How to echo only 1st line from 'curl' command output?
I'm trying to get only first line of the curl command output. (Sorry if this is confusing)
Let's say, for an instance, i run simply:
# curl http://localhost
<!-- This is the hidden line i want to grab. -->
<!DOCTYPE html>
<html>
<head>
..
..
What to do if i want the very first line of the output here, which is:
<!-- This is the hidden line i want to grab. -->
I've tried things like this, but no luck yet:
# curl http://localhost | head -n 1
# curl http://localhost | sed -n '1!p'
.. etc. All gives me rubbish outputs, like this:
% Total % Received % Xferd Average Speed Time Time Time Current
Dload Upload Total Spent Left Speed
0 0 0 0 0 0 0 0 --:--:-- --:--:-- --:--:-- 0<!-- This is the hidden line i want to grab. -->
0 0 0 0 0 0 0 0 --:--:-- --:--:-- --:--:-- 0
curl: (23) Failed writing body (173 != 1763)
It's not the output as expected as mentioned above:
<!-- This is the hidden line i want to grab. -->
Any experts here please =(
Solution 1:
This so called rubbish output is basically a progress meter during operation of downloading the data. You can basically ignore that, since it is by default going into standard error stream which is ignored, so only relevant part is printed out to standard output.
Here is the test:
$ curl http://example.com/ | head -n1 > example.html
% Total % Received % Xferd Average Speed Time Time Time Current
Dload Upload Total Spent Left Speed
100 1270 100 1270 0 0 112k 0 --:--:-- --:--:-- --:--:-- 124k
(23) Failed writing body
$ cat example.html
<!doctype html>
If you still want to silence it, add -s parameter for quiet mode or redirect standard error stream into /dev/null, for example:
$ curl -s http://example.com/ 2> /dev/null | head -n1
<!doctype html>
Or using command substitution:
head -n1 <(curl -s http://example.com/ 2> /dev/null)