How can multiple arguments be passed to shell script?

Shell Programming and Scripting
1

My requirement is that I want to pass similar argument to a shell script and process it in the script. Something like below:

myScript.sh   -c COMPONENT1 -c COMPONENT2  -a APP

Note: -c option can be specified multiple times and -a is optional parameter

I know this can be achieved using parser.add_option in Perl. But can be this be done in Shell?

2

getopts might help you. Have a look at this sample script:

m@home$cat test.sh
#! /bin/bash

#An array to hold the values passed using -c option
declare -a c_arr

#The count of -c options
c_count=0

#Go through all the optins passed
while getopts ":c:a:" o
do
        case "${o}" in
        c)
        c_arr[$c_count]=${OPTARG}
        let c_count=$c_count+1
        ;;

        a)
        a=${OPTARG}
        ;;

        *)
        echo "Wrong usage"
        exit 1
        ;;
        esac
done

if [ $c_count -eq 0 ]
then
        echo "option -c is mandatory"
        exit 1
fi

let i=$c_count-1

while [ $i -ge 0 ]
do
        echo "Got value for c option: ${c_arr[$i]}"
        let i=$i-1
done

if [ ! -z $a ]
then
        echo "Got value for a option: $a"
fi

And the output

m@home$./test.sh -c C1 -c C2 -a A
Got value for c option: C2
Got value for c option: C1
Got value for a option: A
1 Like
3

Thanks a lot chacko193 !!