working with options in bash code

Question

Possible Duplicate:
Using getopts in bash shell script to get long and short command line options

I'm trying to figure out how to make use of flag like -e/--email -h/--help for example.

UPDATE: Current example:

while getopts ":ec:h" OptionArgument; do
case $OptionArgument in

        e ) S1=${OPTARG};;
        c ) S2=${OPTARG};;
        h ) usage;;
        \?) usage;;
        * ) usage;;
esac
done

However, if I leave blank it does nothing. If I add an -h it still runs and will not show usage.

UPDATE2

while [ $1 ]; do
        case $1 in
                '-h' | '--help' | '?' )
                        usage
                        exit
                        ;;
                '--conf' | '-c' )

                        ;;
                '--email' | '-e' )
                        EMAIL=$1
                        ;;
                * )
                        usage
                        exit
                        ;;
        esac
        shift
done

-h/--help works but everything else fails and displays usage.

Answer 1

If you want long GNU like options and short ones, see this script to source in your script http://stchaz.free.fr/getopts_long.sh and an example :

Example :

#!/bin/bash
# ( long_option ) 0 or no_argument,
# 1 or required_argument, 2 or optional_argument).

Help() {
cat<<HELP
Usage :
    $0 [ --install-modules <liste des modules> ] [-h|--help]
HELP
    exit 0
}

. /PATH/TO/getopts_long.sh

OPTLIND=1
while getopts_long :h opt \
    install-modules 1 \
    help 0 "" "$@"
do
    case "$opt" in
        install-modules)
            echo $OPTLARG
        ;;
        h|help)
            Help; exit 0
        ;;
        :)
           printf >&2 '%s: %s\n' "${0##*/}" "$OPTLERR"
           Help
           exit 1
        ;;
    esac
done
shift "$(($OPTLIND - 1))"