Best way to parse command line args in Bash?

Question

After several days of research, I still can\'t figure out the best method for parsing cmdline args in a .sh script. According to my references the getopts cmd is the way to

Answer 1

I find the use of getopt to be the easiest. It provides correct handling of arguments which is tricky otherwise. For example, getopt will know how to handle arguments to a long option specified on the command line as --arg=option or --arg option.

What is useful in parsing any input passed to a shell script is the use of the "$@" variables. See the bash man page for how this differs from $@. It ensures that you can process arguments that include spaces.

Here's an example of how I might write s script to parse some simple command line arguments:

#!/bin/bash

args=$(getopt -l "searchpath:" -o "s:h" -- "$@")

eval set -- "$args"

while [ $# -ge 1 ]; do
        case "$1" in
                --)
                    # No more options left.
                    shift
                    break
                   ;;
                -s|--searchpath)
                        searchpath="$2"
                        shift
                        ;;
                -h)
                        echo "Display some help"
                        exit 0
                        ;;
        esac

        shift
done

echo "searchpath: $searchpath"
echo "remaining args: $*"

And used like this to show that spaces and quotes are preserved:

user@machine:~/bin$ ./getopt_test --searchpath "File with spaces and \"quotes\"."
searchpath: File with spaces and "quotes".
remaining args: other args

Some basic information about the use of getopt can be found here