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Home/ Questions/Q 9132903
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T08:23:38+00:00 2026-06-17T08:23:38+00:00

I am trying to create a bash script which passes its own argument onto

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I am trying to create a bash script which passes its own argument onto a python script. I want it to work like this.

If I run it as this:

script.sh latest

Then within the bash script it runs a python script with the “latest” argument like this:

python script.py latest

Likewise if the bash script is run with the argument 123 then the python script as such:

python script.py 123

Can anyone help me understand how to accomplish this please?

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  1. Editorial Team
    Editorial Team
    2026-06-17T08:23:39+00:00Added an answer on June 17, 2026 at 8:23 am

    In this case the trick is to pass however many arguments you have, including the case where there are none, and to preserve any grouping that existed on the original command line.

    So, you want these three cases to work:

    script.sh                       # no args
    script.sh how now               # some number
    script.sh "how now" "brown cow" # args that need to stay quoted
    

    There isn’t really a natural way to do this because the shell is a macro language, so they’ve added some magic syntax that will just DTRT.

    #!/bin/sh
    
    python script.py "$@"
    
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