BASH: getopts with default parameters value

Question

I\'ve got another bash-script problem I simply couldn\'t solve. It\'s my simplified script showing the problem:

while getopts \"r:\" opt; do
case $opt in

  r)
          


        

Answer 1

This works for me:

#!/bin/bash

while getopts "r" opt; do
case $opt in

  r)
    fold=/dev
    dir=${2:-$fold}

     echo "asdasd"
  ;;
esac
done

Remove the colon (:) in the getopts argument. This caused getopt to expect an argument. (see here for more information about getopt)

Answer 2

There may be other parameters before, don't hard-code the parameter number ($2).

The getopts help says

When an option requires an argument, getopts places that argument into the shell variable OPTARG.
...
[In silent error reporting mode,] if a required argument is not found, getopts places a ':' into NAME and sets OPTARG to the option character found.

So you want:

dir=/dev                            # the default value
while getopts ":r:" opt; do         # note the leading colon
    case $opt in
        r) dir=${OPTARG} ;;
        :) if [[ $OPTARG == "r" ]]; then
               # -r with required argument missing.
               # we already have a default "dir" value, so ignore this error
               :
           fi
           ;;
    esac
done
shift $((OPTIND-1))

a=$(find "$dir" -type b | wc -l)
echo "$a"