How to get arguments with flags in Bash

Question

I know that I can easily get positioned parameters like this in bash:

$0 or $1

I want to be able to use flag options like this to specify for what each parameter is used:

mysql -u user -h host

What is the best way to get -u param value and -h param value by flag instead of by position?

Answer 1

This is the idiom I usually use:

while test $# -gt 0; do
  case "$1" in
    -h|--help)
      echo "$package - attempt to capture frames"
      echo " "
      echo "$package [options] application [arguments]"
      echo " "
      echo "options:"
      echo "-h, --help                show brief help"
      echo "-a, --action=ACTION       specify an action to use"
      echo "-o, --output-dir=DIR      specify a directory to store output in"
      exit 0
      ;;
    -a)
      shift
      if test $# -gt 0; then
        export PROCESS=$1
      else
        echo "no process specified"
        exit 1
      fi
      shift
      ;;
    --action*)
      export PROCESS=`echo $1 | sed -e 's/^[^=]*=//g'`
      shift
      ;;
    -o)
      shift
      if test $# -gt 0; then
        export OUTPUT=$1
      else
        echo "no output dir specified"
        exit 1
      fi
      shift
      ;;
    --output-dir*)
      export OUTPUT=`echo $1 | sed -e 's/^[^=]*=//g'`
      shift
      ;;
    *)
      break
      ;;
  esac
done

Key points are:

• $# is the number of arguments
• while loop looks at all the arguments supplied, matching on their values inside a case statement
• shift takes the first one away. You can shift multiple times inside of a case statement to take multiple values.