How to exit recent getopts options?

Question

My script:

 run() {
     while getopts ":dr" option; do
        case "$option" in
            d) echo "__DEBUG__";;
            r) echo "__RELEASE__";;
        esac
    done
 if [ $option -eq ""]
    then 
      echo "__DEBUG__" ; 
 fi
}

Hi, i'm using zsh. When i call run without any options

$ run

$ __ DEBUG __

in the first time it's ok and show DEBUG

Then i call with option (d/r) and call run with no agrument again it show this error

$ run -r

$ __ RELEASE __

$ run

$ run-9: parse error: condition expected: r

I don't know how to fix this, it seem that the while loop still has effect

Answer 1

You need to reset OPTIND before the getopts call at the start of your function, this will not be done automatically in this case. In bash, it is set to 1 when a shell or shell script starts, but not when a function is called.

This seems to be what POSIX getopts specifies.

Whenever the shell is invoked, OPTIND shall be initialized to 1.

zsh getopts behaves differently (thanks @PesaThe):

OPTIND has an initial value of 1, and is normally set to 1 upon entry to a shell function and restored upon exit.

Also, when getopts has finished parsing, option will be set to ?, which will not help you determine whether a (valid) option was provided.

Try something like this:

run() {
  OPTIND=1
  found=0
  while getopts ":dr" option; do
    case "$option" in
      d) echo "__DEBUG__"; found=1 ;;
      r) echo "__RELEASE__"; found=1 ;;
    esac
  done
  if [ $found -eq 0 ]
  then
    echo "__DEBUG__" ; 
  fi
}

Or a bit simpler (but not identical to yours, to illustrate):

run() {
  OPTIND=1
  mode="__DEFAULT__"
  while getopts ":dr" option; do
    case "$option" in
      d) mode="__DEBUG__" ;;
      r) mode="__RELEASE__" ;;
    esac
  done
  echo "$mode"
}