How to create a function that passes arguments to another bash script?

Question

I want to create a function in bash that executes another script with the same arguments passed to the first function.

This is what I've tried so far (I'm getting syntax errors):

.bashrc

my-fn {
    /home/user/myScript "$@"
}

custom-script

#!/bin/bash

another-defined-fn "$@"

I'm getting the following errors when I try to source my .bashrc

-bash: /home/userx/.bashrc: line 206: syntax error near unexpected token `}'
-bash: /home/userx/.bashrc: line 206: `}'

What is the proper syntax to achieve this?

Extra

How can I include a sudo call inside the function?

Something like:

my-fn {
    sudo /home/user/myScript "$@"
}

Kusalananda · Accepted Answer · 2021-10-01 08:35:05Z

1

You have got the right syntax for the arguments, but not for the function definition. You need to define the function like this:

myfun() {
   ...
}

or this:

function myfun {
   ...
}

and also to call the function with all the arguments (or a list of arguments).

From a file:

myfun "$@"

or

myfun one two three...

Kusalananda♦

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answered Sep 22, 2021 at 20:29

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