CODEWITHSUNDEEP
shellparametersparameter-passingwget
Pedro Santangelo
Pedro Santangelo

Reputation: 153

Shell parameter not being received

I am trying to create a shell script that updates my public ip to the DynU service.

this command works fine in the terminal

wget -O - v4.ident.me 2>/dev/null && echo

The issue comes when I try to integrate the ip obtained with the wget into another wget. I'm trying this script:

#!/bin/bash

myip=wget -O - v4.ident.me 2>/dev/null && echo
echo $myip
echo $0
wget "http://api.dynu.com/nic/update?myip=$myip&username=my-username-comes-here
&password=my-pwd-comes-here"

Don't get confused by the "echo $0" it's there just to test the execution, I will remove it when the script works.

Could you see what I am doing wrong? I appreciate your help.

Upvotes: -1

Views: 21

Answers (1)

Tim Roberts
Tim Roberts

Reputation: 54733

What you have just sets the variable to that string. If you want to capture the result of executing the command, when you do, then you need to tell the shell to execute the command with $(...):

myip=$(wget -O - v4.ident.me 2>/dev/null)

Upvotes: 1

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