how to get multiple arguments from option in bash script

Question

#parse options
while getopts ":d:b:n:" opt; do
  case $opt in
    d)
          DIRS+=("$OPTARG")
          echo $DIRS
          ;;
    b)
          PATHBACKUP=$OPTARG
          echo $PATHBACKUP
          ;;
    n)
          FNAME=$OPTARG
          echo $FNAME
          ;;
    :)
          echo "Option -$OPTARG requires an argument." >&2
          error
          exit 1
          ;;
  esac
done
shift $(( OPTIND - 1 ))

This is my code I am trying to store every argument after -d to $DIRS However, when I echo $DIRS I only get the first argument

Example:

/.script -d /dev /home/work -b /backup
echo $DIRS
echo $PATHBACKUP
> /dev
> /backup

@anubhava i still only get the first argument wether I do "${DIRS[@]}" or echo ${DIRS[*]} — T D
– T D, Commented Nov 4, 2022 at 5:13
@anubhava: Not a good idea. You would have to apply word splitting to the argument afterwards, which breaks if you want to pass directory names which contain a space. — user1934428
– user1934428, Commented Nov 4, 2022 at 7:29

Diego Torres Milano · Accepted Answer · 2022-11-04 05:06:19Z

1

Use multiple -ds

./script -d /dev -d /home/work -b /backup

otherwise, the first non-option (i.e. /home/work) would stop getopts option processing and -b won't be considered.

The other alternative would be to use some delimiter like , and parse it yourself

./script -d /dev,/home/work -b /backup

answered Nov 4, 2022 at 5:06

Diego Torres Milano

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1 Comment

T D Over a year ago

As per the assignment i am not allowed to alter his usage of -d /dev /home. Apparently I somehow loop through the $@ and add it to the DIRS but I cannot find the exact solution as when I do this, it adds every argument of the command line

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