in Figure 8.24. Note the 1:1 relationship between the two tables.
FIGURE
The PRODMASTER and PRODSALES tables
8.24
Database name: CH08_UV
Table name: PRODMASTER
UPDATE
SET
WHERE
Table name: PRODSALES
Using the tables in Figure 8.24, let’s update
the PRODMASTER table by subtracting the
PRODSALES table’s product monthly sales
quantity (PS_QTY) from the PRODMASTER
table’s PROD_QOH. To produce the
required update, the update query would be
written like this:
PRODMASTER, PRODSALES
PRODMASTER.PROD_QOH = PROD_QOH − PS_QTY
PRODMASTER.PROD_ID = PRODSALES.PROD_ID;
Note that the update statement reflects the following sequence of events:
쐌
Join the PRODMASTER and PRODSALES tables.
쐌
Update the PROD_QOH attribute (using the PS_QTY value in the PRODSALES table) for each row of the
PRODMASTER table with matching PROD_ID values in the PRODSALES table.
To be used in a batch update, the PRODSALES data must be stored in a base table rather than in a view. That query
will work fine in Access, but Oracle will return the error message shown in Figure 8.25.
A D V A N C E D
FIGURE
S Q L
The Oracle UPDATE error message
8.25
Oracle produced the error message because Oracle expects to find a single table name in the UPDATE statement. In
fact, you cannot join tables in the UPDATE statement in Oracle. To solve that problem, you have to create an
updatable view. As its name suggests, an updatable view is a view that can be used to update attributes in the base
table(s) that is (are) used in the view. You must realize that not all views are updatable. Actually, several restrictions
govern updatable views, and some of them are vendor-specific.
Note
Keep in mind that the examples in this section are generated in Oracle. To see what restrictions are placed on
updatable views by the DBMS you are using, check the appropriate DBMS documentation.
The most common updatable view restrictions are as follows:
쐌
GROUP BY expressions or aggregate functions cannot be used.
쐌
You cannot use set operators such as UNION, INTERSECT, and MINUS.
쐌
Most restrictions are based on the use of JOINs or group operators in views.
To meet the Oracle limitations, an updatable view named PSVUPD has been created, as shown in Figure 8.26.
FIGURE
8.26
Creating an updatable view in Oracle
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One easy way to determine whether a view can be used to update a base table is to examine the view’s output. If the
primary key columns of the base table you want to update still have unique values in the view, the base table is
updatable. For example, if the PROD_ID column of the view returns the A123 or BX34 values more than once, the
PRODMASTER table cannot be updated through the view.
After creating the updatable view shown in Figure 8.26, you can use the UPDATE command to update the view,
thereby updating the PRODMASTER table. Figure 8.27 shows how the UPDATE command is used and what the final
contents of the PRODMASTER table are after the UPDATE has been executed.
FIGURE
PRODMASTER table update, using an updatable view
8.27
Although the batch update procedure just illustrated meets the goal of updating a master table with data from a transaction
table, the preferred real-world solution to the update problem is to use procedural SQL, which you’ll learn about next.
8.7 PROCEDURAL SQL
Thus far, you have learned to use SQL to read, write, and delete data in the database. For example, you learned to
update values in a record, to add records, and to delete records. Unfortunately, SQL does not support the conditional
execution of procedures that are typically supported by a programming language using the general format:
IF <condition>
THEN <perform procedure>
ELSE <perform alternate procedure>
END IF
A D V A N C E D
S Q L
SQL also fails to support the looping operations in programming languages that permit the execution of repetitive
actions typically encountered in a programming environment. The typical format is:
DO WHILE
<perform procedure>
END DO
Traditionally, if you wanted to perform a conditional (IF-THEN-ELSE) or looping (DO-WHILE) type of operation (that
is, a procedural type of programming), you would use a programming language such as Visual Basic .NET, C#, or
COBOL. That’s why many older (so-called legacy) business applications are based on enormous numbers of COBOL
program lines. Although that approach is still common, it usually involves the duplication of application code in many
programs. Therefore, when procedural changes are required, program modifications must be made in many different
programs. An environment characterized by such redundancies often creates data management problems.
A better approach is to isolate critical code and then have all application programs call the shared code. The advantage
of that modular approach is that the application code is isolated in a single program, thus yielding better maintenance
and logic control. In any case, the rise of distributed databases (see Chapter 12, Distributed Database Management
Systems) and object-oriented databases (see Appendix G in the Premium Website) required that more application code
be stored and executed within the database. To meet that requirement, most RDBMS vendors created numerous
programming language extensions. Those extensions include:
쐌
Flow-control procedural programming structures (IF-THEN-ELSE, DO-WHILE) for logic representation.
쐌
Variable declaration and designation within the procedures.
쐌
Error management.
To remedy the lack of procedural functionality in SQL and to provide some standardization within the many vendor
offerings, the SQL-99 standard defined the use of persistent stored modules. A persistent stored module (PSM) is
a block of code containing standard SQL statements and procedural extensions that is stored and executed at the
DBMS server. The PSM represents business logic that can be encapsulated, stored, and shared among multiple
database users. A PSM lets an administrator assign specific access rights to a stored module to ensure that only
authorized users can use it. Support for persistent stored modules is left to each vendor to implement. In fact, for many
years, some RDBMSs (such as Oracle, SQL Server, and DB2) supported stored procedure modules within the database
before the official standard was promulgated.
MS SQL Server implements persistent stored modules via Transact-SQL and other language extensions, the most
notable of which are the .NET family of programming languages. Oracle implements PSMs through its procedural SQL
language. Procedural SQL (PL/SQL) is a language that makes it possible to use and store procedural code and SQL
statements within the database and to merge SQL and traditional programming constructs, such as variables,
conditional processing (IF-THEN-ELSE), basic loops (FOR and WHILE loops,) and error trapping. The procedural code
is executed as a unit by the DBMS when it is invoked (directly or indirectly) by the end user. End users can use PL/SQL
to create:
쐌
Anonymous PL/SQL blocks.
쐌
Triggers (covered in Section 8.7.1).
쐌
Stored procedures (covered in Section 8.7.2 and Section 8.7.3).
쐌
PL/SQL functions (covered in Section 8.7.4).
Do not confuse PL/SQL functions with SQL’s built-in aggregate functions such as MIN and MAX. SQL built-in
functions can be used only within SQL statements, while PL/SQL functions are mainly invoked within PL/SQL
programs such as triggers and stored procedures. Functions can also be called within SQL statements, provided that
they conform to very specific rules that are dependent on your DBMS environment.
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Note
PL/SQL, triggers, and stored procedures are illustrated within the context of an Oracle DBMS. All examples in
the following sections assume the use of Oracle RDBMS.
Using Oracle SQL*Plus, you can write a PL/SQL code block by enclosing the commands inside BEGIN and END
clauses. For example, the following PL/SQL block inserts a new row in the VENDOR table, as shown in Figure 8.28.
BEGIN
INSERT INTO VENDOR
VALUES (25678,'Microsoft Corp. ', 'Bill Gates','765','546-8484','WA','N');
END;
/
FIGURE
Anonymous PL/SQL block examples
8.28
The PL/SQL block shown in Figure 8.28 is known as an anonymous PL/SQL block because it has not been given
a specific name. (Incidentally, note that the block’s last line uses a forward slash (“/”) to indicate the end of the
command-line entry.) That type of PL/SQL block executes as soon as you press the Enter key after typing the forward
slash. Following the PL/SQL block’s execution, you will see the message “PL/SQL procedure successfully completed.”
A D V A N C E D
S Q L
But suppose that you want a more specific message displayed on the SQL*Plus screen after a procedure is completed,
such as “New Vendor Added.” To produce a more specific message, you must do two things:
1.
At the SQL > prompt, type SET SERVEROUTPUT ON. This SQL*Plus command enables the client console
(SQL*Plus) to receive messages from the server side (Oracle DBMS). Remember, just like standard SQL, the
PL/SQL code (anonymous blocks, triggers, and procedures) are executed at the server side, not at the client
side. (To stop receiving messages from the server, you would enter SET SERVEROUT OFF.)
2.
To send messages from the PL/SQL block to the SQL*Plus console, use the DBMS_OUTPUT.PUT_LINE
function.
The following anonymous PL/SQL block inserts a row in the VENDOR table and displays the message “New Vendor
Added!” (See Figure 8.28.)
BEGIN
INSERT INTO VENDOR
VALUES (25772,'Clue Store', 'Issac Hayes', '456','323-2009', 'VA', 'N');
DBMS_OUTPUT.PUT_LINE('New Vendor Added!');
END;
/
In Oracle, you can use the SQL*Plus command SHOW ERRORS to help you diagnose errors found in PL/SQL blocks.
The SHOW ERRORS command yields additional debugging information whenever you generate an error after
creating or executing a PL/SQL block.
The following example of an anonymous PL/SQL block demonstrates several of the constructs supported by the
procedural language. Remember that the exact syntax of the language is vendor-dependent; in fact, many vendors
enhance their products with proprietary features.
DECLARE
W_P1 NUMBER(3) := 0;
W_P2 NUMBER(3) := 10;
W_NUM NUMBER(2) := 0;
BEGIN
WHILE W_P2 < 300 LOOP
SELECT COUNT(P_CODE) INTO W_NUM FROM PRODUCT
WHERE P_PRICE BETWEEN W_P1 AND W_P2;
DBMS_OUTPUT.PUT_LINE('There are ' || W_NUM || ' Products with price between ' || W_P1
|| ' and ' || W_P2);
W_P1 := W_P2 + 1;
W_P2 := W_P2 + 50;
END LOOP;
END;
/
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The block’s code and execution are shown in Figure 8.29.
FIGURE
Anonymous PL/SQL block with variables and loops
8.29
The PL/SQL block shown in Figure 8.29 has the following characteristics:
쐌
The PL/SQL block starts with the DECLARE section in which you declare the variable names, the data types,
and, if desired, an initial value. Supported data types are shown in Table 8.8.
TABLE
PL/SQL Basic Data Types
8.8
DATA TYPE
CHAR
VARCHAR2
NUMBER
DATE
%TYPE
쐌
DESCRIPTION
Character values of a fixed length; for example:
W_ZIP CHAR(5)
Variable length character values; for example:
W_FNAME VARCHAR2(15)
Numeric values; for example:
W_PRICE NUMBER(6,2)
Date values; for example:
W_EMP_DOB DATE
Inherits the data type from a variable that you declared previously or from an attribute of
a database table; for example:
W_PRICE PRODUCT.P_PRICE%TYPE
Assigns W_PRICE the same data type as the P_PRICE column in the PRODUCT table
A WHILE loop is used. Note the syntax:
WHILE condition LOOP
PL/SQL statements;
END LOOP
쐌
The SELECT statement uses the INTO keyword to assign the output of the query to a PL/SQL variable. You
can use the INTO keyword only inside a PL/SQL block of code. If the SELECT statement returns more than
one value, you will get an error.
A D V A N C E D
쐌
Note the use of the string concatenation symbol “||” to display the output.
쐌
Each statement inside the PL/SQL code must end with a semicolon “;”.
S Q L
Note
PL/SQL blocks can contain only standard SQL data manipulation language (DML) commands such as SELECT,
INSERT, UPDATE, and DELETE. The use of data definition language (DDL) commands is not directly supported
in a PL/SQL block.
The most useful feature of PL/SQL blocks is that they let you create code that can be named, stored, and
executed—either implicitly or explicitly—by the DBMS. That capability is especially desirable when you need to use
triggers and stored procedures, which you will explore next.
8.7.1
Triggers
Automating business procedures and automatically maintaining data integrity and consistency are critical in a modern
business environment. One of the most critical business procedures is proper inventory management. For example,
you want to make sure that current product sales can be supported with sufficient product availability. Therefore, it is
necessary to ensure that a product order be written to a vendor when that product’s inventory drops below its minimum
allowable quantity on hand. Better yet, how about ensuring that the task is completed automatically?
To accomplish automatic product ordering, you first must make sure the product’s quantity on hand reflects an
up-to-date and consistent value. After the appropriate product availability requirements have been set, two key issues
must be addressed:
1.
Business logic requires an update of the product quantity on hand each time there is a sale of that product.
2.
If the product’s quantity on hand falls below its minimum allowable inventory (quantity-on-hand) level, the
product must be reordered.
To accomplish those two tasks, you could write multiple SQL statements: one to update the product quantity on hand
and another to update the product reorder flag. Next, you would have to run each statement in the correct order each
time there was a new sale. Such a multistage process would be inefficient because a series of SQL statements must
be written and executed each time a product is sold. Even worse, that SQL environment requires that somebody must
remember to perform the SQL tasks.
A trigger is procedural SQL code that is automatically invoked by the RDBMS upon the occurrence of a given data
manipulation event. It is useful to remember that:
쐌
A trigger is invoked before or after a data row is inserted, updated, or deleted.
쐌
A trigger is associated with a database table.
쐌
Each database table may have one or more triggers.
쐌
A trigger is executed as part of the transaction that triggered it.
Triggers are critical to proper database operation and management. For example:
쐌
Triggers can be used to enforce constraints that cannot be enforced at the DBMS design and implementation
levels.
쐌
Triggers add functionality by automating critical actions and providing appropriate warnings and suggestions
for remedial action. In fact, one of the most common uses for triggers is to facilitate the enforcement of
referential integrity.
쐌
Triggers can be used to update table values, insert records in tables, and call other stored procedures.
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Triggers play a critical role in making the database truly useful; they also add processing power to the RDBMS and to
the database system as a whole. Oracle recommends triggers for:
쐌
Auditing purposes (creating audit logs).
쐌
Automatic generation of derived column values.
쐌
Enforcement of business or security constraints.
쐌
Creation of replica tables for backup purposes.
To see how a trigger is created and used, let’s examine a simple inventory management problem. For example, if a
product’s quantity on hand is updated when the product is sold, the system should automatically check whether the
quantity on hand falls below its minimum allowable quantity. To demonstrate that process, let’s use the PRODUCT
table in Figure 8.30. Note the use of the minimum order quantity (P_MIN_ORDER) and the product reorder flag
(P_REORDER) columns. The P_MIN_ORDER indicates the minimum quantity for restocking an order. The
P_REORDER column is a numeric field that indicates whether the product needs to be reordered (1 = Yes, 0 = No).
The initial P_REORDER values will be set to 0 (No) to serve as the basis for the initial trigger development.
FIGURE
The PRODUCT table
8.30
Online Content
Oracle users can run the PRODLIST.SQL script file to format the output of the PRODUCT table shown in Figure
8.30. The script file is located in the Premium Website for this book.
Given the PRODUCT table listing shown in Figure 8.30, let’s create a trigger to evaluate the product’s quantity on
hand, P_QOH. If the quantity on hand is below the minimum quantity shown in P_MIN, the trigger will set the
A D V A N C E D
S Q L
P_REORDER column to 1. (Remember that the number 1 in the P_REORDER column represents “Yes.”) The syntax
to create a trigger in Oracle is:
CREATE OR REPLACE TRIGGER trigger_name
[BEFORE / AFTER] [DELETE / INSERT / UPDATE OF column_name] ON table_name
[FOR EACH ROW]
[DECLARE]
[variable_namedata type[:=initial_value] ]
BEGIN
PL/SQL instructions;
..........
END;
As you can see, a trigger definition contains the following parts:
쐌
The triggering timing: BEFORE or AFTER. This timing indicates when the trigger’s PL/SQL code executes;
in this case, before or after the triggering statement is completed.
쐌
The triggering event: the statement that causes the trigger to execute (INSERT, UPDATE, or DELETE).
쐌
The triggering level: There are two types of triggers: statement-level triggers and row-level triggers.
쐌
-
A statement-level trigger is assumed if you omit the FOR EACH ROW keywords. This type of trigger
is executed once, before or after the triggering statement is completed. This is the default case.
-
A row-level trigger requires use of the FOR EACH ROW keywords. This type of trigger is executed once
for each row affected by the triggering statement. (In other words, if you update 10 rows, the trigger
executes 10 times.)
The triggering action: The PL/SQL code enclosed between the BEGIN and END keywords. Each statement
inside the PL/SQL code must end with a semicolon “;”.
In the PRODUCT table’s case, you will create a statement-level trigger that is implicitly executed AFTER an UPDATE
of the P_QOH attribute for an existing row or AFTER an INSERT of a new row in the PRODUCT table. The trigger
action executes an UPDATE statement that compares the P_QOH with the P_MIN column. If the value of P_QOH is
equal to or less than P_MIN, the trigger updates the P_REORDER to 1. To create the trigger, Oracle’s SQL*Plus will
be used. The trigger code is shown in Figure 8.31.
FIGURE
8.31
Creating the TRG_PRODUCT_REORDER trigger
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Online Content
The source code for all of the triggers shown in this section can be found in the Premium Website for this book.
