how to properly parse shell script flags and arguments using getopts

Question

I'm using this :

for example ./imgSorter.sh -d directory -f format

the scripts' content is :

#!/bin/bash
while getopts ":d:f:" opt; do
  case $opt in
    d)
      echo "-d was triggered with $OPTARG" >&2
      ;;
    f)
      echo "-f was triggered with $OPTARG" >&2
      ;;
    \?)
      echo "Invalid option: -$OPTARG" >&2
      exit 1
      ;;
    :)
      echo "Option -$OPTARG requires an argument." >&2
      exit 1
      ;;
  esac
done

use cases :

$ ./imgSorter.sh -d myDir -d was triggered with myDir OK

$ ./imgSorter.sh -d -f myFormat -d was triggered with -f NOK : how is it that a string beginning with - is not detected as a flag ?

Kusalananda · Accepted Answer · 2018-04-26 13:45:40Z

You have told getopts that the -d option should take an argument, and in the command line you use -d -f myformat which clearly (?) says "-f is the argument I'm giving to the -d option".

This is not an error in the code, but in the usage of the script on the command line.

Your code needs to verify that the option-arguments are correct and that all options are set in an appropriate way.

Possibly something like

while getopts "d:f:" opt; do
  case $opt in
    d) dir=$OPTARG      ;;
    f) format=$OPTARG   ;;
    *) echo 'error' >&2
       exit 1
  esac
done

# If -d is *required*
if [ ! -d "$dir" ]; then
    echo 'Option -d missing or designates non-directory' >&2
    exit 1
fi

# If -d is *optional*
if [ -n "$dir" ] && [ ! -d "$dir" ]; then
    echo 'Option -d designates non-directory' >&2
    exit 1
fi

If the -d option is optional, and if you want to use a default value for the variable dir in the code above, you would start by setting dir to that default value before the while loop.

A command line option can not both take and not take an argument.

ok thank you @Kusalananda, I wrongly assumed that getopts took every - for the beginning of a flag. — St3an
– St3an, Commented Apr 26, 2018 at 14:01

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