Bash Shell Script - Check for a flag and grab its value

Question

I am trying to make a shell script which is designed to be run like this:

script.sh -t application 

Firstly, in my script I want to check to see if the script has been run with the -t flag. For example if it has been run without the flag like this I want it to error:

script.sh 

Secondly, assuming there is a -t flag, I want to grab the value and store it in a variable that I can use in my script for example like this:

FLAG="application" 

So far the only progress I've been able to make on any of this is that $@ grabs all the command line arguments but I don't know how this relates to flags, or if this is even possible.

Answer 1

You should read this getopts tutorial.

Example with -a switch that requires an argument :

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

Like greybot said(getopt != getopts) :

The external command getopt(1) is never safe to use, unless you know it is GNU getopt, you call it in a GNU-specific way, and you ensure that GETOPT_COMPATIBLE is not in the environment. Use getopts (shell builtin) instead, or simply loop over the positional parameters.

Answer 2

Use $# to grab the number of arguments, if it is unequal to 2 there are not enough arguments provided:

if [ $# -ne 2 ]; then    usage; fi 

Next, check if $1 equals -t, otherwise an unknown flag was used:

if [ "$1" != "-t" ]; then   usage; fi 

Finally store $2 in FLAG:

FLAG=$2 

Note: usage() is some function showing the syntax. For example:

function usage {    cat << EOF Usage: script.sh -t <application>  Performs some activity EOF    exit 1 }