To test the TRG_PRODUCT_REORDER trigger, let’s update the quantity on hand of product ‘11QER/31’ to 4. After
the UPDATE completes, the trigger is automatically fired and the UPDATE statement (inside the trigger code) sets the
P_REORDER to 1 for all products that are below the minimum. (See Figure 8.32.)
FIGURE
Verifying the TRG_PRODUCT_REORDER trigger execution
8.32
The trigger shown in Figure 8.32 seems to work fine, but what happens if you reduce the minimum quantity of product
‘2232/QWE’? Figure 8.33 shows that when you update the minimum quantity, the quantity on hand of the product
‘2232/QWE’ falls below the new minimum, but the reorder flag is still 0. Why?
FIGURE
The P_REORDER value mismatch after update of the P_MIN attribute
8.33
The answer is simple: you updated the P_MIN column, but the trigger is never executed. TRG_PRODUCT_
REORDER executes only after an update of the P_QOH column! To avoid that inconsistency, you must modify the
trigger event to execute after an update of the P_MIN field, too. The updated trigger code is shown in Figure 8.34.
A D V A N C E D
FIGURE
S Q L
Second version of the TRG_PRODUCT_REORDER trigger
8.34
To test this new trigger version, let’s change the minimum quantity for product ‘23114-AA’ to 10. After that update, the
trigger makes sure that the reorder flag is properly set for all of the products in the PRODUCT table. (See Figure 8.35.)
FIGURE
Successful trigger execution after the P_MIN value is updated
8.35
This second version of the trigger seems to work well, but what happens if you change the P_QOH value for product
‘11QER/31’, as shown in Figure 8.36? Nothing! (Note that the reorder flag is still set to 1.) Why didn’t the trigger
change the reorder flag to 0?
The answer is that the trigger does not consider all possible cases. Let’s examine the second version of the
TRG_PRODUCT_REORDER trigger code (Figure 8.34) in more detail:
쐌
The trigger fires after the triggering statement is completed. Therefore, the DBMS always executes two
statements (INSERT plus UPDATE or UPDATE plus UPDATE). That is, after you do an update of P_MIN or
P_QOH or you insert a new row in the PRODUCT table, the trigger executes another UPDATE statement
automatically.
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FIGURE
8
The P_REORDER value mismatch after increasing the P_QOH value
8.36
쐌
The triggering action performs an UPDATE that updates all of the rows in the PRODUCT table, even if the
triggering statement updates just one row! This can affect the performance of the database. Imagine what
will happen if you have a PRODUCT table with 519,128 rows and you insert just one product. The trigger will
update all 519,129 rows (519,128 original rows plus the one you inserted), including the rows that do not need
an update!
쐌
The trigger sets the P_REORDER value only to 1; it does not reset the value to 0, even if such an action is
clearly required when the inventory level is back to a value greater than the minimum value.
In short, the second version of the TRG_PRODUCT_REORDER trigger still does not complete all of the necessary
steps. Now let’s modify the trigger to handle all update scenarios, as shown in Figure 8.37.
FIGURE
The third version of the TRG_PRODUCT_REORDER trigger
8.37
The trigger in Figure 8.37 sports several new features:
쐌
The trigger is executed before the actual triggering statement is completed. In Figure 8.37, the triggering
timing is defined in line 2, BEFORE INSERT OR UPDATE. This clearly indicates that the triggering statement
is executed before the INSERT or UPDATE completes, unlike the previous trigger examples.
A D V A N C E D
S Q L
쐌
The trigger is a row-level trigger instead of a statement-level trigger. The FOR EACH ROW keywords make the
trigger a row-level trigger. Therefore, this trigger executes once for each row affected by the triggering statement.
쐌
The trigger action uses the :NEW attribute reference to change the value of the P_REORDER attribute.
The use of the :NEW attribute references deserves a more detailed explanation. To understand its use, you must first
consider a basic computing tenet: all changes are done first in primary memory, then transferred to permanent
memory. In other words, the computer cannot change anything directly in permanent storage (disk). It must first read
the data from permanent storage to primary memory, then it makes the change in primary memory, and finally, it
writes the changed data back to permanent memory (disk).
The DBMS operates in the same way, with one addition. Because ensuring data integrity is critical, the DBMS makes
two copies of every row being changed by a DML (INSERT, UPDATE, or DELETE) statement. (You will learn more
about this in Chapter 10, Transaction Management and Concurrency Control.) The first copy contains the original
(“old”) values of the attributes before the changes. The second copy contains the changed (“new”) values of the
attributes that will be permanently saved to the database (after any changes made by an INSERT, UPDATE, or
DELETE). You can use :OLD to refer to the original values; you can use :NEW to refer to the changed values (the values
that will be stored in the table). You can use :NEW and :OLD attribute references only within the PL/SQL code of a
database trigger action. For example:
쐌
IF :NEW.P_QOH < = :NEW.P_MIN compares the quantity on hand with the minimum quantity of a product.
Remember that this is a row-level trigger. Therefore, this comparison is done for each row that is updated by
the triggering statement.
쐌
Although the trigger is a BEFORE trigger, this does not mean that the triggering statement hasn’t executed yet.
To the contrary, the triggering statement has already taken place; otherwise, the trigger would not have fired
and the :NEW values would not exist. Remember, BEFORE means before the changes are permanently saved
to disk, but after the changes are made in memory.
쐌
The trigger uses the :NEW reference to assign a value to the P_REORDER column before the UPDATE or
INSERT results are permanently stored in the table. The assignment is always done to the :NEW value (never
to the :OLD value), and the assignment always uses the := assignment operator. The :OLD values are
read-only values; you cannot change them. Note that :NEW.P_REORDER := 1; assigns the value 1 to the
P_REORDER column and :NEW.P_REORDER := 0; assigns the value 0 to the P_REORDER column.
쐌
This new trigger version does not use any DML statement!
Before testing the new trigger, note that product ‘11QER/31’ currently has a quantity on hand that is above the
minimum quantity, yet the reorder flag is set to 1. Given that condition, the reorder flag must be 0. After creating the
new trigger, you can execute an UPDATE statement to fire it, as shown in Figure 8.38.
Note the following important features of the code in Figure 8.38:
쐌
The trigger is automatically invoked for each affected row—in this case, all rows of the PRODUCT table. If your
triggering statement would have affected only three rows, not all PRODUCT rows would have the correct
P_REORDER value set. That’s the reason the triggering statement was set up as shown in Figure 8.38.
쐌
The trigger will run only if you insert a new product row or update P_QOH or P_MIN. If you update any other
attribute, the trigger won’t run.
You can also use a trigger to update an attribute in a table other than the one being modified. For example, suppose
that you would like to create a trigger that automatically reduces the quantity on hand of a product with every sale. To
accomplish that task, you must create a trigger for the LINE table that updates a row in the PRODUCT table. The
sample code for that trigger is shown in Figure 8.39.
Note that the TRG_LINE_PROD row-level trigger executes after inserting a new invoice’s LINE and reduces the
quantity on hand of the recently sold product by the number of units sold. This row-level trigger updates a row in a
different table (PRODUCT), using the :NEW values of the recently added LINE row.
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FIGURE
8
Execution of the third trigger version
8.38
FIGURE
TRG_LINE_PROD trigger to update the PRODUCT quantity on hand
8.39
A third trigger example shows the use of variables within a trigger. In this case, you want to update the customer
balance (CUS_BALANCE) in the CUSTOMER table after inserting every new LINE row. This trigger code is shown
in Figure 8.40.
Let’s carefully examine the trigger in Figure 8.40.
쐌
The trigger is a row-level trigger that executes after each new LINE row is inserted.
쐌
The DECLARE section in the trigger is used to declare any variables used inside the trigger code.
쐌
You can declare a variable by assigning a name, a data type, and (optionally) an initial value, as in the case of
the W_TOT variable.
A D V A N C E D
FIGURE
S Q L
TRG_LINE_CUS trigger to update the customer balance
8.40
쐌
OThe first step in the trigger code is to get the customer code (CUS_CODE) from the related INVOICE table.
Note that the SELECT statement returns only one attribute (CUS_CODE) from the INVOICE table. Also note
that that attribute returns only one value as specified by the use of the WHERE clause to restrict the query
output to a single value.
쐌
Note the use of the INTO clause within the SELECT statement. You use the INTO clause to assign a value from
a SELECT statement to a variable (W_CUS) used within a trigger.
쐌
The second step in the trigger code computes the total of the line by multiplying :NEW.LINE_UNITS times
:NEW.LINE_PRICE and assigning the result to the W_TOT variable.
쐌
The final step updates the customer balance by using an UPDATE statement and the W_TOT and W_CUS
trigger variables.
쐌
Double dashes “--” are used to indicate comments within the PL/SQL block.
Let’s summarize the triggers created in this section.
쐌
The TRG_PROD_REORDER is a row-level trigger that updates P_REORDER in PRODUCT when a new
product is added or when the P_QOH or P_MIN columns are updated.
쐌
The TRG_LINE_PROD is a row-level trigger that automatically reduces the P_QOH in PRODUCT when a new
row is added to the LINE table.
쐌
TRG_LINE_CUS is a row-level trigger that automatically increases the CUS_BALANCE in CUSTOMER when
a new row is added in the LINE table.
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The use of triggers facilitates the automation of multiple data management tasks. Although triggers are independent
objects, they are associated with database tables. When you delete a table, all its trigger objects are deleted with it.
However, if you needed to delete a trigger without deleting the table, you could use the following command:
DROP TRIGGER trigger_name
Trigger Action Based on Conditional DML Predicates
You could also create triggers whose actions depend on the type of DML statement (INSERT, UPDATE, or DELETE)
that fires the trigger. For example, you could create a trigger that executes after an insert, an update, or a delete on
the PRODUCT table. But how do you know which one of the three statements caused the trigger to execute? In those
cases, you could use the following syntax:
IF INSERTING THEN ... END IF;
IF UPDATING THEN ... END IF;
IF DELETING THEN ... END IF;
8.7.2 Stored Procedures
A stored procedure is a named collection of procedural and SQL statements. Just like database triggers, stored
procedures are stored in the database. One of the major advantages of stored procedures is that they can be used to
encapsulate and represent business transactions. For example, you can create a stored procedure to represent a
product sale, a credit update, or the addition of a new customer. By doing that, you can encapsulate SQL statements
within a single stored procedure and execute them as a single transaction. There are two clear advantages to the use
of stored procedures:
쐌
Stored procedures substantially reduce network traffic and increase performance. Because the procedure is
stored at the server, there is no transmission of individual SQL statements over the network. The use of stored
procedures improves system performance because all transactions are executed locally on the RDBMS, so each
SQL statement does not have to travel over the network.
쐌
Stored procedures help reduce code duplication by means of code isolation and code sharing (creating unique
PL/SQL modules that are called by application programs), thereby minimizing the chance of errors and the
cost of application development and maintenance.
To create a stored procedure, you use the following syntax:
CREATE OR REPLACE PROCEDURE procedure_name [(argument [IN/OUT] data-type, ѧ )]
[IS/AS]
[variable_namedata type[:=initial_value] ]
BEGIN
PL/SQL or SQL statements;
...
END;
Note the following important points about stored procedures and their syntax:
쐌
argument specifies the parameters that are passed to the stored procedure. A stored procedure could have
zero or more arguments or parameters.
쐌
IN/OUT indicates whether the parameter is for input, output, or both.
쐌
data-type is one of the procedural SQL data types used in the RDBMS. The data types normally match those
used in the RDBMS table-creation statement.
쐌
Variables can be declared between the keywords IS and BEGIN. You must specify the variable name, its data
type, and (optionally) an initial value.
A D V A N C E D
S Q L
To illustrate stored procedures, assume that you want to create a procedure (PRC_PROD_DISCOUNT) to assign an
additional 5 percent discount for all products when the quantity on hand is more than or equal to twice the minimum
quantity. Figure 8.41 shows how the stored procedure is created.
FIGURE
Creating the PRC_PROD_DISCOUNT stored procedure
8.41
Online Content
The source code for all of the stored procedures shown in this section can be found in the Premium Website for
this book.
Note in Figure 8.41 that the PRC_PROD_DISCOUNT stored procedure uses the DBMS_OUTPUT.PUT_LINE
function to display a message when the procedure executes. (This action assumes that you previously ran SET
SERVEROUTPUT ON.)
To execute the stored procedure, you must use the following syntax:
EXEC procedure_name[(parameter_list)];
For example, to see the results of running the PRC_PROD_DISCOUNT stored procedure, you can use the EXEC
PRC_PROD_DISCOUNT command shown in Figure 8.42.
Using Figure 8.42 as your guide, you can see how the product discount attribute for all products with a quantity on
hand more than or equal to twice the minimum quantity was increased by 5 percent. (Compare the first PRODUCT
table listing to the second PRODUCT table listing.)
One of the main advantages of procedures is that you can pass values to them. For example, the previous
PRC_PROD_DISCOUNT procedure worked fine, but what if you want to make the percentage increase an input
variable? In that case, you can pass an argument to represent the rate of increase to the procedure. Figure 8.43 shows
the code for that procedure.
Figure 8.44 shows the execution of the second version of the PRC_PROD_DISCOUNT stored procedure. Note that,
if the procedure requires arguments, those arguments must be enclosed in parentheses and they must be separated by
commas.
Stored procedures are also useful to encapsulate shared code to represent business transactions. For example, you can
create a simple stored procedure to add a new customer. By using a stored procedure, all programs can call the stored
procedure by name each time a new customer is added. Naturally, if new customer attributes are added later, you will
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FIGURE
8
Results of the PRC_PROD_DISCOUNT stored procedure
8.42
FIGURE
Second version of the PRC_PROD_DISCOUNT stored procedure
8.43
need to modify the stored procedure. However, the programs that use the stored procedure will not need to know the
name of the newly added attribute and will need to add only a new parameter to the procedure call. (Notice the
PRC_CUS_ADD stored procedure shown in Figure 8.45.)
A D V A N C E D
FIGURE
S Q L
Results of the second version of the PRC_PROD_DISCOUNT stored procedure
8.44
FIGURE
The PRC_CUS_ADD stored procedure
8.45
As you examine Figure 8.45, note these features:
쐌
The PRC_CUS_ADD procedure uses several parameters, one for each required attribute in the CUSTOMER table.
쐌
The stored procedure uses the CUS_CODE_SEQ sequence to generate a new customer code.
쐌
The required parameters—those specified in the table definition—must be included and can be null only when
the table specifications permit nulls for that parameter. For example, note that the second customer addition
was unsuccessful because the CUS_AREACODE is a required attribute and cannot be null.
쐌
The procedure displays a message in the SQL*Plus console to let the user know that the customer was added.
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The next two examples further illustrate the use of sequences within stored procedures. In this case, let’s create two
stored procedures:
1.
The PRC_INV_ADD procedure adds a new invoice.
2.
The PRC_LINE_ADD procedure adds a new product line row for a given invoice.
Both procedures are shown in Figure 8.46. Note the use of a variable in the PRC_LINE_ADD procedure to get the
product price from the PRODUCT table.
FIGURE
The PRC_INV_ADD and PRC_LINE_ADD stored procedures
8.46
To test the procedures shown in Figure 8.46:
1.
Call the PRC_INV_ADD procedure with the new invoice data as arguments.
2.
Call the PRC_LINE_ADD procedure and pass the product line arguments.
That process is illustrated in Figure 8.47.
8.7.3 PL/SQL Processing with Cursors
Until now, all of the SQL statements you have used inside a PL/SQL block (trigger or stored procedure) have returned
a single value. If the SQL statement returns more than one value, you will generate an error. If you want to use an SQL
statement that returns more than one value inside your PL/SQL code, you need to use a cursor. A cursor is a special
construct used in procedural SQL to hold the data rows returned by an SQL query. You can think of a cursor as a
reserved area of memory in which the output of the query is stored, like an array holding columns and rows. Cursors
are held in a reserved memory area in the DBMS server, not in the client computer.
There are two types of cursors: implicit and explicit. An implicit cursor is automatically created in procedural SQL
when the SQL statement returns only one value. Up to this point, all of the examples created an implicit cursor. An
explicit cursor is created to hold the output of an SQL statement that may return two or more rows (but could return
0 or only one row). To create an explicit cursor, you use the following syntax inside a PL/SQL DECLARE section:
CURSOR cursor_name IS select-query;
A D V A N C E D
FIGURE
S Q L
Testing the PRC_INV_ADD and PRC_LINE_ADD procedures
8.47
Once you have declared a cursor, you can use specific PL/SQL cursor processing commands (OPEN, FETCH, and
CLOSE) anywhere between the BEGIN and END keywords of the PL/SQL block. Table 8.9 summarizes the main use
of each of those commands.
