Getting correct output in a shell script from the curl command

Question

I am getting my IP address using a curl command, and I want to save it as a shell variable.

I use the following command to get the ip address

curl ipinfo.io/ip

And I assign the variable thusly:

IPADDR=`curl ipinfo.io/ip`

but when I echo this, I get the following:

% Total    % Received % Xferd  Average Speed   Time    Time     Time  Current
                               Dload  Upload   Total   Spent    Left  Speed
100    14  100    14    0     0     41      0 --:--:-- --:--:-- --:--:--   164
24.18.247.198

All I want is 24.18.247.198. Any thoughts?

leu · Accepted Answer · 2014-01-13 21:14:35Z

3

curl provides an attribute to operate "muted": -s

Hence, you can set your variable in this way:

IPADDR=$(curl -s ipinfo.io/ip)

answered Jan 13, 2014 at 21:14

leu

2,1232 gold badges12 silver badges27 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

janos · Accepted Answer · 2014-01-13 21:00:19Z

1

Do like this:

IPADDR=$(curl ipinfo.io/ip 2>/dev/null)

That is, the "% Total", "% Received" and others are printed on stderr. By redirecting stderr to /dev/null you can get rid of that noise.

Always use $(...) instead of `...` when possible.

answered Jan 13, 2014 at 21:00

janos

126k31 gold badges242 silver badges253 bronze badges

Collectives™ on Stack Overflow

Getting correct output in a shell script from the curl command

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related