Bash script argument not being parsed

Question

I wrote a bash script to help me with the build process of couple web apps. There are different build modes so i tried implementing this using flags (getopts)


    dev=false;
    clean_dest=false;
    all=false;
    
    while getopts c:clean-dest:a:all:d:dev:t:target: flags
    do
        case "${flags}" in
            d) dev=true;;
            dev) dev=true;;
            c) clean_dest=true;;
            clean-dest) clean_dest=true;;
            a) all=true;;
            all) all=true;;
            t) target=${OPTARG};;
            target) target=${OPTARG};;
        esac
    done

If I call it with -d -t "someTarget" it works but when I call it with -c -a -d -t "someTarget" it doesn't take into account '-c' and '-a'.

If I swap around the order of arguments in the while another flag combinations breaks.

What am I doing wrong?

I am thankful in advance.

The colon in `c:` means that `-c` takes an argument. So `-c -a` means that the flag is `-c` and its argument is `-a`. — Barmar, Oct 15 '21 at 17:43
BTW, `getopts` doesn't support multi-character options like `-target`. — Barmar, Oct 15 '21 at 17:43

score 0 · Accepted Answer · answered Oct 15 '21 at 18:19

0

Solved by adhering to getopts's syntax; I had to change the while line to while getopts cadt: flags.

answered Oct 15 '21 at 18:19

fmi21

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Bash script argument not being parsed

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