TABLE
8.9
CURSOR
COMMAND
OPEN
FETCH
CLOSE
Cursor Processing Commands
EXPLANATION
Opening the cursor executes the SQL command and populates the cursor with data, opening the
cursor for processing. The cursor declaration command only reserves a named memory area for
the cursor; it doesn’t populate the cursor with the data. Before you can use a cursor, you need to
open it. For example:
OPEN cursor_name
Once the cursor is opened, you can use the FETCH command to retrieve data from the cursor and
copy it to the PL/SQL variables for processing. The syntax is:
FETCH cursor_name INTO variable1 [, variable2, ...]
The PL/SQL variables used to hold the data must be declared in the DECLARE section and must
have data types compatible with the columns retrieved by the SQL command. If the cursors SQL
statement returns five columns, there must be five PL/SQL variables to receive the data from the
cursor.
This type of processing resembles the one-record-at-a-time processing used in previous database
models. The first time you fetch a row from the cursor, the first row of data from the cursor is copied to the PL/SQL variables; the second time you fetch a row from the cursor, the second row of
data is placed in the PL/SQL variables; and so on.
The CLOSE command closes the cursor for processing.
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Cursor-style processing involves retrieving data from the cursor one row at a time. Once you open a cursor, it becomes
an active data set. That data set contains a “current” row pointer. Therefore, after opening a cursor, the current row
is the first row of the cursor.
When you fetch a row from the cursor, the data from the “current” row in the cursor is copied to the PL/SQL variables.
After the fetch, the “current” row pointer moves to the next row in the set and continues until it reaches the end of
the cursor.
How do you know what number of rows are in the cursor? Or how do you know when you have reached the end of
the cursor data set? You know because cursors have special attributes that convey important information. Table 8.10
summarizes the cursor attributes.
TABLE
Cursor Attributes
8.10
ATTRIBUTE
%ROWCOUNT
%FOUND
%NOTFOUND
%ISOPEN
DESCRIPTION
Returns the number of rows fetched so far. If the cursor is not OPEN, it returns an error. If
no FETCH has been done but the cursor is OPEN, it returns 0.
Returns TRUE if the last FETCH returned a row and FALSE if not. If the cursor is not
OPEN, it returns an error. If no FETCH has been done, it contains NULL.
Returns TRUE if the last FETCH did not return any row and FALSE if it did. If the cursor is
not OPEN, it returns an error. If no FETCH has been done, it contains NULL.
Returns TRUE if the cursor is open (ready for processing) or FALSE if the cursor is closed.
Remember, before you can use a cursor, you must open it.
To illustrate the use of cursors, let’s use a simple stored procedure example that lists all products that have a quantity
on hand greater than the average quantity on hand for all products. The code is shown in Figure 8.48.
As you examine the stored procedure code shown in Figure 8.48, note the following important characteristics:
쐌
Lines 2 and 3 use the %TYPE data type in the variable definition section. As indicated in Table 8.8, the %TYPE
data type is used to indicate that the given variable inherits the data type from a variable previously declared
or from an attribute of a database table. In this case, you are using the %TYPE to indicate that the W_P_CODE
and W_P_DESCRIPT will have the same data type as the respective columns in the PRODUCT table. This
way, you ensure that the PL/SQL variable will have a compatible data type.
쐌
Line 5 declares the PROD_CURSOR cursor.
쐌
Line 12 opens the PROD_CURSOR cursor and populates it.
쐌
Line 13 uses the LOOP statement to loop through the data in the cursor, fetching one row at a time.
쐌
Line 14 uses the FETCH command to retrieve a row from the cursor and place it in the respective PL/SQL
variables.
쐌
Line 15 uses the EXIT command to evaluate when there are no more rows in the cursor (using the
%NOTFOUND cursor attribute) and to exit the loop.
쐌
Line 19 uses the %ROWCOUNT cursor attribute to obtain the total number of rows processed.
쐌
Line 21 issues the CLOSE PROD_CURSOR command to close the cursor.
The use of cursors, combined with standard SQL, makes relational databases very desirable because programmers can
work in the best of both worlds: set-oriented processing and record-oriented processing. Any experienced programmer
knows to use the tool that best fits the job. Sometimes you will be better off manipulating data in a set-oriented
environment; at other times, it might be better to use a record-oriented environment. Procedural SQL lets you have
your proverbial cake and eat it, too. Procedural SQL provides functionality that enhances the capabilities of the DBMS
while maintaining a high degree of manageability.
A D V A N C E D
FIGURE
S Q L
A simple PRC_CURSOR_EXAMPLE
8.48
8.7.4 PL/SQL Stored Functions
Using programmable or procedural SQL, you can also create your own stored functions. Stored procedures and
functions are very similar. A stored function is basically a named group of procedural and SQL statements that
returns a value (indicated by a RETURN statement in its program code). To create a function, you use the following
syntax:
CREATE FUNCTION function_name (argument IN data-type, ѧ ) RETURN data-type [IS]
BEGIN
PL/SQL statements;
...
RETURN (value or expression);
END;
Stored functions can be invoked only from within stored procedures or triggers and cannot be invoked from SQL
statements (unless the function follows some very specific compliance rules). Remember not to confuse built-in SQL
functions (such as MIN, MAX, and AVG) with stored functions.
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8.8 EMBEDDED SQL
There is little doubt that SQL’s popularity as a data manipulation language is in part due to its ease of use and its
powerful data-retrieval capabilities. But in the real world, database systems are related to other systems and programs,
and you still need a conventional programming language such as Visual Basic .NET, C#, or COBOL to integrate
database systems with other programs and systems. If you are developing Web applications, you are most likely familiar
with Visual Studio .NET, Java, ASP, or ColdFusion. Yet, almost regardless of the programming tools you use, if your
Web application or Windows-based GUI system requires access to a database such as MS Access, SQL Server, Oracle,
or DB2, you will likely need to use SQL to manipulate the data in the database.
Embedded SQL is a term used to refer to SQL statements that are contained within an application programming
language such as Visual Basic .NET, C#, COBOL, or Java. The program being developed might be a standard binary
executable in Windows or Linux, or it might be a Web application designed to run over the Internet. No matter what
language you use, if it contains embedded SQL statements, it is called the host language. Embedded SQL is still the
most common approach to maintaining procedural capabilities in DBMS-based applications. However, mixing SQL
with procedural languages requires that you understand some key differences between SQL and procedural languages.
쐌
Run-time mismatch: Remember that SQL is a nonprocedural, interpreted language; that is, each instruction
is parsed, its syntax is checked, and it is executed one instruction at a time.1 All of the processing takes place
at the server side. Meanwhile, the host language is generally a binary-executable program (also known as a
compiled program). The host program typically runs at the client side in its own memory space (which is
different from the DBMS environment).
쐌
Processing mismatch: Conventional programming languages (COBOL, ADA, FORTRAN, PASCAL, C++,
and PL/I) process one data element at a time. Although you can use arrays to hold data, you still process the
array elements one row at a time. This is especially true for file manipulation, where the host language typically
manipulates data one record at a time. However, newer programming environments (such as Visual Studio
.NET) have adopted several object-oriented extensions that help the programmer manipulate data sets in a
cohesive manner.
쐌
Data type mismatch: SQL provides several data types, but some of those data types might not match data
types used in different host languages (for example, the date and varchar2 data types).
To bridge the differences, the Embedded SQL standard2 defines a framework to integrate SQL within several
programming languages. The Embedded SQL framework defines the following:
쐌
A standard syntax to identify embedded SQL code within the host language (EXEC SQL/END-EXEC).
쐌
A standard syntax to identify host variables. Host variables are variables in the host language that receive data
from the database (through the embedded SQL code) and process the data in the host language. All host
variables are preceded by a colon (“:”).
쐌
A communication area used to exchange status and error information between SQL and the host language.
This communications area contains two variables—SQLCODE and SQLSTATE.
Another way to interface host languages and SQL is through the use of a call level interface (CLI),2 in which the
programmer writes to an application programming interface (API). A common CLI in Windows is provided by the
Open Database Connectivity (ODBC) interface.
1 The authors are particularly grateful for the thoughtful comments provided by Emil T. Cipolla.
2 You can obtain more details about the Embedded SQL standard at www.ansi.org, SQL/Bindings is in the SQL Part II–SQL/Foundation section of
the SQL 2003 standard.
A D V A N C E D
S Q L
Online Content
Additional coverage of CLIs and ODBC is found in Appendix F, Client/Server Systems, and Appendix J, Web
Database Development with ColdFusion in the Premium Website for this book.
Before continuing, let’s explore the process required to create and run an executable program with embedded SQL
statements. If you have ever programmed in COBOL or C++, you are familiar with the multiple steps required to
generate the final executable program. Although the specific details vary among language and DBMS vendors, the
following general steps are standard:
1.
The programmer writes embedded SQL code within the host language instructions. The code follows the
standard syntax required for the host language and embedded SQL.
2.
A preprocessor is used to transform the embedded SQL into specialized procedure calls that are DBMS- and
language-specific. The preprocessor is provided by the DBMS vendor and is specific to the host language.
3.
The program is compiled using the host language compiler. The compiler creates an object code module for
the program containing the DBMS procedure calls.
4.
The object code is linked to the respective library modules and generates the executable program. This process
binds the DBMS procedure calls to the DBMS run-time libraries. Additionally, the binding process typically
creates an “access plan” module that contains instructions to run the embedded code at run time.
5.
The executable is run, and the embedded SQL statement retrieves data from the database.
Note that you can embed individual SQL statements or even an entire PL/SQL block. Up to this point in the book,
you have used a DBMS-provided application (SQL*Plus) to write SQL statements and PL/SQL blocks in an interpretive
mode to address one-time or ad hoc data requests. However, it is extremely difficult and awkward to use ad hoc queries
to process transactions inside a host language. Programmers typically embed SQL statements within a host language
that it is compiled once and executed as often as needed. To embed SQL into a host language, follow this syntax:
EXEC SQL
SQL statement;
END-EXEC.
The preceding syntax will work for SELECT, INSERT, UPDATE, and DELETE statements. For example, the following
embedded SQL code will delete employee 109, George Smith, from the EMPLOYEE table:
EXEC SQL
DELETE FROM EMPLOYEE WHERE EMP_NUM = 109;
END-EXEC.
Remember, the preceding embedded SQL statement is compiled to generate an executable statement. Therefore, the
statement is fixed permanently and cannot change (unless, of course, the programmer changes it). Each time the
program runs, it deletes the same row. In short, the preceding code is good only for the first run; all subsequent runs
will likely generate an error. Clearly, this code would be more useful if you could specify a variable to indicate the
employee number to be deleted.
In embedded SQL, all host variables are preceded by a colon (“:”). The host variables may be used to send data from
the host language to the embedded SQL, or they may be used to receive the data from the embedded SQL. To use
a host variable, you must first declare it in the host language. Common practice is to use similar host variable names
as the SQL source attributes. For example, if you are using COBOL, you would define the host variables in the
Working Storage section. Then you would refer to them in the embedded SQL section by preceding them with a colon
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(“:”). For example, to delete an employee whose employee number is represented by the host variable W_EMP_NUM,
you would write the following code:
EXEC SQL
DELETE FROM EMPLOYEE WHERE EMP_NUM = :W_EMP_NUM;
END-EXEC.
At run time, the host variable value will be used to execute the embedded SQL statement. What happens if the
employee you are trying to delete doesn’t exist in the database? How do you know that the statement has been
completed without errors? As mentioned previously, the embedded SQL standard defines a SQL communication area
to hold status and error information. In COBOL, such an area is known as the SQLCA area and is defined in the Data
Division as follows:
EXEC SQL
INCLUDE SQLCA
END-EXEC.
The SQLCA area contains two variables for status and error reporting. Table 8.11 shows some of the main values
returned by the variables and their meaning.
TABLE
8.11
SQL Status and Error Reporting Variables
VARIABLE NAME
SQLCODE
VALUE
0
100
-999
SQLSTATE
00000
EXPLANATION
Old-style error reporting supported for backward compatibility only; returns
an integer value (positive or negative).
Successful completion of command.
No data; the SQL statement did not return any rows or did not select, update,
or delete any rows.
Any negative value indicates that an error occurred.
Added by SQL-92 standard to provide predefined error codes; defined as a
character string (5 characters long).
Successful completion of command.
Multiple values in the format XXYYY where:
XX-> represents the class code.
YYY-> represents the subclass code.
The following embedded SQL code illustrates the use of the SQLCODE within a COBOL program.
EXEC SQL
EXEC SQL
SELECT
EMP_LNAME, EMP_LNAME INTO :W_EMP_FNAME, :W_EMP_LNAME
WHERE EMP_NUM = :W_EMP_NUM;
END-EXEC.
IF SQLCODE = 0 THEN
PERFORM DATA_ROUTINE
ELSE
PERFORM ERROR_ROUTINE
END-IF.
In this example, the SQLCODE host variable is checked to determine whether the query completed successfully. If that
is the case, the DATA_ROUTINE is performed; otherwise, the ERROR_ROUTINE is performed.
A D V A N C E D
S Q L
Just as with PL/SQL, embedded SQL requires the use of cursors to hold data from a query that returns more than one
value. If COBOL is used, the cursor can be declared either in the Working Storage Section or in the Procedure
Division. The cursor must be declared and processed as you learned earlier in Section 8.7.3. To declare a cursor, you
use the syntax shown in the following example:
EXEC SQL
DECLARE PROD_CURSOR FOR
SELECT
P_CODE, P_DESCRIPT FROM PRODUCT
WHERE
P_QOH > (SELECT AVG(P_QOH) FROM PRODUCT);
END-EXEC.
Next, you must open the cursor to make it ready for processing:
EXEC SQL
OPEN PROD_CURSOR;
END-EXEC.
To process the data rows in the cursor, you use the FETCH command to retrieve one row of data at a time and place
the values in the host variables. The SQLCODE must be checked to ensure that the FETCH command completed
successfully. This section of code typically constitutes part of a routine in the COBOL program. Such a routine is
executed with the PERFORM command. For example:
EXEC SQL
FETCH PROD_CURSOR INTO :W_P_CODE, :W_P_DESCRIPT;
END-EXEC.
IF SQLCODE = 0 THEN
PERFORM DATA_ROUTINE
ELSE
PERFORM ERROR_ROUTINE
END-IF.
When all rows have been processed, you close the cursor as follows:
EXEC SQL
CLOSE PROD_CURSOR;
END-EXEC.
Thus far, you have seen examples of embedded SQL in which the programmer used predefined SQL statements and
parameters. Therefore, the end users of the programs are limited to the actions that were specified in the application
programs. That style of embedded SQL is known as static SQL, meaning that the SQL statements will not change
while the application is running. For example, the SQL statement might read like this:
SELECT
FROM
WHERE
P_CODE, P_DESCRIPT, P_QOH, P_PRICE
PRODUCT
P_PRICE > 100;
Note that the attributes, tables, and conditions are known in the preceding SQL statement. Unfortunately, end users
seldom work in a static environment. They are more likely to require the flexibility of defining their data access
requirements on the fly. Therefore, the end user requires that SQL be as dynamic as the data access requirements.
Dynamic SQL is a term used to describe an environment in which the SQL statement is not known in advance;
instead, the SQL statement is generated at run time. At run time in a dynamic SQL environment, a program can
generate the SQL statements that are required to respond to ad hoc queries. In such an environment, neither the
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programmer nor the end user is likely to know precisely what kind of queries are to be generated or how those queries
are to be structured. For example, a dynamic SQL equivalent of the preceding example could be:
SELECT
FROM
WHERE
:W_ATTRIBUTE_LIST
:W_TABLE
:W_CONDITION;
Note that the attribute list and the condition are not known until the end user specifies them. W_TABLE,
W_ATTRIBUTE_LIST, and W_CONDITION are text variables that contain the end-user input values used in the query
generation. Because the program uses the end-user input to build the text variables, the end user can run the same
program multiple times to generate varying outputs. For example, in one instance, the end user might want to know
what products have a price less than $100; in another case, the end user might want to know how many units of a
given product are available for sale at any given moment.
Although dynamic SQL is clearly flexible, such flexibility carries a price. Dynamic SQL tends to be much slower than
static SQL. Dynamic SQL also requires more computer resources (overhead). Finally, you are more likely to find
inconsistent levels of support and incompatibilities among DBMS vendors.
A D V A N C E D
S Q L
S u m m a r y
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◗
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◗
◗
◗
◗
◗
◗
◗
SQL provides relational set operators to combine the output of two queries to generate a new relation. The UNION
and UNION ALL set operators combine the output of two (or more) queries and produce a new relation with all
unique (UNION) or duplicate (UNION ALL) rows from both queries. The INTERSECT relational set operator selects
only the common rows. The MINUS set operator selects only the rows that are different. UNION, INTERSECT,
and MINUS require union-compatible relations.
Operations that join tables can be classified as inner joins and outer joins. An inner join is the traditional join in
which only rows that meet a given criteria are selected. An outer join returns the matching rows as well as the rows
with unmatched attribute values for one table or both tables to be joined.
