In a bash script, I have an array of arguments I want to pass in order to another bash script. How would I do this? [duplicate]

Question

Basically, I wrote a bash script to act as a wrapper script for other scripts. The idea is to pass the name of the script you want to run with its arguments, then my script sets some global vars and makes sure you are allowed to use the given inputs, then calls the script with the given args. So how would I invoke the script given as an argument with the given parameters in bash code?

BTW, to really copy content in an array, see stackoverflow.com/a/1063367/14122. The linked duplicates are for passing through the argument list (with "$@"), instead of actually expanding an array onto it. — Charles Duffy
– Charles Duffy, Commented Sep 3, 2021 at 17:22

chepner · Accepted Answer · 2021-09-03 17:19:58Z

1

The arguments you receive are in "$@"; you can use that to pass them on.

#!/bin/bash

script=$1
shift

for x in "$@"; do
    # validate remaining arguments
done

"$script" "$@"

answered Sep 3, 2021 at 17:19

chepner

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2 Comments

Charles Duffy Over a year ago

...already have two linked duplicates that are precisely on-point, no?

chepner Over a year ago

Subtle difference (IMO, anyway): one of the arguments is the script to run itself, rather than an an argument to pass on to a hard-coded command.

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