Diff for "BashFAQ/035"

Anchor(faq35)

How can I handle command-line arguments to my script easily?

Well, that depends a great deal on what you want to do with them. Here's a general template that might help for the simple cases:

    # Bash
    while [[ $1 == -* ]]; do
        case "$1" in
          -h|--help|-\?) show_help; exit 0;;
          -v) verbose=1; shift;;
          -f) output_file=$2; shift 2;;
          --) shift; break;;
        esac
    done

Now all of the remaining arguments are the filenames which followed the optional switches. You can process those with for i or "$@".

For more complex/generalized cases, or if you want things like "-xvf" to be handled as three separate flags, you can use getopts. (NEVER use getopt(1)!)

Here is a simplistic getopts example:

    # POSIX
    x=1         # Avoids an error if we get no options at all.
    while getopts "abcf:g:h:" opt; do
      case "$opt" in
        a) echo "You said a";;
        b) echo "You said b";;
        c) echo "You said c";;
        f) echo "You said f, with argument $OPTARG";;
        g) echo "You said g, with argument $OPTARG";;
        h) echo "You said h, with argument $OPTARG";;
      esac
      x=$OPTIND
    done
    shift $(($x-1))
    echo "Left overs: $@"

If your prefer to check options with if statements, then a function like this one may be useful:

# Bash
HaveOpt() {
  local needle=$1
  shift
  while [[ $1 == -* ]]; do
    case "$1" in
      --) return 1; # by convention, -- is end of options
      $needle) return 0;;
    esac
    shift
  done
  return 1
}

if HaveOpt --quick "$@"; then echo "Option quick is set"; fi

and it will work if script is run as:

• YES: ./script --quick
• YES: ./script -other --quick

but will stop on first argument with no "-" in front (or on --):

• NO: ./script -bar foo --quick
• NO: ./script -bar -- --quick

Of course, this approach (iterating over the argument list every time you want to check for one) is far less efficient than just iterating once and setting flag variables.