A natural join returns all rows with matching values in the matching columns and eliminates duplicate columns. This
style of query is used when the tables share a common attribute with a common name. One important difference
between the syntax for a natural join and for the “old-style” join is that the natural join does not require the use
of a table qualifier for the common attributes.
Joins may use keywords such as USING and ON. If the USING clause is used, the query will return only the rows
with matching values in the column indicated in the USING clause; that column must exist in both tables. If the ON
clause is used, the query will return only the rows that meet the specified join condition.
Subqueries and correlated queries are used when it is necessary to process data based on other processed data.
That is, the query uses results that were previously unknown and that are generated by another query. Subqueries
may be used with the FROM, WHERE, IN, and HAVING clauses in a SELECT statement. A subquery may return
a single row or multiple rows.
Most subqueries are executed in a serial fashion. That is, the outer query initiates the data request, and then the inner
subquery is executed. In contrast, a correlated subquery is a subquery that is executed once for each row in the outer
query. That process is similar to the typical nested loop in a programming language. A correlated subquery is so named
because the inner query is related to the outer query—the inner query references a column of the outer subquery.
SQL functions are used to extract or transform data. The most frequently used functions are date and time
functions. The results of the function output can be used to store values in a database table, to serve as the basis
for the computation of derived variables, or to serve as a basis for data comparisons. Function formats can be
vendor-specific. Aside from time and date functions, there are numeric and string functions as well as conversion
functions that convert one data format to another.
Oracle sequences may be used to generate values to be assigned to a record. For example, a sequence may be used
to number invoices automatically. MS Access uses an AutoNumber data type to generate numeric sequences. MS
SQL Server uses the Identity column property to designate the column that will have sequential numeric values
automatically assigned to it. There can only be one Identity column per SQL Server table.
Procedural SQL (PL/SQL) can be used to create triggers, stored procedures, and PL/SQL functions. A trigger is
procedural SQL code that is automatically invoked by the DBMS upon the occurrence of a specified data
manipulation event (UPDATE, INSERT, or DELETE). Triggers are critical to proper database operation and
management. They help automate various transaction and data management processes, and they can be used to
enforce constraints that are not enforced at the DBMS design and implementation levels.
A stored procedure is a named collection of SQL statements. Just like database triggers, stored procedures are
stored in the database. One of the major advantages of stored procedures is that they can be used to encapsulate
and represent complete business transactions. Use of stored procedures substantially reduces network traffic and
increases system performance. Stored procedures help reduce code duplication by creating unique PL/SQL
modules that are called by the application programs, thereby minimizing the chance of errors and the cost of
application development and maintenance.
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When SQL statements are designed to return more than one value inside the PL/SQL code, a cursor is needed.
You can think of a cursor as a reserved area of memory in which the output of the query is stored, like an array
holding columns and rows. Cursors are held in a reserved memory area in the DBMS server, rather than in the
client computer. There are two types of cursors: implicit and explicit.
Embedded SQL refers to the use of SQL statements within an application programming language such as Visual Basic
.NET, C#, COBOL, or Java. The language in which the SQL statements are embedded is called the host language.
Embedded SQL is still the most common approach to maintaining procedural capabilities in DBMS-based applications.
K e y
T e r m s
anonymous PL/SQL block, 338
host language, 358
set-oriented, 298
batch update routine, 334
implicit cursor, 354
statement-level trigger, 343
correlated subquery, 321
inner join, 306
static SQL, 361
cross join, 306
outer join, 306
stored function, 357
cursor, 354
stored procedure, 350
dynamic SQL, 361
persistent stored
module (PSM), 337
embedded SQL, 358
procedural SQL (PL/SQL), 337
union-compatible, 298
explicit cursor, 354
row-level trigger, 343
updatable view, 335
trigger, 341
Online Content
Answers to selected Review Questions and Problems for this chapter are contained in the Premium Website for
this book.
R e v i e w
Q u e s t i o n s
1. What does it mean to say that SQL operators are set-oriented?
2. The relational set operators UNION, INTERSECT, and MINUS work properly only when the relations are
union-compatible. What does union-compatible mean, and how would you check for this condition?
3. What is the difference between UNION and UNION ALL? Write the syntax for each.
4. Suppose you have two tables: EMPLOYEE and EMPLOYEE_1. The EMPLOYEE table contains the records for
three employees: Alice Cordoza, John Cretchakov, and Anne McDonald. The EMPLOYEE_1 table contains the
records for employees John Cretchakov and Mary Chen. Given that information, list the query output for the
UNION query.
5. Given the employee information in Question 4, list the query output for the UNION ALL query.
6. Given the employee information in Question 4, list the query output for the INTERSECT query.
7. Given the employee information in Question 4, list the query output for the MINUS query of EMPLOYEE to
EMPLOYEE_1.
8. Why does the order of the operands (tables) matter in a MINUS query but not in a UNION query?
9. What is a CROSS JOIN? Give an example of its syntax.
10. What three join types are included in the OUTER JOIN classification?
11. Using tables named T1 and T2, write a query example for each of the three join types you described in Question 10.
Assume that T1 and T2 share a common column named C1.
A D V A N C E D
S Q L
12. What is a subquery, and what are its basic characteristics?
13. What are the three types of results that a subquery can return?
14. What is a correlated subquery? Give an example.
15. Explain the difference between a regular subquery and a correlated subquery.
16. What MS Access/SQL Server function should you use to calculate the number of days between your birth date
and the current date?
17. What Oracle function should you use to calculate the number of days between your birth date and the current date?
18. Suppose a PRODUCT table contains two attributes, PROD_CODE and VEND_CODE. Those two attributes have
values of ABC, 125, DEF, 124, GHI, 124, and JKL, 123, respectively. The VENDOR table contains a single
attribute, VEND_CODE, with values 123, 124, 125, and 126, respectively. (The VEND_CODE attribute in the
PRODUCT table is a foreign key to the VEND_CODE in the VENDOR table.) Given that information, what
would be the query output for:
a. A UNION query based on the two tables?
b. A UNION ALL query based on the two tables?
c. An INTERSECT query based on the two tables?
d. A MINUS query based on the two tables?
19. What string function should you use to list the first three characters of a company’s EMP_LNAME values? Give
an example using a table named EMPLOYEE. Provide examples for Oracle and SQL Server.
20. What is an Oracle sequence? Write its syntax.
21. What is a trigger, and what is its purpose? Give an example.
22. What is a stored procedure, and why is it particularly useful? Give an example.
23. What is embedded SQL, and how is it used?
24. What is dynamic SQL, and how does it differ from static SQL?
P r o b l e m s
Use the database tables in Figure P8.1 as the basis for Problems 1−18.
FIGURE
Ch08_SimpleCo database tables
P8.1
Database name: CH08_SimpleCo
Table name: CUSTOMER
Table name: CUSTOMER_2
Table name: INVOICE
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Online Content
The Ch08_SimpleCo database is located in the Premium Website for this book, as are the script files to
duplicate this data set in Oracle.
1. Create the tables. (Use the MS Access example shown in Figure P8.1 to see what table names and attributes to use.)
2. Insert the data into the tables you created in Problem 1.
3. Write the query that will generate a combined list of customers (from the tables CUSTOMER and CUSTOMER_2)
that do not include the duplicate customer records. (Note that only the customer named Juan Ortega shows up
in both customer tables.)
4. Write the query that will generate a combined list of customers to include the duplicate customer records.
5. Write the query that will show only the duplicate customer records.
6. Write the query that will generate only the records that are unique to the CUSTOMER_2 table.
7. Write the query to show the invoice number, the customer number, the customer name, the invoice date, and the
invoice amount for all customers with a customer balance of $1,000 or more.
8. Write the query that will show (for all the invoices) the invoice number, the invoice amount, the average invoice
amount, and the difference between the average invoice amount and the actual invoice amount.
9. Write the query that will write Oracle sequences to produce automatic customer number and invoice number
values. Start the customer numbers at 1000 and the invoice numbers at 5000.
10. Modify the CUSTOMER table to included two new attributes: CUST_DOB and CUST_AGE. Customer 1000
was born on March 15, 1979, and customer 1001 was born on December 22, 1988.
11. Assuming you completed Problem 10, write the query that will list the names and ages of your customers.
12. Assuming the CUSTOMER table contains a CUST_AGE attribute, write the query to update the values in that
attribute. (Hint: Use the results of the previous query.)
13. Write the query that lists the average age of your customers. (Assume that the CUSTOMER table has been
modified to include the CUST_DOB and the derived CUST_AGE attribute.)
14. Write the trigger to update the CUST_BALANCE in the CUSTOMER table when a new invoice record is entered.
(Assume that the sale is a credit sale.) Test the trigger, using the following new INVOICE record:
8005, 1001, ‘27-APR-10’, 225.40
Name the trigger trg_updatecustbalance.
15. Write a procedure to add a new customer to the CUSTOMER table. Use the following values in the new record:
1002, ‘Rauthor’, ‘Peter’, 0.00
Name the procedure prc_cust_add. Run a query to see if the record has been added.
16. Write a procedure to add a new invoice record to the INVOICE table. Use the following values in the new record:
8006, 1000, ‘30-APR-10’, 301.72
Name the procedure prc_invoice_add. Run a query to see if the record has been added.
17. Write a trigger to update the customer balance when an invoice is deleted. Name the trigger trg_
updatecustbalance2.
18. Write a procedure to delete an invoice, giving the invoice number as a parameter. Name the procedure
prc_inv_delete. Test the procedure by deleting invoices 8005 and 8006.
Use the Ch08_SaleCo2 database to work Problems 19−22, shown in Figure P8.19.
A D V A N C E D
FIGURE
S Q L
Ch08_SaleCo2 database tables
P8.19
Database name: CH08_SaleCo2
Table name: CUSTOMER
Table name: INVOICE
Table name: LINE
Table name: PRODUCT
Table name: VENDOR
Online Content
The Ch08_SaleCo2 database used in Problems 19−22 is located in the Premium Website for this book, as are
the script files to duplicate this data set in Oracle.
19. Create a trigger named trg_line_total to write the LINE_TOTAL value in the LINE table every time you add a
new LINE row. (The LINE_TOTAL value is the product of the LINE_UNITS and the LINE_PRICE values.)
20. Create a trigger named trg_line_prod that will automatically update the quantity on hand for each product sold
after a new LINE row is added.
21. Create a stored procedure named prc_inv_amounts to update the INV_SUBTOTAL, INV_TAX, and
INV_TOTAL. The procedure takes the invoice number as a parameter. The INV_SUBTOTAL is the sum of the
LINE_TOTAL amounts for the invoice, the INV_TAX is the product of the INV_SUBTOTAL and the tax rate
(8%), and the INV_TOTAL is the sum of the INV_SUBTOTAL and the INV_TAX.
22. Create a procedure named prc_cus_balance_update that will take the invoice number as a parameter and
update the customer balance. (Hint: You can use the DECLARE section to define a TOTINV numeric variable
that holds the computed invoice total.)
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Use the Ch08_AviaCo database to work Problems 23−34, shown in Figure P8.23.
FIGURE
Ch08_AviaCo database tables
P8.23
Table name: CHARTER
Database name: CH08_AviaCo
Table name: EARNEDRATING
Table name: CREW
Table name: CUSTOMER
Table name: RATING
Table name: EMPLOYEE
Table name: MODEL
Table name: AIRCRAFT
Table name: PILOT
Online Content
The Ch08_AviaCo database used for Problems 23−34 is located in the Premium Website for this book, as are
the script files to duplicate this data set in Oracle.
23. Modify the MODEL table to add the attribute and insert the values shown in the following table.
ATTRIBUTE NAME
ATTRIBUTE DESCRIPTION
ATTRIBUTE TYPE
ATTRIBUTE VALUES
MOD_WAIT_CHG
Waiting charge per hour for each
model
Numeric
$100 for C-90A
$50 for PA23-250
$75 for PA31-350
24. Write the queries to update the MOD_WAIT_CHG attribute values based on Problem 23.
A D V A N C E D
S Q L
25. Modify the CHARTER table to add the attributes shown in the following table.
ATTRIBUTE NAME
ATTRIBUTE DESCRIPTION
CHAR_WAIT_CHG
CHAR_FLT_CHG_HR
Waiting charge for each model (copied from the MODEL table)
Flight charge per mile for each model (copied from the MODEL table
using the MOD_CHG_MILE attribute)
Flight charge (calculated by CHAR_HOURS_FLOWN x CHAR_FLT_
CHG_HR)
CHAR_FLT_CHG x tax rate (8%)
CHAR_FLT_CHG + CHAR_TAX_CHG
Amount paid by customer
Balance remaining after payment
CHAR_FLT_CHG
CHAR_TAX_CHG
CHAR_TOT_CHG
CHAR_PYMT
CHAR_BALANCE
ATTRIBUTE
TYPE
Numeric
Numeric
Numeric
Numeric
Numeric
Numeric
Numeric
26. Write the sequence of commands required to update the CHAR_WAIT_CHG attribute values in the CHARTER
table. (Hint: Use either an updatable view or a stored procedure.)
27. Write the sequence of commands required to update the CHAR_FLT_CHG_HR attribute values in the
CHARTER table. (Hint: Use either an updatable view or a stored procedure.)
28. Write the command required to update the CHAR_FLT_CHG attribute values in the CHARTER table.
29. Write the command required to update the CHAR_TAX_CHG attribute values in the CHARTER table.
30. Write the command required to update the CHAR_TOT_CHG attribute values in the CHARTER table.
31. Modify the PILOT table to add the attribute shown in the following table.
ATTRIBUTE NAME
ATTRIBUTE DESCRIPTION
PIL_PIC_HRS
Pilot in command (PIC) hours; updated by adding the CHARTER table’s
CHAR_HOURS_FLOWN to the PIL_PIC_HRS when the CREW table shows the
CREW_JOB to be pilot
ATTRIBUTE
TYPE
Numeric
32. Create a trigger named trg_char_hours that will automatically update the AIRCRAFT table when a new
CHARTER row is added. Use the CHARTER table’s CHAR_HOURS_FLOWN to update the AIRCRAFT table’s
AC_TTAF, AC_TTEL, and AC_TTER values.
33. Create a trigger named trg_pic_hours that will automatically update the PILOT table when a new CREW row
is added and the CREW table uses a ‘pilot’ CREW_JOB entry. Use the CHARTER table’s CHAR_HOURS_
FLOWN to update the PILOT table’s PIL_PIC_HRS only when the CREW table uses a ‘pilot’ CREW_JOB entry.
34. Create a trigger named trg_cust_balance that will automatically update the CUSTOMER table’s CUST_
BALANCE when a new CHARTER row is added. Use the CHARTER table’s CHAR_TOT_CHG as the update
source. (Assume that all charter charges are charged to the customer balance.)
C a s e s
The following problems expand on the TinyVideo case from Chapter 7. To complete the following problems, it is
necessary to have first completed the table creation and data entry requirements specified in Problems 44 and 45 in
Chapter 7.
35. Alter the DETAILRENTAL table to include a derived attribute named DETAIL_DAYSLATE to store integers up
to 3 digits. The attribute should accept null values.
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36. Alter the VIDEO table to include an attribute named VID_STATUS to store character data up to 4 characters
long. The attribute should not accept null values. The attribute should have a constraint to enforce the domain
(“IN”, “OUT”, and “LOST”) and have a default value of “IN”.
37. Update the VID_STATUS attribute of the VIDEO table using a subquery to set the VID_STATUS to “OUT” for
all videos that have a null value in the DETAIL_RETURNDATE attribute of the DETAILRENTAL table.
38. Alter the PRICE table to include an attribute named PRICE_RENTDAYS to store integers up to 2 digits. The
attribute should not accept null values, and should have a default value of 3.
39. Update the PRICE table to place the values shown in the following table in the PRICE_RENTDAYS attribute.
PRICE_CODE
1
2
3
4
PRICE_RENTDAYS
5
3
5
7
40. Create a trigger named trg_late_return that will write the correct value to DETAIL_DAYSLATE in the
DETAILRENTAL table whenever a video is returned. The trigger should execute as a BEFORE trigger when the
DETAIL_RETURNDATE or DETAIL_DUEDATE attributes are updated. The trigger should satisfy the following
conditions.
a. If the return date is null, then the days late should also be null.
b. If the return date is not null, then the days late should determine if the video is returned late.
c. If the return date is noon of the day after the due date or earlier, then the video is not considered late, and
the days late should have a value of zero (0).
d. If the return date is past noon of the day after the due date, then the video is considered late so the number
of days late must be calculated and stored.
