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Home/ Questions/Q 6251341
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T13:35:46+00:00 2026-05-24T13:35:46+00:00

I have 2 shell scripts, namely script A and script B. I have both

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I have 2 shell scripts, namely script A and script B.
I have both of them “set -e”, telling them to stop upon error.

However, when script A call script B, and script B had an error and stopped, script A didn’t stop.

What can I stop the mother script when the child script dies?

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  1. Editorial Team
    Editorial Team
    2026-05-24T13:35:47+00:00Added an answer on May 24, 2026 at 1:35 pm

    It should work as you’d expect. For example:

    In mother.sh:

    #!/bin/bash
    set -ex
    ./child.sh
    echo "you should not see this (a.sh)"
    

    In child.sh:

    #!/bin/bash
    set -ex
    ls &> /dev/null # good cmd
    ls /path/that/does/not/exist &> /dev/null # bad cmd
    echo "you should not see this (b.sh)"
    

    Calling mother.sh:

    [me@home]$ ./mother.sh
    ++ ./child.sh
    +++ ls
    +++ ls /path/that/does/not/exist
    

    Why is it not working for you?

    One possible situation where it won’t work as expected is if you specified -e in the shabang line (#!/bin/bash -e) and passed the script directly to bash which will treat that as a comment.

    For example, if we change mother.sh to:

    #!/bin/bash -ex
    ./child.sh
    echo "you should not see this (a.sh)"
    

    Notice how it behaves differently depending on how you call it:

    [me@home]$ ./mother.sh
    + ./child.sh
    + ls
    + ls /path/that/does/not/exist
    
    [me@home]$ bash mother.sh  
    + ls
    + ls /path/that/does/not/exist
    you should not see this (a.sh)
    

    Explicitly calling set -e within the script will solve this problem.

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