Demonstrates how to parse arguments in bash

bash-argument-parsing.sh

#!/bin/bash
set -euo pipefail
IFS=$'\n\t'

# -e: immediately exit if any command has a non-zero exit status
# -o: prevents errors in a pipeline from being masked
# IFS new value is less likely to cause confusing bugs when looping arrays or arguments (e.g. $@)

function usage()
{
    echo "Usage: $0 args ..."
    echo "  -w, --workload              <string> [required]"
    echo "  -l, --location              <string> [default='australiaeast']"
    echo "  -h, --help                  <flag>   [optional]"
    echo "  e.g.: $ $0 -w workload"
    exit 1
}

# set help flag
declare help=0

# declare required params
declare workload=""             # "someValue"

# decalre optional params
declare location="australiaeast"

# parse arguments
# https://gist.github.com/hfossli/4368aa5a577742c3c9f9266ed214aa58
# https://stackoverflow.com/questions/192249/how-do-i-parse-command-line-arguments-in-bash
# best answer by Inanc Gumus edited by Michael
while [[ "$#" > 0 ]]; do case $1 in
    -w|--workload) workload="$2"; shift;;
    -l|--location) location="$2"; shift;;
    -h|--help) help=1;;
    *) echo "Unknown parameter passed: $1\n"; usage;;
esac; shift; done

# check for -h, --help flag and exit if present
if [ "$help" == 1 ]; then usage; fi;

# check required arguments
if [ "$workload" == "" ]; then
    echo "-w, --workload is not set"
    echo ""
    usage
fi;