title: Shell Script Argument without value
body: <p>I'm creating a very simple sh file to do something, but I want to pass an argument without a value, for example:</p>
  
  <pre><code>./fuim -l
  </code></pre>
  
  <p>But I receive the following message:</p>
  
  <pre><code>./fuim: option requires an argument -- l
  </code></pre>
  
  <p>If I pass a random value, like <code>./fuim -l 1</code>, it works perfectly. How can I do that?</p>
  
  <p>Here's what I have so far:</p>
  
  <pre><code>while getopts e:f:l:h OPT
  do
      case "$OPT" in
          h) print_help ;;
          e) EXT=$OPTARG ;;
          f) PROJECT_FOLDER=$OPTARG ;;
          l) LIST_FILES=1 ;;
          ?) print_help ;;
      esac
  done
  
  shift $((OPTIND-1))
  
  if [ -z "$EXT" ] || [ -z "$PROJECT_FOLDER" ]; then
      print_help
  fi
  </code></pre>