Bash pass variable name as argument

Question

I have a simple script (let's call it script.sh):

#!/bin/bash

source variables
echo $1

What I want to achieve is scripts output:

255.255.255.255

When executed like:

./script.sh VARIABLE

where VARIABLE is defined in variables file:

VARIABLE=255.255.255.255

Is it possible?

thanasisp · Accepted Answer · 2017-12-12 09:19:12Z

25

example:

#!/bin/bash

variable=255.255.255.255
param=$1
echo ${!param}

then execute:

bash script.sh variable
255.255.255.255

The exclamation mark makes param to get the value of the variable with that name. See about parameter indirection.

answered Dec 12, 2017 at 9:19

thanasisp

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2 Comments

could we use echo ${!$1}? nop

@jon_eldiablo: echo ${!1} shoud do fine.

RavinderSingh13 · Accepted Answer · 2017-12-12 09:10:02Z

4

Could you please try following it may help you in same.

./script.sh "$VARIABLE"

answered Dec 12, 2017 at 9:10

RavinderSingh13

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Niquit · Accepted Answer · 2017-12-12 09:13:15Z

0

You have to add $ before variable name. For example: ./script.sh $VARIABLE

Is good to put argument in double quotes like: ./script.sh „$VARIABLE”

Writing from mobile phone, sorry for no using syntax.

answered Dec 12, 2017 at 9:13

Niquit

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