Aborting a shell script if any command returns a non
I have a Bash shell script that invokes a number of commands. I would like to have the shell script automatically exit with a return value of 1 if any of the commands return a non-zero value.
Is this possible without explicitly checking the result of each command?
eg
dosomething1
if [[ $? -ne 0 ]]; then
exit 1
fi
dosomething2
if [[ $? -ne 0 ]]; then
exit 1
fi
Add this to the beginning of the script:
set -e
This will cause the shell to exit immediately if a simple command exits with a nonzero exit value. A simple command is any command not part of an if, while, or until test, or part of an && or || list.
See the bash(1) man page on the "set" internal command for more details.
I personally start almost all shell scripts with "set -e". It's really annoying to have a script stubbornly continue when something fails in the middle and breaks assumptions for the rest of the script.
To add to the accepted answer:
Bear in mind that set -e sometimes is not enough, specially if you have pipes.
For example, suppose you have this script
#!/bin/bash
set -e
./configure > configure.log
make
... which works as expected: an error in configure aborts the execution.
Tomorrow you make a seemingly trivial change:
#!/bin/bash
set -e
./configure | tee configure.log
make
... and now it does not work. This is explained here, and a workaround (Bash only) is provided:
#!/bin/bash set -e set -o pipefail ./configure | tee configure.log make
The if statements in your example are unnecessary. Just do it like this:
dosomething1 || exit 1
If you take Ville Laurikari's advice and use set -e then for some commands you may need to use this:
dosomething || true
The || true || true will make the command pipeline have a true return value even if the command fails so the the -e option will not kill the script.
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