41. Create a trigger named trg_mem_balance that will maintain the correct value in the membership balance in the
MEMBERSHIP table when videos are returned late. The trigger should execute as an AFTER trigger when
the due date or return date attributes are updated in the DETAILRENTAL table. The trigger should satisfy the
following conditions.
a. Calculate the value of the late fee prior to the update that triggered this execution of the trigger. The value
of the late fee is the days late times the daily late fee. If the previous value of the late fee was null, then treat
it as zero (0).
b. Calculate the value of the late fee after the update that triggered this execution of the trigger. If the value of
the late fee is now null, then treat it as zero (0).
c. Subtract the prior value of the late fee from the current value of the late fee to determine the change in late
fee for this video rental.
d. If the amount calculated in part c is not zero (0), then update the membership balance by the amount
calculated for the membership associated the rental that this detail is a part of.
42. Create a sequence named rent_num_seq to start with 1100, increment by 1, and do not cache any values.
43. Create a stored procedure named prc_new_rental to insert new rows in the RENTAL table. The procedure should
satisfy the following conditions.
a. The membership number will be provided as a parameter.
b. Use a Count() function to verify that the membership number exists in the MEMBERSHIP table. If it does not
exist, then a message should be displayed stating that the membership does not exist and no data should be
written to the database.
A D V A N C E D
S Q L
c. If the membership does exist, then retrieve the membership balance and display a message stating the
balance amount as the previous balance. (For example, if the membership has a balance of $5.00, then
display “Previous balance: $5.00”.)
d. Insert a new row in the rental table using the sequence created in #42 above to generate the value for
RENT_NUM, the current system date for the value for RENT_DATE, and the membership number provided
as the value for MEM_NUM.
44. Create a stored procedure named prc_new_detail to insert new rows in the DETAILRENTAL table. The
procedure should satisfy the following requirements.
a. The video number will be provided as a parameter.
b. Verify that the video number exists in the VIDEO table. If it does not exist, then display a message that the
video does not exist, and do not write any data to the database.
c. If the video number does exist, then verify that the VID_STATUS for that video is “IN”. If the status is not
“IN”, then display a message that the return of the video must be entered before it can be rented again, and
do not write any data to the database.
d. If the status is “IN”, then retrieve the values of PRICE_RENTFEE, PRICE_DAILYLATEFEE, and PRICE_
RENTDAYS associated with the video from the PRICE table.
e. Calculate the due date for the video rental by adding the number of days found in PRICE_RENTDAYS above
to 11:59:59PM (hours:minutes:seconds) on the current system date.
f.
Insert a new row in the DETAILRENTAL table using the previous value returned by rent_num_seq as the
RENT_NUM, the video number provided in the parameter as the VID_NUM, the PRICE_RENTFEE as the
value for DETAIL_FEE, the due date calculated above for the DETAIL_DUEDATE, PRICE_DAILYLATEFEE
as the value for DETAIL_DAILYLATEFEE, and null for the DETAIL_RETURNDATE.
45. Create a stored procedure named prc_return_video to enter data about the return of videos that have been
rented. The procedure should satisfy the following requirements.
a. The video number will be provided as a parameter.
b. Verify the video number exists in the VIDEO table. If it does not exist, display a message that the video
number provided was not found and do not write any data to the database.
c. If the video number does exist, then use a Count() function to ensure that the video has only one record in
DETAILRENTAL for which it does not have a return date. If more than one row in DETAILRENTAL indicates
that the video is rented but not returned, display an error message that the video has multiple outstanding
rentals and do not write any data to the database.
d. If the video does not have any outstanding rentals, then update the video status for the video in the VIDEO
table to “IN”, and display a message that the video had no outstanding rentals but it is now available for rental.
If the video has only one outstanding rental, then update the return date to the current system date, and
update the video status for that video in the VIDEO table to “IN”. Then display a message stating that the
video was successfully returned.
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9
Database Design
In this chapter, you will learn:
쐍 That successful database design must reflect the information system of which the database is
a part
쐍 That successful information systems are developed within a framework known as the
Systems Development Life Cycle (SDLC)
쐍 That within the information system, the most successful databases are subject to frequent
evaluation and revision within a framework known as the Database Life Cycle (DBLC)
쐍 How to conduct evaluation and revision within the SDLC and DBLC frameworks
쐍 About database design strategies: top-down vs. bottom-up design and centralized vs.
decentralized design
Databases are a part of a larger picture called an information system. Database designs that
fail to recognize that the database is part of this larger whole are not likely to be successful.
That is, database designers must recognize that the database is a critical means to an end
rather than an end in itself. (Managers want the database to serve their management needs,
but too many databases seem to require that managers alter their routines to fit the
database requirements.)
Information systems don’t just happen; they are the product of a carefully staged
development process. Systems analysis is used to determine the need for an information
system and to establish its limits. Within systems analysis, the actual information system is
created through a process known as systems development.
The creation and evolution of information systems follows an iterative pattern called the
Systems Development Life Cycle (SDLC), a continuous process of creation, maintenance,
enhancement, and replacement of the information system. A similar cycle applies to
databases.The database is created, maintained, and enhanced.When even enhancement can
no longer stretch the database’s usefulness and the database can no longer perform its
functions adequately, it might have to be replaced. The Database Life Cycle (DBLC) is
carefully traced in this chapter and is shown in the context of the larger Systems
Development Life Cycle.
At the end of the chapter, you will be introduced to some classical approaches to database
design: top-down vs. bottom-up and centralized vs. decentralized.
P
review
D A T A B A S E
D E S I G N
9.1 THE INFORMATION SYSTEM
Basically, a database is a carefully designed and constructed repository of facts. The database is a part of a larger whole
known as an information system, which provides for data collection, storage, and retrieval. The information system
also facilitates the transformation of data into information, and it allows for the management of both data and
information. Thus, a complete information system is composed of people, hardware, software, the database(s),
application programs, and procedures. Systems analysis is the process that establishes the need for and the extent
of an information system. The process of creating an information system is known as systems development.
One key characteristic of current information systems is the strategic value of information in the age of global business.
Therefore, information systems should always be aligned with the strategic business goals; the view of isolated and
independent information systems is no longer valid. Current information systems should always be integrated with the
company’s enterprise-wide information systems architecture.
Note
This chapter does not mean to cover all aspects of systems analysis and development—these are usually covered
in a separate course or book. However, this chapter should help you develop a better understanding of the
issues associated with database design, implementation, and management that are affected by the information
system in which the database is a critical component.
Within the framework of systems development, applications transform data into the information that forms the basis
for decision making. Applications usually produce formal reports, tabulations, and graphic displays designed to
produce insight into the information. Figure 9.1 illustrates that every application is composed of two parts: the data
and the code (program instructions) by which the data are transformed into information. The data and the code work
together to represent real-world business functions and activities. At any given moment, physically stored data
represent a snapshot of the business. But the picture is not complete without an understanding of the business activities
that are represented by the code.
FIGURE
Generating information for decision making
9.1
Information
Data
Application
code
90
80
70
60
50
40
30
20
10
0
Decisions
East
West
North
South
1st Qtr
2nd Qtr 3rd Qtr 4th Qtr
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9
The performance of an information system depends on three factors:
쐌
Database design and implementation.
쐌
Application design and implementation.
쐌
Administrative procedures.
This book emphasizes the database design and implementation segment of the triad—arguably the most important of
the three. However, failure to address the other two segments will likely yield a poorly functioning information system.
Creating a sound information system is hard work: systems analysis and development require much planning to ensure
that all of the activities will interface with each other, that they will complement each other, and that they will be
completed on time.
In a broad sense, the term database development describes the process of database design and implementation.
The primary objective in database design is to create complete, normalized, nonredundant (to the extent possible), and
fully integrated conceptual, logical, and physical database models. The implementation phase includes creating the
database storage structure, loading data into the database, and providing for data management.
To make the procedures discussed in this chapter broadly applicable, the chapter focuses on the elements that are
common to all information systems. Most of the processes and procedures described in this chapter do not depend on
the size, type, or complexity of the database being implemented. However, the procedures that would be used to design
a small database, such as one for a neighborhood shoe store, do not precisely scale up to the procedures that would
be needed to design a database for a large corporation or even a segment of such a corporation. To use an analogy,
building a small house requires a blueprint, just as building the Golden Gate Bridge does, but the bridge requires more
complex and farther-ranging planning, analysis, and design than the house.
The next sections will trace the overall Systems Development Life Cycle and the related Database Life Cycle. Once you
are familiar with those processes and procedures, you will learn about various approaches to database design, such as
top-down vs. bottom-up and centralized vs. decentralized design.12
Note
The Systems Development Life Cycle (SDLC) is a general framework through which you can track and
understand the activities required to develop and maintain information systems. Within that framework, there
are several ways to complete various tasks specified in the SDLC. For example, this book focuses on ER
modeling and on relational database design and implementation issues, and that focus is maintained in this
chapter. However, there are alternative methodologies, such as:
• Unified Modeling Language (UML) provides object-oriented tools to support the tasks associated with the
development of information systems. UML is covered in Appendix H, Unified Modeling Language
(UML), in the Premium Website for this book.
• Rapid Application Development (RAD)1 is an iterative software development methodology that uses
prototypes, CASE tools, and flexible management to develop application systems. RAD started as an
alternative to traditional structured development, which suffered from long deliverable times and unfulfilled requirements.
• Agile Software Development2 is a framework for developing software applications that divides the work to
be done in smaller subprojects to obtain valuable deliverables in shorter times and with better cohesion.
This method emphasizes close communication among all users and continuous evaluation with the
purpose of increasing customer satisfaction.
Although the development methodologies may change, the basic framework within which those methodologies are used does not change.
1See Rapid Application Development, James Martin, Prentice-Hall, Macmillan College Division, 1991.
2Further information about Agile Software Development can be found online at www.agilealliance.org.
D A T A B A S E
D E S I G N
9.2 THE SYSTEMS DEVELOPMENT LIFE CYCLE (SDLC)
The Systems Development Life Cycle (SDLC) traces the history (life cycle) of an information system. Perhaps
more important to the system designer, the SDLC provides the big picture within which the database design and
application development can be mapped out and evaluated.
As illustrated in Figure 9.2, the traditional SDLC is divided into five phases: planning, analysis, detailed systems design,
implementation, and maintenance. The SDLC is an iterative rather than a sequential process. For example, the details
of the feasibility study might help refine the initial assessment, and the details discovered during the user requirements
portion of the SDLC might help refine the feasibility study.
FIGURE
The Systems Development Life Cycle (SDLC)
9.2
Phase
Action(s)
Section
Planning
Initial assessment
Feasibility study
9.2.1
User requirements
Existing system evaluation
Logical system design
9.2.2
Analysis
Detailed system specification
9.2.3
Coding, testing, and debugging
9.2.4
Detailed
systems design
Implementation
Maintenance
Installation, fine-tuning
Evaluation
Maintenance
Enhancement
9.2.5
Because the Database Life Cycle (DBLC) fits into and resembles the Systems Development Life Cycle (SDLC), a brief
description of the SDLC is in order.
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9.2.1 Planning
The SDLC planning phase yields a general overview of the company and its objectives. An initial assessment of the
information flow-and-extent requirements must be made during this discovery portion of the SDLC. Such an
assessment should answer some important questions:
쐌
Should the existing system be continued? If the information generator does its job well, there is no point in
modifying or replacing it. To quote an old saying, “If it ain’t broke, don’t fix it.”
쐌
Should the existing system be modified? If the initial assessment indicates deficiencies in the extent and flow
of the information, minor (or even major) modifications might be in order. When considering modifications, the
participants in the initial assessment must keep in mind the distinction between wants and needs.
쐌
Should the existing system be replaced? The initial assessment might indicate that the current system’s flaws
are beyond fixing. Given the effort required to create a new system, a careful distinction between wants and
needs is perhaps even more important in this case than it is when modifying the system.
Participants in the SDLC’s initial assessment must begin to study and evaluate alternative solutions. If it is decided that
a new system is necessary, the next question is whether it is feasible. The feasibility study must address the following:
쐌
The technical aspects
specific, but they must
computer, mainframe,
systems, database type
쐌
The system cost. The admittedly mundane question, “Can we afford it?” is crucial. (And the answer to that
question might force a careful review of the initial assessment.) It bears repeating that a million-dollar solution
to a thousand-dollar problem is not defensible.
쐌
The operational cost. Does the company possess the human, technical, and financial resources to keep the
system operational? Do we include the cost of the management and end-user support needed to put in place
the operational procedures to ensure the success of this system?
of hardware and software requirements. The decisions might not (yet) be vendoraddress the nature of the hardware requirements (desktop computer, multiprocessor
or supercomputer) and the software requirements (single- or multiuser operating
and software, programming languages to be used by the applications, and so on).
9.2.2 Analysis
Problems defined during the planning phase are examined in greater detail during the analysis phase. A macroanalysis
must be made of both individual needs and organizational needs, addressing questions such as:
쐌
What are the requirements of the current system’s end users?
쐌
Do those requirements fit into the overall information requirements?
The analysis phase of the SDLC is, in effect, a thorough audit of user requirements.
The existing hardware and software systems are also studied during the analysis phase. The result of analysis should
be a better understanding of the system’s functional areas, actual and potential problems, and opportunities.
End users and the system designer(s) must work together to identify processes and to uncover potential problem areas.
Such cooperation is vital to defining the appropriate performance objectives by which the new system can be judged.
Along with a study of user requirements and the existing systems, the analysis phase also includes the creation of a
logical systems design. The logical design must specify the appropriate conceptual data model, inputs, processes, and
expected output requirements.
When creating a logical design, the designer might use tools such as data flow diagrams (DFDs), hierarchical input
process output (HIPO) diagrams, and entity relationship (ER) diagrams. The database design’s data-modeling activities
take place at this point to discover and describe all entities and their attributes and the relationships among the entities
within the database.
D A T A B A S E
D E S I G N
Defining the logical system also yields functional descriptions of the system’s components (modules) for each process
within the database environment. All data transformations (processes) are described and documented, using such
systems analysis tools as DFDs. The conceptual data model is validated against those processes.
9.2.3 Detailed Systems Design
In the detailed systems design phase, the designer completes the design of the system’s processes. The design includes
all the necessary technical specifications for the screens, menus, reports, and other devices that might be used to help
make the system a more efficient information generator. The steps are laid out for conversion from the old to the new
system. Training principles and methodologies are also planned and must be submitted for management’s approval.
Note
Because attention has been focused on the details of the systems design process, the book has not until this
point explicitly recognized the fact that management approval is needed at all stages of the process. Such
approval is needed because a GO decision requires funding. There are many GO/NO GO decision points along
the way to a completed systems design!
9.2.4 Implementation
During the implementation phase, the hardware, DBMS software, and application programs are installed, and the
database design is implemented. During the initial stages of the implementation phase, the system enters into a cycle
of coding, testing, and debugging until it is ready to be delivered. The actual database is created, and the system is
customized by the creation of tables and views, user authorizations, and so on.
The database contents might be loaded interactively or in batch mode, using a variety of methods and devices:
쐌
Customized user programs.
쐌
Database interface programs.
쐌
Conversion programs that import the data from a different file structure, using batch programs, a database
utility, or both.
The system is subjected to exhaustive testing until it is ready for use. Traditionally, the implementation and testing of
a new system took 50 to 60 percent of the total development time. However, the advent of sophisticated application
generators and debugging tools has substantially decreased coding and testing time. After testing is concluded, the final
documentation is reviewed and printed and end users are trained. The system is in full operation at the end of this
phase but will be continuously evaluated and fine-tuned.
9.2.5 Maintenance
Almost as soon as the system is operational, end users begin to request changes in it. Those changes generate system
maintenance activities, which can be grouped into three types:
쐌
Corrective maintenance in response to systems errors.
쐌
Adaptive maintenance due to changes in the business environment.
쐌
Perfective maintenance to enhance the system.
Because every request for structural change requires retracing the SDLC steps, the system is, in a sense, always at
some stage of the SDLC.
Each system has a predetermined operational life span. The actual operational life span of a system depends on its
perceived utility. There are several reasons for reducing the operational life of certain systems. Rapid technological
change is one reason, especially for systems based on processing speed and expandability. Another common reason
is the cost of maintaining a system.
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If the system’s maintenance cost is high, its value becomes suspect. Computer-aided systems engineering (CASE)
tools, such as System Architect or Visio Professional, helps make it possible to produce better systems within a
reasonable amount of time and at a reasonable cost. In addition, CASE-produced applications are more structured,
better documented, and especially standardized, which tends to prolong the operational life of systems by making
them easier and cheaper to update and maintain.
9.3 THE DATABASE LIFE CYCLE (DBLC)
Within the larger information system, the database, too, is subject to a life cycle. The Database Life Cycle (DBLC)
contains six phases, as shown in Figure 9.3: database initial study, database design, implementation and loading,
testing and evaluation, operation, and maintenance and evolution.
