I have this code, and when I call it in command line, it shows the echo in option/the one I have in each option/ but $OPTARG is undefined. And I have no idea why.

    while getopts "ml:" opt
    case "$opt" in
    echo "m was triggered"
echo "l was triggered"
   shift "$((OPTIND-1))"

and my second question is: OPTIND=2 can this be used to start with second argument? I mean it works but wheater it is legit way to do that. for example:

./project first -l second 

so it would skip the first agument and started with -l?

  • 1
    I would usually store the first argument in a variable, then shift. That way I don't tinker around with the value of OPTIND. Furthermore, is there a reason why you do optm=$OPTARG? According to getopts "ml:", the -m flag does not take any parameter. Apart from that, I don't think I understood your question; the slashes are quite confusing: which echo does it show? How do you see that $OPTARG is undefined? – ayekat Mar 19 '15 at 11:55

You've got this line

while getopts "ml:" opt

This tells the shell that m may exist but takes no parameter, and that l may appear with exactly one parameter.

I suspect you meant to use getopts 'm:l' opt.

  • Just noticed @ayekat has alluded to the same issue. Kudos there. – roaima Mar 19 '15 at 12:35

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