How to print help if no argument passed in bash [duplicate]

Question

I have created a script in order to use arguments while running that script. How to check if there were no arguments provided ? It must shows print help if no argument passes.

while test -n "$1"; do
         case "$1" in
            -help|-h)
            print_help
            exit $ST_UK
            ;;
        --version|-v)
            print_version $PROGNAME $VERSION
            exit $ST_UK
            ;;
        --activeusers|-a)
            opt_var=$2
            au
            shift;;
        --dailyusers|-d)
            opt_var1=$2
            dau
            shift;;
        *)
    echo "Unknown argument: $1"
        print_help
        exit $ST_UK
        ;;
    esac
    shift
done

Answer 1

You can do it the same way you would for any POSIX shell, by testing the $# (number of arguments) magic variable:

if [ "$#" -eq 0 ]
then
    usage >&2
    exit 1
fi