FIGURE
The Database Life Cycle (DBLC)
9.3
Phase
Action(s)
Section
Analyze the company situation
Define problems and constraints
Define objectives
Define scope and boundaries
9.3.1
Database initial
study
9.3.2
Database design
Create the conceptual design
DBMS software selection
Create the logical design
Create the physical design
Implementation
and loading
Install the DBMS
Create the database(s)
Load or convert the data
9.3.3
Testing and
evaluation
Test the database
9.3.4
Fine-tune the database
Evaluate the database and its application programs
Operation
Produce the required information flow
9.3.5
Introduce changes
Make enhancements
9.3.6
Maintenance and
evolution
9.3.1 The Database Initial Study
If a designer has been called in, chances are the current system has failed to perform functions deemed vital by the
company. (You don’t call the plumber unless the pipes leak.) So, in addition to examining the current system’s
operation within the company, the designer must determine how and why the current system fails. That means
D A T A B A S E
D E S I G N
spending a lot of time talking with (but mostly listening to) end users. Although database design is a technical business,
it is also people-oriented. Database designers must be excellent communicators, and they must have finely tuned
interpersonal skills.
Depending on the complexity and scope of the database environment, the database designer might be a lone operator
or part of a systems development team composed of a project leader, one or more senior systems analysts, and one
or more junior systems analysts. The word designer is used generically here to cover a wide range of design team
compositions.
The overall purpose of the database initial study is to:
쐌
Analyze the company situation.
쐌
Define problems and constraints.
쐌
Define objectives.
쐌
Define scope and boundaries.
Figure 9-4 depicts the interactive and iterative processes required to complete the first phase of the DBLC successfully.
As you examine Figure 9.4, note that the database initial study phase leads to the development of the database system
objectives. Using Figure 9.4 as a discussion template, let’s examine each of its components in greater detail.
FIGURE
A summary of activities in the database initial study
9.4
Analysis of the
company situation
Company objectives
Company operations
Company structure
Definition of
problems and constraints
Database system
specifications
Objectives
Scope
Boundaries
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Analyze the Company Situation
The company situation describes the general conditions in which a company operates, its organizational structure,
and its mission. To analyze the company situation, the database designer must discover what the company’s
operational components are, how they function, and how they interact.
These issues must be resolved:
쐌
What is the organization’s general operating environment, and what is its mission within that
environment? The design must satisfy the operational demands created by the organization’s mission. For
example, a mail-order business is likely to have operational requirements involving its database that are quite
different from those of a manufacturing business.
쐌
What is the organization’s structure? Knowing who controls what and who reports to whom is quite useful
when you are trying to define required information flows, specific report and query formats, and so on.
Define Problems and Constraints
The designer has both formal and informal sources of information. If the company has existed for any length of time,
it already has some kind of system in place (either manual or computer-based). How does the existing system function?
What input does the system require? What documents does the system generate? By whom and how is the system
output used? Studying the paper trail can be very informative. In addition to the official version of the system’s
operation, there is also the more informal, real version; the designer must be shrewd enough to see how these differ.
The process of defining problems might initially appear to be unstructured. Company end users are often unable to
describe precisely the larger scope of company operations or to identify the real problems encountered during
company operations. Often the managerial view of a company’s operation and its problems is different from that of
the end users, who perform the actual routine work.
During the initial problem definition process, the designer is likely to collect very broad problem descriptions. For
example, note these concerns expressed by the president of a fast-growing transnational manufacturing company:
Although the rapid growth is gratifying, members of the management team are concerned that such growth is
beginning to undermine the ability to maintain a high customer service standard and, perhaps worse, to diminish
manufacturing standards control.
The problem definition process quickly leads to a host of general problem descriptions. For example, the marketing
manager comments:
I’m working with an antiquated filing system. We manufacture more than 1,700 specialty machine parts. When
a regular customer calls in, we can’t get a very quick inventory scan. If a new customer calls in, we can’t do a
current parts search by using a simple description, so we often do a machine setup for a part that we have in
inventory. That’s wasteful. And of course, some new customers get irritated when we can’t give a quick response.
The production manager comments:
At best, it takes hours to generate the reports I need for scheduling purposes. I don’t have hours for quick
turnarounds. It’s difficult to manage what I don’t have information about.
I don’t get quick product request routing. Take machine setup. Right now I’ve got operators either waiting for
the right stock or getting it themselves when a new part is scheduled for production. I can’t afford to have an
operator doing chores that a much lower-paid worker ought to be doing. There’s just too much waiting around
with the current scheduling. I’m losing too much time, and my schedules back up. Our overtime bill is ridiculous.
D A T A B A S E
D E S I G N
I sometimes produce parts that are already in inventory because we don’t seem to be able to match what we’ve
got in inventory with what we have scheduled. Shipping yells at me because I can’t turn out the parts, and often
they’ve got them in inventory one bay down. That’s costing us big bucks sometimes.
New reports can take days or even weeks to get to my office. And I need a ton of reports to schedule personnel,
downtime, training, etc. I can’t get new reports that I need NOW. What I need is the ability to get quick updates
on percent defectives, percent rework, the effectiveness of training, you name it. I need such reports by shift, by
date, by any characteristic I can think of to help me manage scheduling, training, you name it.
A machine operator comments:
It takes a long time to set my stuff up. If I get my schedule banged up because John doesn’t get the paperwork
on time, I wind up looking for setup specs, startup material, bin assignments, and other stuff. Sometimes I spend
two or three hours just setting up. Now you know why I can’t meet schedules. I try to be productive, but I’m
spending too much time getting ready to do my job.
After the initial declarations, the database designer must continue to probe carefully in order to generate additional
information that will help define the problems within the larger framework of company operations. How does the
problem of the marketing manager’s customer fit within the broader set of marketing department activities? How does
the solution to the customer’s problem help meet the objectives of the marketing department and the rest of the
company? How do the marketing department’s activities relate to those of the other departments? That last question
is especially important. Note that there are common threads in the problems described by the marketing and
production department managers. If the inventory query process can be improved, both departments are likely to find
simple solutions to at least some of the problems.
Finding precise answers is important, especially concerning the operational relationships among business units. If a
proposed system will solve the marketing department’s problems but exacerbate those of the production department,
not much progress will have been made. Using an analogy, suppose that your home water bill is too high. You have
determined the problem: the faucets leak. The solution? You step outside and cut off the water supply to the house.
Is that an adequate solution? Or would the replacement of faucet washers do a better job of solving the problem? You
might find the leaky faucet scenario simplistic, yet almost any experienced database designer can find similar instances
of so-called database problem solving (admittedly more complicated and less obvious).
Even the most complete and accurate problem definition does not always lead to the perfect solution. The real world
usually intrudes to limit the design of even the most elegant database by imposing constraints. Such constraints include
time, budget, personnel, and more. If you must have a solution within a month and within a $12,000 budget, a solution
that takes two years to develop at a cost of $100,000 is not a solution. The designer must learn to distinguish
between what’s perfect and what’s possible.
Define Objectives
A proposed database system must be designed to help solve at least the major problems identified during the problem
discovery process. As the list of problems unfolds, several common sources are likely to be discovered. In the previous
example, both the marketing manager and the production manager seem to be plagued by inventory inefficiencies. If
the designer can create a database that sets the stage for more efficient parts management, both departments gain.
The initial objective, therefore, might be to create an efficient inventory query and management system.
Note
When trying to develop solutions, the database designer must look for the source of the problems. There are
many cases of database systems that failed to satisfy the end users because they were designed to treat the
symptoms of the problems rather than their source.
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Note that the initial study phase also yields proposed problem solutions. The designer’s job is to make sure that the
database system objectives, as seen by the designer, correspond to those envisioned by the end user(s). In any case,
the database designer must begin to address the following questions:
쐌
What is the proposed system’s initial objective?
쐌
Will the system interface with other existing or future systems in the company?
쐌
Will the system share the data with other systems or users?
Define Scope and Boundaries
The designer must recognize the existence of two sets of limits: scope and boundaries. The system’s scope defines
the extent of the design according to operational requirements. Will the database design encompass the entire
organization, one or more departments within the organization, or one or more functions of a single department? The
designer must know the “size of the ballpark.” Knowing the scope helps in defining the required data structures, the
type and number of entities, the physical size of the database, and so on.
The proposed system is also subject to limits known as boundaries, which are external to the system. Has any
designer ever been told, “We have all the time in the world” or “Use an unlimited budget and use as many people as
needed to make the design come together”? Boundaries are also imposed by existing hardware and software. Ideally,
the designer can choose the hardware and software that will best accomplish the system goals. In fact, software
selection is an important aspect of the Systems Development Life Cycle. Unfortunately, in the real world, a system
must often be designed around existing hardware. Thus, the scope and boundaries become the factors that force the
design into a specific mold, and the designer’s job is to design the best system possible within those constraints. (Note
that problem definitions and the objectives must sometimes be reshaped to meet the system scope and boundaries.)
9.3.2 Database Design
The second phase focuses on the design of the database model that will support company operations and objectives.
This is arguably the most critical DBLC phase: making sure that the final product meets user and system requirements.
In the process of database design, you must concentrate on the data characteristics required to build the database
model. At this point, there are two views of the data within the system: the business view of data as a source of
information and the designer’s view of the data structure, its access, and the activities required to transform the data
into information. Figure 9.5 contrasts those views. Note that you can summarize the different views by looking at the
terms what and how. Defining data is an integral part of the DBLC’s second phase.
As you examine the procedures required to complete the design phase in the DBLC, remember these points:
쐌
The process of database design is loosely related to the analysis and design of a larger system. The data
component is only one element of a larger information system.
쐌
The systems analysts or systems programmers are in charge of designing the other system components. Their
activities create the procedures that will help transform the data within the database into useful information.
쐌
The database design does not constitute a sequential process. Rather, it is an iterative process that provides
continuous feedback designed to trace previous steps.
D A T A B A S E
FIGURE
D E S I G N
Two views of data: business manager and database designer
9.5
Company
Engineering
Purchasing
Manufacturing
Manager’s view
Shared information
What are the problems?
What are the solutions?
What information is needed to
implement the solutions?
What data are required to
generate the desired information?
Designer’s view
C o m pa
n y D a t ab a s e
How must the data be structured?
How will the data be accessed?
How are the data transformed
into information?
The database design process is depicted in Figure 9.6.
Looking at Figure 9.6, you can see that there are four essential stages: conceptual, logical, and physical design, plus
the DBMS selection stage, which is critical to determine the type of logical and physical design to be performed. The
design process starts with conceptual design and moves to the logical and physical design stages. At each stage, more
details about the data model design are determined and documented. You could think of the conceptual design as the
overall data as seen by the end user, the logical design as the data as seen by the DBMS, and the physical design as
the data as seen by the operating system’s storage management devices.
It is important to note that the overwhelming majority of database designs and implementations are based on the
relational model and, therefore, use the relational model constructs and techniques. At the completion of the database
design activities, you will have a complete database design ready to be implemented.
Database design activities will be covered in detail in Sections 9.4 (Conceptual Design), 9.5 (DBMS Software
Selection), 9.6 (Logical Design), and 9.7 (Physical Design).
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FIGURE
Database design process
9.6
Section
9.4
Stage
Conceptual
Design
Steps
Activities
• Data analysis and requirements
• Determine end-user views, outputs and
transaction requirements
• Entity Relationship modeling and normalization
• Define entities, attributes, domains and relationships
• Draw ER diagrams. Normalize entity attributes
• Data model verification
• Identify ER modules and validate insert, update and delete rules
• Validate reports, queries,views integrity, access and security
• Distributed database design*
• Define DBMS and data model to use
DBMS and Hardware Independent
9.5
9.6
9.7
DBMS
Selection
• Select the DBMS data model
• Determine DBMS and data model to use
• Map conceptual model to logical model components
• Define tables, columns, relationships, and constraints
• Validate logical model using normalization
• Normalized set of tables
• Validate logical modeling integrity constraints
• Ensure entity and referential integrity, Define column constraints
• Validate logical model against user requirements
• Ensure the model supports user requirements
• Define data storage organization
• Define tables, indexes, views physical organizations
• Define integrity and security measures
• Define users, security groups, roles, and access controls
• Determine performance measures+
• Define database and query execution parameters
DBMS Dependent
Logical
Design
Hardware Dependent
Physical
Design
* See Chapter 12, Distributed Database Management Systems
+ See Chapter 11, Database Performance Tuning and Query Optimization
Online Content
Appendixes B and C in the Premium Website, The University Lab: Conceptual Design and The University
Lab: Conceptual Design Verification, Logical Design, and Implementation, respectively, provide a concise
example of a simple real-world database development process.
9.3.3 Implementation and Loading
The output of the database design phase is a series of instructions detailing the creation of tables, attributes, domains,
views, indexes, security constraints, and storage and performance guidelines. In this phase, you actually implement all
these design specifications.
Install the DBMS
This step is required only when a new dedicated instance of the DBMS is necessary for the system. In many cases, the
organization will have standardized on a particular DBMS in order to leverage investments in the technology and the
skills that employees have already developed. The DBMS may be installed on a new server or it may be installed on
existing servers. One current trend is called virtualization. Virtualization is a technique that creates logical
representations of computing resources that are independent of the underlying physical computing resources. This
technique is used in many areas of computing such as the creation of virtual servers, virtual storage, and virtual private
networks. In a database environment, database virtualization refers to the installation of a new instance of the DBMS
on a virtual server running on shared hardware. This is normally a task that involves system and network administrators
to create appropriate user groups and services in the server configuration and network routing.
D A T A B A S E
D E S I G N
Create the Database(s)
In most modern relational DBMSs a new database implementation requires the creation of special storage-related
constructs to house the end-user tables. The constructs usually include the storage group (or file groups), the table
spaces, and the tables. Figure 9.7 depicts the fact that a storage space can contain more than one table space and that
a table space can contain more than one table.
FIGURE
Physical organization of a DB2 database environment
9.7
Storage group
Database
Table
Table
Table
Table
Table
Table
Table space
Table space
Table space
Table
Table space
Table space
For example, the implementation of the logical design in IBM’s DB2 would require that you:
1.
Create the database storage group. This step (done by the system administrator or SYSADM) is mandatory for
such mainframe databases as DB2. Other DBMS software may create equivalent storage groups automatically
when a database is created. (See Step 2.) Consult your DBMS documentation to see if you must create a
storage group and, if so, what the command syntax must be.
2.
Create the database within the storage group (also done by the SYSADM).
3.
Assign the rights to use the database to a database administrator (DBA).
4.
Create the table space(s) within the database (usually done by a DBA).
5.
Create the table(s) within the table space(s) (also usually done by a DBA).
6.
Assign access rights to the table spaces and to the tables within specified table spaces (another DBA duty).
Access rights may be limited to views rather than to whole tables. The creation of views is not required for
database access in the relational environment, but views are desirable from a security standpoint. For example,
access rights to a table named PROFESSOR may be granted to the user Shannon Scott whose identification
code is SSCOTT by typing:
GRANT SELECT ON PROFESSOR TO USER SSCOTT;
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Load or Convert the Data
After the database has been created, the data must be loaded into the database tables. Typically, the data will have to be
migrated from the prior version of the system. Often, data to be included in the system must be aggregated from multiple
sources. In a best-case scenario, all of the data will be in a relational database so that it can be readily transferred to the
new database. Unfortunately, this is not always the case. Data may have to be imported from other relational databases,
nonrelational databases, flat files, legacy systems, or even manual paper-and-pencil systems. If the data format does not
support direct importing into the new database, conversion programs may have to be created to reformat the data so that
it can be imported. In a worst-case scenario, much of the data may have to be manually entered into the database. Once
the data has been loaded, the DBA works with the application developers to test and evaluate the database.
9.3.4 Testing and Evaluation
In the design phase, decisions were made to ensure integrity, security, performance, and recoverability of the database.
During implementation and loading, these plans were put into place. In testing and evaluation, the DBA tests and
fine-tunes the database to ensure that it performs as expected. This phase occurs in conjunction with applications
programming. Programmers use database tools to prototype the applications during the coding of the programs. Tools
such as report generators, screen painters, and menu generators are especially useful to the applications programmers.
Test the Database
During this step, the DBA tests the database to ensure that it maintains the integrity and security of the data. Data
integrity is enforced by the DBMS through the proper use of primary and foreign key rules. Many DBMS also support
the creation of domain constraints, and database triggers. Testing will ensure that these constraints were properly
designed and implemented. In addition, data integrity is also the result of properly implemented data management
policies. Such policies are part of a comprehensive data administration framework. For a more detailed study of this
topic, see The DBA’s Managerial Role section in Chapter 15, Database Administration and Security.
Previously, users and roles were created to grant users access to the data. In this stage, not only must those privileges
be tested, but also the broader view of data privacy and security must be addressed. Data stored in the company
database must be protected from access by unauthorized users. (It does not take much imagination to predict the likely
results if students have access to a student database or if employees have access to payroll data!) Consequently, you
must test for (at least) the following:
쐌
Physical security allows only authorized personnel physical access to specific areas. Depending on the type of
database implementation, however, establishing physical security might not always be practical. For example,
a university student research database is not a likely candidate for physical security.
