Calling a different bash script from a running script based on passed arguments [duplicate]

Question

I am tasked with making a user menu that will have 2 different variations, one for the novice user and the others for an expert.

I got it calling all my novice scripts, but in order to call my expert scripts, my main script needs to detect if a user has passed any of the arguments like "help, man, status" etc. And if they do I need my main script to call expert scripts in that case.

I am very new to bash scripting so I hope I can get some guidance here.

#!/bin/bash
 echo -n "Name please? "
 read name
 echo "Menu for $name"
 echo "===================================================="
 echo -e "Enter 
 ( 1 ) File and Directory Managment Commands\nEnter
 ( 2 ) Text Processing Commands\nEnter
 ( 3 ) System Status Commands\nEnter
 ( 4 ) Exit"
  echo "Enter your choice :"
 read en
 case "$en" in
 1)sh ./fileNovice.sh;;
 2)sh ./textNovice.sh;;
 3)sh ./systemStatusNovice.sh;;
 4)echo "Good Bye..."
 sleep 0;;
 *)echo "Wrong Option selected" ;;
 esac

Answer 1

Do not reinvent the wheel and use Bash's select:

#!/usr/bin/env bash

declare -a level_names=([0]='Novice' [1]='Expert')
declare -i level=0

while :; do
  printf 'Current mode : %s\n' "${level_names[level]}"
  PS3='Enter your choice : '
  select _ in \
    'File and Directory Managment Commands' \
    'Text Processing Commands' \
    'System Status Commands' \
    "Switch to ${level_names[level ^ 1]} mode" \
    'Exit'; do
    case "$REPLY" in
      1) bash "./file${level_names[level]}.sh"; exit ;;
      2) bash "./text${level_names[level]}.sh"; exit ;;
      3) bash "./systemStatus${level_names[level]}.sh"; exit ;;
      4) level=$((level ^ 1)); break ;;
      5) exit ;;
      *) echo 'Wrong Option selected'; break ;;
    esac
  done
done