In a Bash script, how can I exit the entire script if a certain condition occurs?

Question

I'm writing a script in Bash to test some code. However, it seems silly to run the tests if compiling the code fails in the first place, in which case I'll just abort the tests.

Is there a way I can do this without wrapping the entire script inside of a while loop and using breaks? Something like a dun dun dun goto?

Answer 1

Use set -e

#!/bin/bash  set -e  /bin/command-that-fails /bin/command-that-fails2 

The script will terminate after the first line that fails (returns nonzero exit code). In this case, command-that-fails2 will not run.

If you were to check the return status of every single command, your script would look like this:

#!/bin/bash  # I'm assuming you're using make  cd /project-dir make if [[ $? -ne 0 ]] ; then     exit 1 fi  cd /project-dir2 make if [[ $? -ne 0 ]] ; then     exit 1 fi 

With set -e it would look like:

#!/bin/bash  set -e  cd /project-dir make  cd /project-dir2 make 

Any command that fails will cause the entire script to fail and return an exit status you can check with $?. If your script is very long or you're building a lot of stuff it's going to get pretty ugly if you add return status checks everywhere.

Answer 2

Try this statement:

exit 1 

Replace 1 with appropriate error codes. See also Exit Codes With Special Meanings.