getopts when first option is a path in bash - bash

Question

I'm having an issue with getopts in a bash script. Basically my script would have to be called with something like:
./myScript /path/to/a/folder -a -b
What I have at the top of my code is this:
while getopts ":ab" opt; do
case $opt in
a)
variable=a
;;
b)
variable=b
;;
\?)
echo "invalid option -$OPTARG"
exit 0
esac
done
echo "$variable was chosen"
Now, this works as long as I call my script without /path/to/a/folder… How can I make it to work with it instead?
Thanks a lot

Answer 1

If you MUST put a path before the arguments, use a shift command to pop the first positional argument off, and leave the rest for getopts.
# Call as ./myScript /path/to/a/folder -a -b
path_argument="$1"
shift # Shifts away one argument by default
while getopts ":ab" opt; do
case $opt in
a)
variable=a
;;
b)
variable=b
;;
\?)
echo "invalid option -$OPTARG"
exit 0
esac
done
echo "$variable was chosen, path argument was $path_argument"
The more-standard answer, as Etan mentioned, is to put the non-option arguments AFTER the options. Prefer this style, as it makes your script more consistent with built-in POSIX option parsing.
# Call as ./myScript -a -b /path/to/a/folder
while getopts ":ab" opt; do
case $opt in
a)
variable=a
;;
b)
variable=b
;;
\?)
echo "invalid option -$OPTARG"
exit 0
esac
done
shift $((OPTIND - 1)) # shifts away every option argument,
# leaving your path as $1, and every
# positional argument as $#
path_argument="$1"
echo "$variable was chosen, path argument was $path_argument"