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Home/ Questions/Q 8563541
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T16:58:59+00:00 2026-06-11T16:58:59+00:00

I have a bash script with -e option set, which fails the whole script

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I have a bash script with -e option set, which fails the whole script on the very first error.

In the script, I am trying to do an ls on a directory. But that path may or may not exist. If the path does not exist, the ls command fails, since the -e flag is set.

Is there a way by which I can prevent the script from failing?

As a side note, I have tried the trick to do an set +e and set -e before and after that command and it works. But I am looking for some better solution.

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  1. Editorial Team
    Editorial Team
    2026-06-11T16:59:00+00:00Added an answer on June 11, 2026 at 4:59 pm

    You can “catch” the error using || and a command guaranteed to exit with 0 status:

    ls $PATH || echo "$PATH does not exist"
    

    Since the compound command succeeds whether or not $PATH exists, set -e is not triggered and your script will not exit.

    To suppress the error silently, you can use the true command:

    ls $PATH || true
    

    To execute multiple commands, you can use one of the compound commands:

    ls $PATH || { command1; command2; }
    

    or

    ls $PATH || ( command1; command2 )
    

    Just be sure nothing fails inside either compound command, either. One benefit of the second example is that you can turn off immediate-exit mode inside the subshell without affecting its status in the current shell:

    ls $PATH || ( set +e; do-something-that-might-fail )
    
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