(Bash) echo not output stored variables

Question

I'm new with bash and doing my first steps in it I want to learn How to store argument localy in script for later use and wrote this script

#!/bin/bash

while getopts ":d" opt;do
     case "${opt}" in
         d)
           d =$OPTARG
           echo "-d was triggered ! storing ${d}" >&2
           ;;
     esac
done

this is my output :

$ -d was triggered! storing

what is my problem? : I don't store the data correctly or I don't print it correctly or both :)

chepner · Accepted Answer · 2014-10-22 15:48:13Z

2

Keep : after option name in getopts command :

#!/bin/bash

while getopts "d:" opt; do
     case "${opt}" in
         d)
           d="$OPTARG"
           echo "-d was triggered ! storing ${d}" >&2
           ;;
     esac
done

You also need to remove the spaces around = and better to use quotes in assignment.

Now when you run:

./script.sh -d 1234
-d was triggered ! storing 1234

edited Oct 22, 2014 at 15:48

chepner

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answered Oct 19, 2014 at 10:47

anubhava

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1 Comment

LordTitiKaka Over a year ago

this is exactly what I was missing !!

Cyrus · Accepted Answer · 2014-10-19 10:45:12Z

2

Remove whitespace before =$OPTARG. Take a look at http://www.shellcheck.net/

answered Oct 19, 2014 at 10:45

Cyrus

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1 Comment

LordTitiKaka Over a year ago

thanks ! but @anubhava's answer is closer to solve my problem

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