how to call a function with 2 arguments which under the option of "getopts"

Question

New in Linux bash script. Here I tried to create some files with getopts. For example I'd like to create 3 files called xyzfile, in command line ./createfiles -n xyzfile 3should be given (2 arguments after the option -n). The result should be 3 files with the names xyzfile_1, xyzfile_2 and xyzfile_3.

I tried to put my createfile() function outside the while-loop and as well as inside the while-loop. But the option -n doesn't work. I also tried to create another function called foo() with included the function createfile(), but still something wrong there. I have no idea anymore what I can do. Hope I can get some advices from you guys. Thank you very much!

#!/bin/bash

    while getopts :n:bc opt; do
        case $opt in
            n) echo test 3333333
                 createfile() {
                    echo "$OPTARG"
                    sum=$2

                 for((i=1;i<=sum;i++))
                    do
                    touch "$OPTARG_${i}"
                done
                 }
                 createfile $OPTARG ${2};;
            b) echo "test 1111111";;
            c) echo "test 2222222";;
            *) echo error!;;
        esac
    done

Answer 1

Use a separate option for the count, and create your files after the option processing.

Something like:

while getopts "n:c:" opt; do
    case $opt in
        n) name="$OPTARG";;
        c) count=$OPTARG;;
        # other options...
    esac
done

shift $((OPTIND -1))

while (( count > 0 )); do
    touch "${name}_$count"
    (( count-- ))
    # ...
done