쐌
Password security allows the assignment of access rights to specific authorized users. Password security is
usually enforced at login time at the operating system level.
쐌
Access rights can be established through the use of database software. The assignment of access rights may
restrict operations (CREATE, UPDATE, DELETE, and so on) on predetermined objects such as databases,
tables, views, queries, and reports.
쐌
Audit trails are usually provided by the DBMS to check for access violations. Although the audit trail is an
after-the-fact device, its mere existence can discourage unauthorized use.
쐌
Data encryption can be used to render data useless to unauthorized users who might have violated some of
the database security layers.
쐌
Diskless workstations allow end users to access the database without being able to download the information
from their workstations.
For a more detailed discussion of security issues, please refer to Chapter 15, Database Administration and Security.
D A T A B A S E
D E S I G N
Fine-Tune the Database
Although database performance can be difficult to evaluate because there are no standards for database performance
measures, it is typically one of the most important factors in database implementation. Different systems will place
different performance requirements on the database. Systems to support rapid transactions will require the database
to be implemented in such a way so as to provide superior performance during high volumes of inserts, updates, and
deletes. Other systems, like decision support systems, may require superior performance on complex data retrieval
tasks. Many factors can impact the database’s performance on various tasks. Environmental factors, such as the
hardware and software environment in which the database exists, can have a significant impact on database
performance. Naturally, the characteristics and volume of the data in the database also affect database performance:
a search of 10 tuples will be faster than a search of 100,000 tuples. Other important factors in database performance
include system and database configuration parameters such as data placement, access path definition, the use of
indexes, and buffer size. For a more in-depth discussion of database performance issues, see Chapter 11, Database
Performance Tuning and Query Optimization.
Evaluate the Database and Its Application Programs
As the database and application programs are created and tested, the system must also be evaluated from a more
holistic approach. Testing and evaluation of the individual components should culminate in a variety of broader system
tests to ensure that all of the components interact properly to meet the needs of the users. At this time, integration
issues and deployment plans are refined, user training is conducted, and system documentation is finalized. Once the
system receives final approval, it must be a sustainable resource for the organization. To ensure that the data contained
in the database are protected against loss, backup and recovery plans are tested.
Timely data availability is crucial for almost every database. Unfortunately, the database can be subject to data loss
through unintended data deletion, power outages, and other causes. Data backup and recovery procedures create a
safety valve, ensuring the availability of consistent data. Typically, database vendors encourage the use of fault-tolerant
components such as uninterruptible power supply (UPS) units, RAID storage devices, clustered servers, and data
replication technologies to ensure the continuous operation of the database in case of a hardware failure. Even with
these components, backup and restore functions constitute a very important component of daily database operations.
Some DBMSs provide functions that allow the database administrator to schedule automatic database backups to
permanent storage devices such as disks, DVDs, tapes, and online storage. Database backups can be performed at
different levels:
쐌
A full backup of the database, or dump of the entire database. In this case, all database objects are backed
up in their entirety.
쐌
A differential backup of the database, in which only the last modifications to the database (when compared
with a previous full backup copy) are copied. In this case, only the objects that have been updated since the
last full backup are backed up.
쐌
A transaction log backup, which backs up only the transaction log operations that are not reflected in a
previous backup copy of the database. In this case, only the transaction log is backed up; no other database
objects are backed up. (For a complete explanation of the use of the transaction log see Chapter 10,
Transaction Management and Concurrency Control.)
The database backup is stored in a secure place, usually in a different building from the database itself, and is protected
against dangers such as fire, theft, flood, and other potential calamities. The main purpose of the backup is to
guarantee database restoration following system (hardware/software) failures.
Failures that plague databases and systems are generally induced by software, hardware, programming exemptions,
transactions, or external factors. Table 9.1 summarizes the most common sources of database failure.
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TABLE
9.1
Common Sources of Database Failure
SOURCE
Software
DESCRIPTION
Software-induced failures may be traceable to the operating system,
the DBMS software, application programs, or viruses.
Hardware
Hardware-induced failures may include memory chip errors, disk
crashes, bad disk sectors, and “disk full” errors.
Programming
exemptions
Application programs or end users may roll back transactions when
certain conditions are defined. Programming exemptions can also
be caused by malicious or improperly tested code that can be
exploited by hackers.
The system detects deadlocks and aborts one of the transactions.
(See Chapter 10.)
Transactions
External factors
Backups are especially important when a system suffers complete
destruction from fire, earthquake, flood, or other natural disaster.
EXAMPLE
The SQL.Slammer
worm affected many
unpatched MS SQL
Server systems in 2003
causing damages valued in millions of
dollars.
A bad memory module
or a multiple hard disk
failure in a database
system can bring a
database system to an
abrupt stop.
Hackers constantly
searching for exploits
in unprotected Web
database systems.
Deadlock occurs when
executing multiple
simultaneous
transactions.
In 2005, Hurricane
Katrina in New Orleans
caused data losses in
the millions of dollars.
Depending on the type and extent of the failure, the recovery process ranges from a minor short-term inconvenience
to a major long-term rebuild. Regardless of the extent of the required recovery process, recovery is not possible without
a usable backup.
The database recovery process generally follows a predictable scenario. First, the type and extent of the required
recovery are determined. If the entire database needs to be recovered to a consistent state, the recovery uses the most
recent backup copy of the database in a known consistent state. The backup copy is then rolled forward to restore all
subsequent transactions by using the transaction log information. If the database needs to be recovered but the
committed portion of the database is still usable, the recovery process uses the transaction log to “undo” all of the
transactions that were not committed (see Chapter 10, Transaction Management and Concurrency Control).
At the end of this phase, the database completes an iterative process of continuous testing, evaluation, and
modification that continues until the system is certified as ready to enter the operational phase.
D A T A B A S E
D E S I G N
9.3.5 Operation
Once the database has passed the evaluation stage, it is considered to be operational. At that point, the database, its
management, its users, and its application programs constitute a complete information system.
The beginning of the operational phase invariably starts the process of system evolution. As soon as all of the targeted
end users have entered the operations phase, problems that could not have been foreseen during the testing phase
begin to surface. Some of the problems are serious enough to warrant emergency “patchwork,” while others are
merely minor annoyances. For example, if the database design is implemented to interface with the Web, the sheer
volume of transactions might cause even a well-designed system to bog down. In that case, the designers have to
identify the source(s) of the bottleneck(s) and produce alternative solutions. Those solutions may include using
load-balancing software to distribute the transactions among multiple computers, increasing the available cache for the
DBMS, and so on. In any case, the demand for change is the designer’s constant concern, which leads to phase 6,
maintenance and evolution.
9.3.6 Maintenance and Evolution
The database administrator must be prepared to perform routine maintenance activities within the database. Some of
the required periodic maintenance activities include:
쐌
Preventive maintenance (backup).
쐌
Corrective maintenance (recovery).
쐌
Adaptive maintenance (enhancing performance, adding entities and attributes, and so on).
쐌
Assignment of access permissions and their maintenance for new and old users.
쐌
Generation of database access statistics to improve the efficiency and usefulness of system audits and to
monitor system performance.
쐌
Periodic security audits based on the system-generated statistics.
쐌
Periodic (monthly, quarterly, or yearly) system-usage summaries for internal billing or budgeting purposes.
The likelihood of new information requirements and the demand for additional reports and new query formats require
application changes and possible minor changes in the database components and contents. Those changes can be
easily implemented only when the database design is flexible and when all documentation is updated and online.
Eventually, even the best-designed database environment will no longer be capable of incorporating such evolutionary
changes and then the whole DBLC process begins anew.
As you can see, many of the activities described in the Database Life Cycle (DBLC) are similar to those in the Systems
Development Life Cycle (SDLC). This should not be surprising, because the SDLC is the framework within which the
DBLC activities take place. A summary of the parallel activities that take place within the SDLC and the DBLC is
shown in Figure 9.8.
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9
Parallel activities in the DBLC and the SDLC
9.8
DBLC
System
design
SDLC
Database initial
study
Database design
Analysis
Conceptual
Logical
Physical
Implementation
and loading
Detailed design
Screens
Reports
Procedures
Coding
Prototyping
Testing and
evaluation
Debugging
Creation
Loading
Fine-tuning
System
implementation
Testing and
evaluation
Operation
Database maintenance
and evolution
Application program
maintenance
9.4 CONCEPTUAL DESIGN
Recall that the second phase of the DBLC is database design, and that database design comprises four stages:
conceptual design, DBMS selection, logical design, and physical design. Conceptual design is the first stage in the
database design process. The goal at this stage is to design a database that is independent of database software and
physical details. The output of this process is a conceptual data model that describes the main data entities, attributes,
relationships, and constraints of a given problem domain. This design is descriptive and narrative in form. That is, it
is generally composed of a graphical representation as well as textual descriptions of the main data elements,
relationships, and constraints.
In this stage, data modeling is used to create an abstract database structure that represents real-world objects in the
most realistic way possible. The conceptual model must embody a clear understanding of the business and its functional
areas. At this level of abstraction, the type of hardware and/or database model to be used might not have been
identified yet. Therefore, the design must be software and hardware independent so that the system can be set up
within any hardware and software platform chosen later.
Keep in mind the following minimal data rule:
All that is needed is there, and all that is there is needed.
D A T A B A S E
D E S I G N
In other words, make sure that all data needed are in the model and that all data in the model are needed. All data
elements required by the database transactions must be defined in the model, and all data elements defined in the
model must be used by at least one database transaction.
However, as you apply the minimal data rule, avoid an excessive short-term bias. Focus not only on the immediate data
needs of the business but also on the future data needs. Thus, the database design must leave room for future
modifications and additions, ensuring that the business’s investment in information resources will endure.
The conceptual design has four steps, which are depicted in Table 9.2.
TABLE
9.2
STEP
1
2
3
4
Conceptual Design Steps
ACTIVITY
Data analysis and requirements
Entity relationship modeling and normalization
Data model verification
Distributed database design
The following sections cover these steps in more detail.
9.4.1 Data Analysis and Requirements
The first step in conceptual design is to discover the characteristics of the data elements. An effective database is an
information factory that produces key ingredients for successful decision making. Appropriate data element characteristics are those that can be transformed into appropriate information. Therefore, the designer’s efforts are focused on:
쐌
Information needs. What kind of information is needed—that is, what output (reports and queries) must be
generated by the system, what information does the current system generate, and to what extent is that
information adequate?
쐌
Information users. Who will use the information? How is the information to be used? What are the various
end-user data views?
쐌
Information sources. Where is the information to be found? How is the information to be extracted once it
is found?
쐌
Information constitution. What data elements are needed to produce the information? What are the data
attributes? What relationships exist among the data? What is the data volume? How frequently are the data
used? What data transformations are to be used to generate the required information?
The designer obtains the answers to those questions from a variety of sources in order to compile the necessary
information. Note these sources:
쐌
Developing and gathering end-user data views. The database designer and the end user(s) interact to jointly
develop a precise description of end-user data views. In turn, the end-user data views will be used to help
identify the database’s main data elements.
쐌
Directly observing the current system: existing and desired output. The end user usually has an existing
system in place, whether it’s manual or computer-based. The designer reviews the existing system to identify
the data and their characteristics. The designer examines the input forms and files (tables) to discover the data
type and volume. If the end user already has an automated system in place, the designer carefully examines the
current and desired reports to describe the data required to support the reports.
쐌
Interfacing with the systems design group. As noted earlier in this chapter, the database design process is part
of the Systems Development Life Cycle (SDLC). In some cases, the systems analyst in charge of designing the
new system will also develop the conceptual database model. (This is usually true in a decentralized
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environment.) In other cases, the database design is considered part of the database administrator’s job. The
presence of a database administrator (DBA) usually implies the existence of a formal data-processing
department. The DBA designs the database according to the specifications created by the systems analyst.
To develop an accurate data model, the designer must have a thorough understanding of the company’s data types and
their extent and uses. But data do not, by themselves, yield the required understanding of the total business. From a
database point of view, the collection of data becomes meaningful only when business rules are defined. Remember
from Chapter 2, Data Models, that a business rule is a brief and precise description of a policy, procedure, or principle
within a specific organization’s environment. Business rules, derived from a detailed description of an organization’s
operations, help to create and enforce actions within that organization’s environment. When business rules are written
properly, they define entities, attributes, relationships, connectivities, cardinalities, and constraints.
To be effective, business rules must be easy to understand and they must be widely disseminated to ensure that every
person in the organization shares a common interpretation of the rules. Using simple language, business rules describe
the main and distinguishing characteristics of the data as viewed by the company. Examples of business rules are as
follows:
쐌
A customer may make many payments on an account.
쐌
Each payment on an account is credited to only one customer.
쐌
A customer may generate many invoices.
쐌
Each invoice is generated by only one customer.
Given their critical role in database design, business rules must not be established casually. Poorly defined or inaccurate
business rules lead to database designs and implementations that fail to meet the needs of the organization’s end users.
Ideally, business rules are derived from a formal description of operations, which is a document that provides a
precise, up-to-date, and thoroughly reviewed description of the activities that define an organization’s operating
environment. (To the database designer, the operating environment is both the data sources and the data users.)
Naturally, an organization’s operating environment is dependent on the organization’s mission. For example, the
operating environment of a university would be quite different from that of a steel manufacturer, an airline, or a nursing
home. Yet no matter how different the organizations may be, the data analysis and requirements component of the
database design process is enhanced when the data environment and data use are described accurately and precisely
within a description of operations.
In a business environment, the main sources of information for the description of operations—and, therefore, of
business rules—are company managers, policy makers, department managers, and written documentation such as
company procedures, standards, and operations manuals. A faster and more direct source of business rules is direct
interviews with end users. Unfortunately, because perceptions differ, the end user can be a less reliable source when
it comes to specifying business rules. For example, a maintenance department mechanic might believe that any
mechanic can initiate a maintenance procedure, when actually only mechanics with inspection authorization should
perform such a task. Such a distinction might seem trivial, but it has major legal consequences. Although end users are
crucial contributors to the development of business rules, it pays to verify end-user perceptions. Often interviews with
several people who perform the same job yield very different perceptions of their job components. While such a
discovery might point to “management problems,” that general diagnosis does not help the database designer. Given
the discovery of such problems, the database designer’s job is to reconcile the differences and verify the results of the
reconciliation to ensure that the business rules are appropriate and accurate.
Knowing the business rules enables the designer to fully understand how the business works and what role the data
plays within company operations. Consequently, the designer must identify the company’s business rules and analyze
their impact on the nature, role, and scope of data.
D A T A B A S E
D E S I G N
Business rules yield several important benefits in the design of new systems:
쐌
They help standardize the company’s view of data.
쐌
They constitute a communications tool between users and designers.
쐌
They allow the designer to understand the nature, role, and scope of the data.
쐌
They allow the designer to understand business processes.
쐌
They allow the designer to develop appropriate relationship participation rules and foreign key constraints.
(See Chapter 4, Entity Relationship (ER) Modeling.)
The last point is especially noteworthy: whether a given relationship is mandatory or optional is usually a function of
the applicable business rule.
9.4.2 Entity Relationship Modeling and Normalization
Before creating the ER model, the designer must communicate and enforce appropriate standards to be used in the
documentation of the design. The standards include the use of diagrams and symbols, documentation writing style,
layout, and any other conventions to be followed during documentation. Designers often overlook this very important
requirement, especially when they are working as members of a design team. Failure to standardize documentation
often means a failure to communicate later, and communications failures often lead to poor design work. In contrast,
well-defined and enforced standards make design work easier and promise (but do not guarantee) a smooth integration
of all system components.
Because the business rules usually define the nature of the relationship(s) among the entities, the designer must
incorporate them into the conceptual model. The process of defining business rules and developing the conceptual
model using ER diagrams can be described using the steps shown in Table 9.3.3
TABLE
9.3
STEP
1
2
3
4
5
6
7
8
Developing the Conceptual Model Using ER Diagrams
ACTIVITY
Identify, analyze, and refine the business rules.
Identify the main entities, using the results of Step 1.
Define the relationships among the entities, using the results of Steps 1 and 2.
Define the attributes, primary keys, and foreign keys for each of the entities.
Normalize the entities. (Remember that entities are implemented as tables in an RDBMS.)
Complete the initial ER diagram.
Validate the ER model against the end users’ information and processing requirements.
Modify the ER model, using the results of Step 7.
Some of the steps listed in Table 9.3 take place concurrently. And some, such as the normalization process, can
generate a demand for additional entities and/or attributes, thereby causing the designer to revise the ER model. For
example, while identifying two main entities, the designer might also identify the composite bridge entity that
represents the many-to-many relationship between those two main entities.
