How to exit recent getopts options? - bash

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"
}