How to write getopt in bash script [duplicate] - bash

Question

This question already has an answer here:
How to support both short and long options at the same time in bash? [duplicate]
(1 answer)
Closed 7 years ago.
I have a function with a few parameters. For example:
makeUser{
login
email
password}
I want to make flags like -l|--login, -e|--email and -p|--password but I don't know how.
The sample for it should look like the code below:
./script.sh --mode makeUser --login test --email test#test.com -p testxx
How can I achieve that result? I know only how to make it with getopts (short flags).
Should it look like code below?
while true ; do
case "$1" in
-m|--mode)
makeUser)
case "$2" in
-l|--login)
makeUser "$OPTARG"
case "$3" in
-e|--email)
makeUser "$OPTARG"
case "$4" in
-p|--password)
makeUser "$OPTARG"
exit $?
esac ;;
exit $?
esac ;;
exit $?
esac ;;
makeProject)...
makeSite)...
exit $?
esac ;;
done

Answer 1

Using while and shift makes a clean solution for getopt-alike behaviour in bash:
while [ $# -gt 0 ]; do
case "$1" in
-h|"-?"|--help)
shift
echo "usage: $0 [-v] [--mode MODE] [-l login] [-e email] [...]"
exit 0
;;
--mode)
MODE=$2
shift; shift;
;;
-l|--login)
LOGIN=$2
shift; shift;
;;
-e|--email)
EMAIL=$2
shift; shift;
;;
-v|--verbose)
VERBOSE=1
shift;
;;
*)
echo "Error: unknown option '$1'"
exit 1
esac
done
# example function makeUser
makeUser()
{
login=$1
email=$2
echo "function makeUser with login=${login} and email=${email}"
}
if [ "$MODE" == "makeUser" ]; then
makeUser $LOGIN $EMAIL # ... and so on
fi