Shell getopts

getopts_example.sh

#!/bin/sh

test=false
host=localhost
port=6600

while getopts ":th:p:" opt; do
    case $opt in
        t)
            test=true
            ;;
        h)
            host="$OPTARG"
            ;;
        p)
            port="$OPTARG"
            ;;
        \?)
            echo "usage: $0 [-t] [-h host] [-p port] arg1" && exit 1
            ;;
    esac
done

shift $((OPTIND-1))

echo "test: $test"
echo "host: $host"
echo "port: $port"
echo "arg1: $1"

What does shift $((OPTIND-1)) do?

A shell script doesn't differentiate between options and arguments by default, so all the options will be in the argument list as well. This is impractical when you want to use the arguments. An example:

script_without_shift_optind.sh -t -p 6000 arg1 arg2

When running this the argument variables will get these values

$0 script_without_shift_optind.sh
$1 -t
$2 -p
$3 6000
$4 arg1
$5 arg2

Since getopts has already processed the options, we are not interested in those in the argument list anymore. Therefore, we shift the argument list by the same number of places that getopts process (which it stores in the $OPTIND variable) to remove them. After that we end up with:

$0 script_with_shift_optind.sh
$1 arg1
$2 arg2