Automatic exit from Bash shell script on error [duplicate]

Question

I've been writing some shell script and I would find it useful if there was the ability to halt the execution of said shell script if any of the commands failed. See below for an example:

#!/bin/bash  cd some_dir  ./configure --some-flags  make  make install 

So in this case, if the script can't change to the indicated directory, then it would certainly not want to do a ./configure afterwards if it fails.

Now I'm well aware that I could have an if check for each command (which I think is a hopeless solution), but is there a global setting to make the script exit if one of the commands fails?

Answer 1

Use the set -e builtin:

#!/bin/bash set -e # Any subsequent(*) commands which fail will cause the shell script to exit immediately 

Alternatively, you can pass -e on the command line:

bash -e my_script.sh 

You can also disable this behavior with set +e.

You may also want to employ all or some of the the -e -u -x and -o pipefail options like so:

set -euxo pipefail 

-e exits on error, -u errors on undefined variables, and -o (for option) pipefail exits on command pipe failures. Some gotchas and workarounds are documented well here.

(*) Note:

The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test following the if or elif reserved words, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command's return value is being inverted with !

(from man bash)

Answer 2

To exit the script as soon as one of the commands failed, add this at the beginning:

set -e 

This causes the script to exit immediately when some command that is not part of some test (like in a if [ ... ] condition or a && construct) exits with a non-zero exit code.