Passing Curl result to a variable in a Bash script

Entropy1024 · 10-29-2015, 03:21 PM

I'm attempting to get my external IP as a variable (myextip) using curl in a bash script called extip.sh, however I can't quite work out the syntax. I have read a few examples and my best guess to make it work is below. But when I run it it returns:
extip.sh: 5: pcstatus.sh: -s: not found

Code:

#!/bin/bash
myextip= $(curl -s checkip.dyndns.org | sed -e 's/.*Current IP Address: //' -e 's/<.*$//')

Can you please help me find the correct syntax?
Many thanks
Tim

HMW · 10-29-2015, 03:44 PM

I don't have curl installed, but here's how I would do it with wget:

Code:

MY_EXT_IP=$(wget -qO - http://icanhazip.com/)

I see that there is a space in your assignment:

Code:

myextip= $...

This will always fail. You cannot use spaces when assigning a variable in Bash. It has to be like this:

Code:

myextip=$...

Best regards,
HMW

Entropy1024 · 10-29-2015, 03:50 PM

Quote:

Originally Posted by HMW

I don't have curl installed, but here's how I would do it with wget:

Code:

MY_EXT_IP=$(wget -qO - http://icanhazip.com/)

I see that there is a space in your assignment:

Code:

myextip= $...

This will always fail. You cannot use spaces when assigning a variable in Bash. It has to be like this:

Code:

myextip=$...

Best regards,
HMW

Thank you very much.
Tim