To review, suppose that you are creating a conceptual model for the JollyGood Movie Rental Corporation, whose end
users want to track customers’ movie rentals. The simple ER diagram presented in Figure 9.9 shows a composite entity
that helps track customers and their video rentals. Business rules define the optional nature of the relationships between
3See “Linking Rules to Models,” Alice Sandifer and Barbara von Halle, Database Programming and Design, 4(3), March 1991, pp. 13−16. Although
the source seems dated, it remains the current standard. The technology has changed substantially, but the process has not.
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the entities VIDEO and CUSTOMER depicted in Figure 9.9. (For example, customers are not required to check out a
video. A video need not be checked out in order to exist on the shelf. A customer may rent many videos, and a video
may be rented by many customers.) In particular, note the composite RENTAL entity that connects the two main entities.
FIGURE
JollyGood Movie Rental ER
9.9
As you will likely discover, the initial ER model may be subjected to several revisions before it meets the system’s
requirements. Such a revision process is quite natural. Remember that the ER model is a communications tool as well as
a design blueprint. Therefore, when you meet with the proposed system users, the initial ER model should give rise to
questions such as, “Is this really what you meant?” For example, the ERD shown in Figure 9.9 is far from complete.
Clearly, many more attributes must be defined and the dependencies must be checked before the design can be
implemented. In addition, the design cannot yet support the typical video rental transaction environment. For example,
each video is likely to have many copies available for rental purposes. However, if the VIDEO entity shown in Figure 9.9
is used to store the titles as well as the copies, the design triggers the data redundancies shown in Table 9.4.
TABLE
9.4
Data Redundancies in the VIDEO Table
VIDEO_ID
SF-12345FT-1
SF-12345FT-2
SF-12345FT-3
WE-5432GR-1
WE-5432GR-2
VIDEO_TITLE
Adventures on Planet III
Adventures on Planet III
Adventures on Planet III
TipToe Canu and Tyler 2:
A Journey
TipToe Canu and Tyler 2:
A Journey
VIDEO_COPY
1
2
3
1
VIDEO_CHG
$4.50
$4.50
$4.50
$2.99
VIDEO_DAYS
1
1
1
2
2
$2.99
2
The initial ERD shown in Figure 9.9 must be modified to reflect the answer to the question, “Is more than one copy
available for each title?” Also, payment transactions must be supported. (You will have an opportunity to modify this
initial design in Problem 5 at the end of the chapter.)
From the preceding discussion, you might get the impression that ER modeling activities (entity/attribute definition,
normalization, and verification) take place in a precise sequence. In fact, once you have completed the initial ER model,
chances are you will move back and forth among the activities until you are satisfied that the ER model accurately
represents a database design that is capable of meeting the required system demands. The activities often take place
in parallel, and the process is iterative. Figure 9.10 summarizes the ER modeling process interactions. Figure 9.11
summarizes the array of design tools and information sources that the designer can use to produce the
conceptual model.
D A T A B A S E
FIGURE
ER modeling is an iterative process based on many activities
9.10
Database initial study
Data analysis
User views and
business rules
DBLC
processes and
database transactions
Initial ER model
Verification
Attributes
Normalization
Final ER model
FIGURE
Conceptual design tools and information sources
9.11
Information sources
Business rules and
data constraints
Design tools
Conceptual model
ER diagram
ERD
Data flow diagrams
DFD*
Normalization
Process functional
descriptions (FD)*
(user views)
Data dictionary
* Output generated by the systems analysis and design activities
Definition
and
validation
D E S I G N
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All objects (entities, attributes, relations, views, and so on) are defined in a data dictionary, which is used in tandem with
the normalization process to help eliminate data anomalies and redundancy problems. During this ER modeling
process, the designer must:
쐌
Define entities, attributes, primary keys, and foreign keys. (The foreign keys serve as the basis for the
relationships among the entities.)
쐌
Make decisions about adding new primary key attributes to satisfy end-user and/or processing requirements.
쐌
Make decisions about the treatment of composite and multivalued attributes.
쐌
Make decisions about adding derived attributes to satisfy processing requirements.
쐌
Make decisions about the placement of foreign keys in 1:1 relationships.
쐌
Avoid unnecessary ternary relationships.
쐌
Draw the corresponding ER diagram.
쐌
Normalize the entities.
쐌
Include all data element definitions in the data dictionary.
쐌
Make decisions about standard naming conventions.
The naming conventions requirement is important, yet it is frequently ignored at the designer’s risk. Real database
design is generally accomplished by teams. Therefore, it is important to ensure that the team members work in an
environment in which naming standards are defined and enforced. Proper documentation is crucial to the successful
completion of the design. Adherence to the naming conventions serves database designers well. In fact, a common
refrain from users seems to be this: “I didn’t know why you made such a fuss over naming conventions, but now that
I’m doing this stuff for real, I’ve become a true believer.”
9.4.3 Data Model Verification
The data model verification step is one of the last steps in the conceptual design stage, and it is also one of the most
critical ones. In this step, the ER model must be verified against the proposed system processes in order to corroborate
that the intended processes can be supported by the database model. Verification requires that the model be run
through a series of tests against:
쐌
End-user data views.
쐌
All required transactions: SELECT, INSERT, UPDATE, and DELETE operations.
쐌
Access rights and security.
쐌
Business-imposed data requirements and constraints.
Because real-world database design is generally done by teams, it is very likely the database design is divided into major
components known as modules. A module is an information system component that handles a specific business
function, such as inventory, orders, payroll, and so on. Under these conditions, each module is supported by an
ER segment that is a subset or fragment of an enterprise ER model. Working with modules accomplishes several
important ends:
쐌
The modules (and even the segments within them) can be delegated to design groups within teams, greatly
speeding up the development work.
쐌
The modules simplify the design work. The large number of entities within a complex design can be daunting.
Each module contains a more manageable number of entities.
쐌
The modules can be prototyped quickly. Implementation and applications programming trouble spots can be
identified more readily. (Quick prototyping is also a great confidence builder.)
쐌
Even if the entire system can’t be brought online quickly, the implementation of one or more modules will
demonstrate that progress is being made and that at least part of the system is ready to begin serving the
end users.
D A T A B A S E
D E S I G N
As useful as modules are, they represent a loose collection of ER model fragments that if left unchecked could break
havoc in the database. For example, the ER model fragments:
쐌
Might present overlapping, duplicated or conflicting views of the same data.
쐌
Might not be able to support all system’s modules processes.
To avoid these problems, it is better if the modules’ ER fragments are merged into a single enterprise ER model. This
process starts by selecting a central ER model segment and iteratively adding additional ER model segments one at a
time. At each stage, for each new entity added to the model, you need to validate that the new entity doesn’t overlap
or conflict with a previously identified entity in the enterprise ER model.
Merging the ER model segments into an enterprise ER model triggers a careful reevaluation of the entities, followed
by a detailed examination of the attributes that describe those entities. This process serves several important purposes:
쐌
The emergence of the attribute details might lead to a revision of the entities themselves. Perhaps some of the
components first believed to be entities will, instead, turn out to be attributes within other entities. Or what was
originally considered to be an attribute might turn out to contain a sufficient number of subcomponents to
warrant the introduction of one or more new entities.
쐌
The focus on attribute details can provide clues about the nature of relationships as they are defined by the
primary and foreign keys. Improperly defined relationships lead to implementation problems first and to
application development problems later.
쐌
To satisfy processing and/or end-user requirements, it might be useful to create a new primary key to replace
an existing primary key. For example, in the example illustrated in Figure 9.9, a surrogate primary key (i.e.
RENTAL_ID) could be introduced to replace the original primary key composed of VIDEO_ID and
CUST_NUMBER.
쐌
Unless the entity details (the attributes and their characteristics) are precisely defined, it is difficult to evaluate
the extent of the design’s normalization. Knowledge of the normalization levels helps guard against undesirable
redundancies.
쐌
A careful review of the rough database design blueprint is likely to lead to revisions. Those revisions will help
ensure that the design is capable of meeting end-user requirements.
After finishing the merging process, the resulting enterprise ER model is verified against each of the modules processes
The ER model verification process is detailed in Table 9.5.
TABLE
9.5
STEP
1
2
3
4
5
6
The ER Model Verification Process
ACTIVITY
Identify the ER model’s central entity.
Identify each module and its components.
Identify each module’s transaction requirements:
Internal: Updates/Inserts/Deletes/Queries/Reports
External: Module interfaces
Verify all processes against system requirements.
Make all necessary changes suggested in Step 4.
Repeat Steps 2−5 for all modules.
Keep in mind that the verification process requires the continuous verification of business transactions as well as system
and user requirements. The verification sequence must be repeated for each of the system’s modules. Figure 9.12
illustrates the iterative nature of the process.
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9
Iterative ER model verification process
9.12
Identify central entity,
module and components
Define processes and
transaction steps
Verify ER model
Does ER
require changes
Yes
Make changes
to ER model
No
ER model verified
The verification process starts with selecting the central (most important) entity. The central entity is defined in terms
of its participation in most of the model’s relationships, and it is the focus for most of the system’s operations. In other
words, to identify the central entity, the designer selects the entity involved in the greatest number of relationships. In
the ER diagram, it is the entity that has more lines connected to it than any other.
The next step is to identify the module or subsystem to which the central entity belongs and to define that module’s
boundaries and scope. The entity belongs to the module that uses it most frequently. Once each module is identified,
the central entity is placed within the module’s framework to let you focus your attention on the module’s details.
Within the central entity/module framework, you must:
쐌
Ensure the module’s cohesivity. The term cohesivity describes the strength of the relationships found
among the module’s entities. A module must display high cohesivity—that is, the entities must be strongly
related, and the module must be complete and self-sufficient.
쐌
Analyze each module’s relationships with other modules to address module coupling. Module coupling
describes the extent to which modules are independent of one another. Modules must display low coupling,
indicating that they are independent of other modules. Low coupling decreases unnecessary intermodule
dependencies, thereby allowing the creation of a truly modular system and eliminating unnecessary relationships among entities.
D A T A B A S E
D E S I G N
Processes may be classified according to their:
쐌
Frequency (daily, weekly, monthly, yearly, or exceptions).
쐌
Operational type (INSERT or ADD, UPDATE or CHANGE, DELETE, queries and reports, batches, maintenance, and backups).
All identified processes must be verified against the ER model. If necessary, appropriate changes are implemented. The
process verification is repeated for all of the model’s modules. You can expect that additional entities and attributes will
be incorporated into the conceptual model during its validation.
At this point, a conceptual model has been defined as hardware- and software-independent. Such independence
ensures the system’s portability across platforms. Portability can extend the database’s life by making it possible to
migrate to another DBMS and/or another hardware platform.
9.4.4 Distributed Database Design
Although not a requirement for most databases, sometimes a database may need to be distributed among multiple
geographically disperse locations. Processes that access the database may also vary from one location to another. For
example, a retail process and a warehouse storage process are likely to be found in different physical locations. If the
database data and processes are to be distributed across the system, portions of a database, known as database
fragments, may reside in several physical locations. A database fragment is a subset of a database that is stored at
a given location. The database fragment may be composed of a subset of rows or columns from one or multiple tables.
Distributed database design defines the optimum allocation strategy for database fragments in order to ensure database
integrity, security, and performance. The allocation strategy determines how to partition the database and where to
store each fragment. The design implications introduced by distributed processes are examined in detail in Chapter 12,
Distributed Database Management Systems.
9.5 DBMS SOFTWARE SELECTION
The selection of DBMS software is critical to the information system’s smooth operation. Consequently, the
advantages and disadvantages of the proposed DBMS software should be carefully studied. To avoid false expectations,
the end user must be made aware of the limitations of both the DBMS and the database.
Although the factors affecting the purchasing decision vary from company to company, some of the most
common are:
쐌
Cost. This includes the original purchase price, along with maintenance, operational, license, installation,
training, and conversion costs.
쐌
DBMS features and tools. Some database software includes a variety of tools that facilitate the application
development task. For example, the availability of query by example (QBE), screen painters, report generators,
application generators, data dictionaries, and so on, helps to create a more pleasant work environment for
both the end user and the application programmer. Database administrator facilities, query facilities, ease of
use, performance, security, concurrency control, transaction processing, and third-party support also influence
DBMS software selection.
쐌
Underlying model. This can be hierarchical, network, relational, object/relational, or object-oriented.
쐌
Portability. A DBMS can be portable across platforms, systems, and languages.
쐌
DBMS hardware requirements. Items to consider include processor(s), RAM, disk space, and so on.
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9.6 LOGICAL DESIGN
Logical design is the second stage in the database design process. The logical design goal is to design an
enterprise-wide database based on a specific data model but independent of physical-level details. Logical design
requires that all objects in the conceptual model be mapped to the specific constructs used by the selected database
model. For example, the logical design for a relational DBMS includes the specifications for the relations (tables),
relationships, and constraints (i.e., domain definitions, data validations, and security views).
The logical design is generally performed in four steps, which are depicted in Table 9.6.
TABLE
9.6
STEP
1
2
3
4
Logical Design Steps
ACTIVITY
Map conceptual model to logical model components
Validate logical model using normalization
Validate logical model integrity constraints
Validate logical model against user requirements
Such steps, like most of the data-modeling process, are not necessarily performed sequentially, but in an iterative
fashion. The following sections cover these steps in more detail.
9.6.1 Map the Conceptual Model to the Logical Model
The first step in creating the logical design is to map the conceptual model to the chosen database constructs. Because
this book deals primarily with relational databases and because most current database design projects are based on the
relational database model, we will focus on logical design using relational constructs. In the real world, logical design
generally involves translating the ER model into a set of relations (tables), columns, and constraints definitions. The
process of translating the conceptual model into a set of relations is depicted in Table 9.7.
TABLE
9.7
STEP
1
2
3
4
5
Mapping the Conceptual Model to the Relational Model
ACTIVITY
Map strong entities
Map supertype/subtype relationships
Map weak entities
Map binary relationships
Map higher degree relationships
To illustrate this process, we will use the example of the SimpleCollege ER model shown in Figure 9.13. Remember,
the steps indicated in Table 9.7 are not sequential but iterative.
As indicated in Table 9.7, the first step in the logical design stage is to map strong entities to tables. Recall from
Chapter 4 that a strong entity is one that resides in the 1 side of all its relationships, that is, an entity that does not
have any mandatory attribute that is a foreign key to another table. Therefore, the first entities to be translated into
tables would be the EMPLOYEE and COURSE entities. In this case, you define what would be the table name, what
D A T A B A S E
FIGURE
D E S I G N
The SimpleCollege conceptual model
9.13
would be its columns and their characteristics. For example, the relation definitions for the strong entities on
SimpleCollege would be:
COURSE (CRS_CODE, CRS_TITLE, CRS_DESCRIPT, CRS_CREDIT)
PRIMARY KEY:
CRS_CODE
EMPLOYEE (EMP_NUM, EMP_LNAME, EMP_FNAME, EMP_INITIAL, EMP_E_MAIL)
PRIMARY KEY:
EMP_NUM
Once all strong entities are mapped, you are ready to map any entities involved in a supertype/subtype relationship
or any weak entities. In the case of SimpleCollege, you have the PROFESSOR entity that is a subtype of the
EMPLOYEE entity. PROFESSOR is also a weak entity because it inherits its primary key from EMPLOYEE and it is
existence-dependent on EMPLOYEE. At this time, you could also start defining the relationships between supertype
and subtype entities. For example:
PROFESSOR (EMP_NUM, PROF_SPECIALTY, PROF_RANK)
PRIMARY KEY:
EMP_NUM
FOREIGN KEY:
EMP_NUM REFERENCES PROFESSOR
Next, you start mapping all binary relationships. Note that in the previous example you have already defined the
supertype/subtype relationship between EMPLOYEE and PROFESSOR. This is an instance that demonstrates the
iterative nature of this process. Continuing with the SimpleCollege ER model, you would define the CLASS relation
and define its 1:M relationships with PROFESSOR and COURSE:
CLASS (CLASS_CODE, EMP_NUM, CLASS_TIME, CLASS_DAYS, CRS_CODE)
PRIMARY KEY:
CLASS_CODE
FOREIGN KEYS:
EMP_NUM REFERENCES PROFESSOR
CRS_CODE REFERENCES COURSE
Next, you will proceed with all relationships between three or more entities until all relationships in the model are
clearly defined. The logical design’s tables must correspond to the entities (EMPLOYEE, PROFESSOR, COURSE, and
CLASS) shown in the conceptual design of Figure 9.13, and the table columns must correspond to the attributes
specified in the conceptual design. The final outcome of this process is a list of relations, attributes, and relationships
that will be the basis for the next step.
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9.6.2 Validate the Logical Model Using Normalization
The logical design should contain only properly normalized tables. The process of mapping the conceptual model to
the logical model may unveil some new attributes or the discovery of new multivalued or composite attributes.
Therefore, it’s very likely that new attributes may be adde