My option parsing snippet - bash

Question

Here's a snippet that I use to parse options in Bash:
#!/bin/bash
PROGNAME=${0##*/}
PROGVERSION=0.1
wW='-4.5.5-double'
reName=
usage()
{
cat << EO
Script purpose goes here.
EO
cat <<EO | column -s\& -t
-h, --help & show this output
-r, --rename & renames confout to Your gro
-v, --version & show version information
-w, --workWith & gromax exec suffix
EO
}
SHORTOPTS="hvw:r"
LONGOPTS="help,version,workWith:rename"
ARGS=$(getopt -s bash --options $SHORTOPTS --longoptions $LONGOPTS --name $PROGNAME -- "$#")
eval set -- "$ARGS"
while true; do
case $1 in
-h|--help)
usage; exit 0;;
-v|--version)
echo "$PROGVERSION"; exit 0;;
-w|--workWith)
wW=$2; shift;;
-r|--rename)
reName="true"; shift;;
--)
shift; break;;
*)
shift; break;;
esac
shift
done
# ====================
## finally the script:
echo "rename:" $reName
echo ' wW:' $wW
This snippet parses option (-w) only if it is before the trigger (-r):
~/wrk/mlDn/vas/res/bbst: test.bash -w 'dfdff' -r
rename: true
wW: dfdff
~/wrk/mlDn/vas/res/bbst: test.bash -r -w 'dfdff'
rename: true
wW: -4.5.5-double
How can this be fixed? What is wrong with my snippet?

Answer 1

Remove the call to shift from the -r processing. It's removing the -w from ARGS.

Answer 2

The problem is that you are shifting twice: once when you see the -r option and then again when you exit the case statement. This causes you to jump over the -w option. Therefore, you should only shift once. Remove the shift from outside the case statement and do all the shifting inside it.
Change your code to:
while true; do
case $1 in
-h|--help)
usage; exit 0;;
-v|--version)
echo "$PROGVERSION"; exit 0;;
-w|--workWith)
wW=$2; shift 2;; # shift twice here
-r|--rename)
reName="true"; shift;;
--)
shift; break;;
*)
shift; break;;
esac
#shift # this shift is not required
done
echo "rename:" $reName
echo ' wW:' $wW