Bash keyword argument with multiple values

Question

I'm using this to get keyword arguments for my shell script

while [[ $# -gt 0 ]]; do
  key="$1"
  case $key in
    -u) 
      USERID="$2"
      shift # past argument
      shift # past value
      ;;
    -n)
      JOBNAME="$2"
      shift # past argument
      shift # past value
      ;;
    -j) 
      job="$2"
      shift # past argument
      shift # past value
      ;;
      *) echo "Improper usage"; exit 1;;
  esac
done

Now the job variable usually is multi-word. I can give the whole value for that argument inside a quote and it works perfectly. But is it possible to provide the value without any quote but still store inside that single variable. For example, this works

./myscript -j "my long string"

and it stores "my long string" to the variable job. But I want this to work too similarly,

./myscript -j my long string

Answer 1

You could pick the remaining parameters with "$*", but I really wouldn't recommend it. The standard expectation is that option flags should take no more than a single parameter value.

#!/bin/bash
#
progname="${0##*/}"
usage="Usage: $progname -u <userid> -n <jobname> -j <job>"

while getopts ':u:n:j:' OPT
do
    case "$OPT" in
        u)      userid="$OPTARG" ;;
        n)      jobname="$OPTARG" ;;
        j)      job="$OPTARG" ;;
        ?)      echo "$usage" >&2; exit 1 ;;
    esac
done
shift $(($OPTIND - 1))

# Append any remaining arguments to the job
[[ $# -gt 0 ]] && job="${job:+$job }$*"

printf "userid=%s, jobname=%s, job=%s.\n" "$userid" "$jobname" "$job"
exit 0

To stay within standard expectations, do not include the line that appends remaining arguments to the job.