BASH script and parameters with/out options

Question

I want to create script that will have two types of arguments/options.

For example:

./script.sh --date 15.05.05 --host localhost

but I also want to be able to run it with parameters without value:

./script --date 15.0.50.5 --host localhost --logs --config

For now I have something like:

while [[ $# -gt 0 ]]; do
  case $1 in
    --date|-d ) DATE="$2" ;;
    --host|-h ) HOST="$2" ;;
    --logs ) LOGS=true ;;
    --config ) CONFIG=true ;;
#    --all ) LOGS=true ; PROCS=true ; CONFIG=true ;;
    * ) usage; exit 1 ;;
  esac
  shift 2
done

However, when I'm using it like that, I have to put a value after --logs and --config to prevent the shift from taking the next valid parameter, like this:

./script.sh --date 15.05.05 --logs 1 --config 1

Is there any other way?

Answer 1

What about this simple solution?

while [[ $# -gt 0 ]]; do
  case $1 in
    --date|-d ) DATE="$2" ; shift 2 ;;
    --host|-h ) HOST="$2" ; shift 2 ;;
    --logs ) LOGS=true ; shift 1 ;;
    --config ) CONFIG=true ; shift 1 ;;
    * ) usage; exit 1 ;;
  esac
done

Or you can use getopts (which only supports short parameters though), approximately like this:

while getopts ":d:h:lc" OPT; do
  case $opt in
    -d ) DATE="$OPTARG" ;;
    -h ) HOST="$OPTARG" ;;
    -l ) LOGS=true ;;
    -c ) CONFIG=true ;;
    * ) usage; exit 1 ;;
  esac
done