Bash : Parse options after arguments with getopts - bash

Question

In a script which request some arguments (arg) and options (-a), I would like to let the script user the possibility to place the options where he wants in the command line.
Here is my code :
while getopts "a" opt; do
echo "$opt"
done
shift $((OPTIND-1))
echo "all_end : $*"
With this order, I have the expected behaviour :
./test.sh -a arg
a
all_end : arg
I would like to get the same result with this order :
./test.sh arg -a
all_end : arg -a

Answer 1

The getopt command (part of the util-linux package and different from getopts) will do what you want. The bash faq has some opinions about using that, but honestly these days most systems will have the modern version of getopt.
Consider the following example:
#!/bin/sh
options=$(getopt -o o: --long option: -- "$#")
eval set -- "$options"
while :; do
case "$1" in
-o|--option)
shift
OPTION=$1
;;
--)
shift
break
;;
esac
shift
done
echo "got option: $OPTION"
echo "remaining args are: $#"
We can call this like this:
$ ./options.sh -o foo arg1 arg2
got option: foo
remaining args are: arg1 arg2
Or like this:
$ ./options.sh arg1 arg2 -o foo
got option: foo
remaining args are: arg1 arg2

Answer 2

You can still go with argument parsing by looking at each of them
#!/bin/bash
for i in "$#"
do
case $i in
-a)
ARG1="set"
shift
;;
*)
# the rest (not -a)
ARGS="${ARGS} $i"
;;
esac
done
if [ -z "$ARG1" ]; then
echo "You haven't passed -a"
else
echo "You have passed -a"
fi
echo "You have also passed: ${ARGS}"
and then you will get:
> ./script.sh arg -a
You have passed -a
You have also passed: arg
> ./script.sh -a arg
You have passed -a
You have also passed: arg
> ./script.sh -a
You have passed -a
You have also passed:
> ./script.sh arg
You haven't passed -a
You have also passed: arg

Answer 3

Consider this approach
#!/bin/bash
opt(){
case $1 in
-o|--option) option="$2";;
-h|--help ) echo "$help"; exit 0;;
*) echo "unknown option: $1"; exit 1;;
esac
}
while [[ $# ]]; do
case $1 in
arg1) var1=$1 ;;
arg2) var2=$1 ;;
-*) opt "$#"; shift;;
*) echo "unknown option: $1"; exit 1;;
esac
shift
done
echo args: $var1 $var2
echo opts: $option

Answer 4

In the many years of bash programming, I've found it useful to get rid of the bone in my head that says functions should look like f(x,y), especially since bash requires clumsy/inefficient code to handle command line arguments.
Option arguments often have default values, and can be more readily passed as environmental variables whose scope is limited to the called script, and save arguments for what must be provided.
Applying this to your example, The script would look like:
OPTION=${OPTION:="fee"}
echo "option: $OPTION"
echo "remaining args are: $*"
And would be called with:
OPTION="foo" ./options.sh arg1 arg2