shell - BASH script : usage displays even if you provide a switch without argument -

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i've tried many ways , i'm stumped. when try sudo admpanel -a without argument script displays message usage() method instead of error msg i've built it. on other hand if sudo admpanel -a firstname displays me error msg portion. if enter sudo admpanel -a firstname lastname functions way supposed to. below script :

#!/bin/bash  # program name : admpanel.sh  progname=${0##*/} version="1.0" script_shell=${shell}  usage() {   echo "usage: ${progname} [-h|--help ] [-a|--add] [-r|--remove] [-d|--display] [-c|--check] [argument_1] [argument_2]" }  help_message() {     cat <<- _eof_     ${progname} ${version}     add/remove/display user(s) on admin panel. please note password hardcoded.     please consult system administrator if don't know it.      $(usage)      options:      -h, --help      displays help.     -a, --add       add user     -r, --remove    remove user     -d, --display   display users     -c, --check     check specific user      example:      ${progname} -a firstname lastname     ${progname} --add firstname lastname      ${progname} -r firstname lastname     ${progname} --remove firstname lastname      ${progname} -d     ${progname} -display      ${progname} -c firstname lastname     ${progname} --check firstname lastname  _eof_ }  verify() {     echo "are sure want this?"     read answer }  add_employee() {     echo adding user }  remove_employee() {     echo removing user }  display_users() {     echo displaying users }  check_user() {     echo checking users }   if [ "$#" -le 1 ];     usage     exit 1 elif [[ $user != "root" ]];     echo "this script must run root!"     exit 1 elif [[ ($1 == "-h" ) || ($1 == "-help") || ($1 == "--help") ]];     help_message     exit 0 elif [[ ($1 == "-a") || ($1 == "-add") || ($1 == "--add") ]];     if [[ -z "$3" ]];         echo "error: first name , last name wasn't provided"         echo "usage: ${progname} -a firstname lastname"          exit 1       elif [[ (-z "$2") || (-z "$3") ]];         echo "error: either first name or last name wasn't provided or both"         echo "usage: ${progname} -a firstname lastname"         exit 1     else         read -p "are sure want continue? <y/n> " prompt         if [[ $prompt == "y" || $prompt == "y" || $prompt == "yes" || $prompt == "yes" ]];             add_employee             exit 0         else             exit 0         fi     fi elif [[ ($1 == "-r") || ($1 == "-remove") || ($1 == "--remove") ]];     remove_employee     exit 0 elif [[ ($1 == "-d") || ($1 == "-display") || ($1 == "--display") ]];     display_users     exit 0 elif [[ ($1 == "-c") || ($1 == "-check") || ($1 == "--check") ]];     check_user     exit 0 else     echo went wrong try again...     exit 1 fi

please advise. -thanks in advance.

it's because of line:

if [ "$#" -le 1 ];

it says "less or equal to 1". therefore, message triggers when give 1 argument, namely -a.

you should instead change "strictly less than":

if [ "$#" -lt 1 ];

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