How do you read multiple arguments with getopts? The following is a test case:

$ cat test.sh
#!/bin/bash
ECHO=/bin/echo
while getopts "hmq?" opt; do
case "$opt" in
"m") MYWAY=1;
$ECHO "MYWAY"
shift;
;;
"q") QUIET=1;
$ECHO "QUIET"
shift;
;;
"h" | "?")
$ECHO "Usage: $0 [-m| -q]"
exit 0
;;
esac
done
shift $(($OPTIND - 1))
$ ./test.sh -m
MYWAY
$ ./test.sh -q
QUIET
$ ./test.sh -q -m <-- Shouldn't it read -m too?
QUIET
$ ./test.sh -t <-- Why is this caught
./test.sh: illegal option -- t
Usage: ./test.sh [-m| -q]
$ ./test.sh - <-- And not this?
$

I was getting a fair bit confused with getopt and getopts earlier in the week, I found this article really useful - http://aplawrence.com/Unix/getopts.html and ended up migrating from getopt to getopts with the help of the really simple example he gives.

For your specific questions, firstly I think as your shifting AND using the options as a while input you're jumping over it. secondly there's nothing wrong with no options by default, it's merely an empty set of data to work on.

You don't need the "shift"s in the getopts loop; getopts does it all for you.

When posting code, it's easier to read if you put it in code tags.

Yep, thanks I shifted from a case to getopts as I now have more than a few
switches.

I was wondering where my formatting went!