parse arguments after getopts - bash

Question

I want to call a bash script like this
$ ./scriptName -o -p -t something path/to/file
This is as far as I get
#!/bin/bash
o=false
p=false
while getopts ":opt:" options
do
case $options in
o ) opt1=true
;;
p ) opt2=true
;;
t ) opt3=$OPTARG
;;
esac
done
but how do I get the path/to/file?

Answer 1

You can do something like:
shift $(($OPTIND - 1))
first_arg=$1
second_arg=$2
after the loop has run.

Answer 2

To capture all the remaining parameters after the getopts processing, a good solution is to shift (remove) all the processed parameters (variable $OPTIND) and assign the remaining parameters ($#) to a specific variable.
Short answer:
shift $(($OPTIND - 1))
remaining_args="$#"
Long example:
#!/bin/bash
verbose=false
function usage () {
cat <<EOUSAGE
$(basename $0) hvr:e:
show usage
EOUSAGE
}
while getopts :hvr:e: opt
do
case $opt in
v)
verbose=true
;;
e)
option_e="$OPTARG"
;;
r)
option_r="$option_r $OPTARG"
;;
h)
usage
exit 1
;;
*)
echo "Invalid option: -$OPTARG" >&2
usage
exit 2
;;
esac
done
echo "Verbose is $verbose"
echo "option_e is \"$option_e\""
echo "option_r is \"$option_r\""
echo "\$# pre shift is \"$#\""
shift $((OPTIND - 1))
echo "\$# post shift is \"$#\""
This will output
$ ./test-getopts.sh -r foo1 -v -e bla -r foo2 remain1 remain2
Verbose is true
option_e is "bla"
option_r is " foo1 foo2"
$# pre shift is "-r foo1 -v -e bla -r foo2 remain1 remain2"
$# post shift is "remain1 remain2"