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Bash: get directory argument

I am writing a bash script that uses the call syntax

script [options] [dir]

To get a set of parameters and parse them, I use getopts. But how can I get the dir argument? In general, if I receive the last argument as $ {@: $ {# @}}, it does not have to be a director, it can still be an option or a value.

The code I'm using for getopts:

DIR="."
RECURSIVE=
FILTER=
while getopts "hnf:" OPTION
do
case $OPTION in
    h)
        usage
        exit 1
        ;;
    n)
        RECURSIVE="-maxdepth 1"
        ;;
    f)
        FILTER=$OPTARG
        ;;
    \?)
         exit 1
         ;;
    :)
         exit 1
         ;;
esac
done

You can help?

+3


source to share


2 answers


OPTIND

saves the position of the processed parameter. After the loop, do a:

shift $((OPTIND-1))


Now the directories are in $@

, the first directory is in $1

.

+2


source 


DIR=
RECURSIVE=
FILTER=
while getopts ':hnf:' OPTION ;do
  case $OPTION in
    h)  usage; exit 1 ;;
    n)  RECURSIVE="-maxdepth 1" ;;
    f)  FILTER=$OPTARG ;;
    *)  echo "ERROR: invalid opion: -$OPTARG" 1>&2; exit 1 ;;
  esac
done
# remove the options from the positional parameters
shift $((OPTIND-1))
DIR="$1"
echo "FILTER=$FILTER"
echo "DIR=$DIR"

Command line example

script -n -f 'my.*filter' 'my/directory'


Output example

FILTER=my.*filter
DIR=my/directory
0


source






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