Advanced Bash-Scripting Guide
# Uses the 'more' filter.
# May substitute 'less' if desired.
exit $? # Script returns exit status of pipe.
# Actually "exit $?" is unnecessary, as the script will, in any case,
#+ return the exit status of the last command executed.
compound comparison
-a logical and
-o
exp1 -a exp2
returns true if both exp1 and exp2 are true.
logical or
exp1 -o exp2
returns true if either exp1 or exp2 is true.
These are similar to the Bash comparison operators && and ||, used within double brackets.
[[ condition1 && condition2 ]]
The -o and -a operators work with the test command or occur within single test brackets.
if [ "$expr1" -a "$expr2" ] then
echo "Both expr1 and expr2 are true." else
echo "Either expr1 or expr2 is false." fi
But, as rihad points out:
[ 1 -eq 1 ] && [ -n "`echo true 1>&2`" ] # true
[ 1 -eq 2 ] && [ -n "`echo true 1>&2`" ] # (no output)
# ^^^^^^^ False condition. So far, everything as expected.
# However ...
[ 1 -eq 2 -a -n "`echo true 1>&2`" ] # true
# ^^^^^^^ False condition. So, why "true" output?
# Is it because both condition clauses within brackets evaluate?
[[ 1 -eq 2 && -n "`echo true 1>&2`" ]] # (no output)
# No, that's not it.
# Apparently && and || "short-circuit" while -a and -o do not.
Refer to Example 8-3, Example 27-17, and Example A-29 to see compound comparison operators in action.
7.4. Nested
if/then
Condition Tests
Condition tests using the
if/then
construct may be nested. The net result is equivalent to using the &&
compound comparison operator.
a=3
Chapter 7. Tests 70
