Returning different exit code than what command returns in bash script

Question

I have a bash script that run some tool that on some use cases returns an error code of 137. I use this script as part of my CI pipeline and and as far as i am concerned this is suceessful state. My question is how can i catch this exit code and return an exit code of 0 instead(like trap command, only the I do not think it support custom exit codes, only predefined one).

Thanks in advance, Alon

JNevill · Accepted Answer · 2021-09-01 14:19:06Z

1

After you run the command/tool in your script:

command

You can see what the exit code and do some logic based on that return:

command
return=$?
if [ $return -eq 137 ]; then
    exit 0
else # @chepner's very good suggestion
    exit $return 
fi

That will end your bash script with an exit code of 0 if the command you ran (your tool) returns an exit code of 137.

edited Sep 1, 2021 at 14:19

answered Sep 1, 2021 at 14:14

JNevill

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8 Comments

chepner Over a year ago

Be sure to have an explicit exit $return in an else clause. Otherwise, where the exit status of the script was implicitly set by the exit status of command before, now the exit status would be the exit status of if.

chepner Over a year ago

Don't use set -e if you want control over the exit status of your script.

JNevill Over a year ago

Check out some of these answers for set -e: stackoverflow.com/questions/26675681/…

Alon Tsaraf Over a year ago

The problem is that i can't use it because i use "set -e" in my script(because it is in CI pipeline i want that each failure that is not 137 will fail the pipeline). trap command was great for handling it with predefined signals such as SIGINT or SIGTERM, this time i want to handle a custom exit code returned by some tool.

k314159 Over a year ago

@AlonTsaraf put this answer's code into a new Bash script, add set +e at the top to temporarily disable the set -e in this script, and then call this wrapper script from your main script instead of the real command.

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