title: from the shellscript one option has been chosen OR none of them
body: <p>am not expert on bash script and am looking help on. I would like to find a way at the end from the while loop condition that the script can be executed unless one option (<strong>[-1|-2|-3]</strong>) has been chosen OR none of them. </p>
  
  <p>What the best way to do? I have absolutely no idea how to.
  Many Thanks.</p>
  
  <pre><code>#!/bin/bash
  
  echo "  $0 [-e &lt;option_e&gt;] [-f &lt;option_f&gt;]  [-1|-2|-3] [user@]fqdn"
  
  OPTION1=""
  OPTION2=""
  OPTION3=""
  
  while (( "$#" )); do
      if [ "$1" == "-1" ]; then
          OPTION1=1
      elif [ "$1" == "-2" ]; then
          OPTION2=1
      elif [ "$1" == "-3" ]; then
          OPTION3=1
      fi
  
     shift
  done
  </code></pre>
  
  <p>I forgot to mention. this is a part of the script. further, there are several conditions that can conflict if there more then 2 options chosen.</p>
  
  <p>EDIT: the valid selections either exactly one option or zero, but not two or more!</p>