Pass multiple values using one argument in bash

Question

I need to pass more than one value to an argument (the number of values I'm passing is different each time), but how can I do it? Here's my code:


usage() { echo "Usage: $0 [-i <string>]" 1>&2; exit 1; }

while getopts ":i:" o; do
    case "${o}" in
        i)
            i=${OPTARG}
            ;;
        *)
            usage
            ;;
    esac
done

use shift: here an example: unix.stackexchange.com/questions/174566/… — Incrivel Monstro Verde
– Incrivel Monstro Verde, Commented Oct 1, 2021 at 13:46
consider updating the question with some command line examples of passing more than one value to your script; also provide an explanation of what the script is supposed to do with the multiple values (eg, store in an array? append to a variable? something else?) — markp-fuso
– markp-fuso, Commented Oct 1, 2021 at 14:04
#1 something liike this: scripts.sh PARAM1=value1,value2,value3 ? or #2 scripts.sh PARAM1=value1 PARAM1=value2 — JRichardsz
– JRichardsz, Commented Oct 2, 2021 at 16:32

Dmitrii Kudriashov · Accepted Answer · 2022-10-13 09:41:06Z

1

I used this workaround with parsing values from the string Use this as a delimter in a script

IFS=','

Then inside of an option when you assigning OPTARG to it:

OPTION_STRING=${OPTARG}
# Check if there is multiple values passed with IFS delimeter
read -ra my_array <<<$OPTION_STRING

Then for the logic check the lenght of array and if it's more than 1 do a cycle like this

if [ "${#my_array[@]}" -gt 1 ]; then
        # Do your something before the cycle
        # Cycling array for values
        for i in "${my_array[@]}"; do
            # Do what you need based on $i
        done

I guess it's too late for initial post, but maybe this may help someone else.

edited Oct 13, 2022 at 9:41

answered Oct 13, 2022 at 9:38

Dmitrii Kudriashov

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