How to print first line of the response and status code in curl-CodePudding

curl -o /dev/null -s -w "HTTPCode=%{http_code}_TotalTime%{time_total}s\n" http://test.com

just gives me the status code

I also want the body of the response, How can i do that in the command?

CodePudding user response：

Don't send it to /dev/null? -o is throwing it away.

curl -sw "HTTPCode=%{http_code}_TotalTime%{time_total}s\n" http://test.com

If you ONLY want the first line, as your title suggests, filter it.

curl -sw "HTTPCode=%{http_code}_TotalTime%{time_total}s\n" http://test.com | sed -n '1p; $p;'

This tells sed to print the first and last lines, because you asked for the first one, and -w prints after completetion. My test:

$: curl -s -w "HTTPCode=%{http_code}_TotalTime%{time_total}s\n" google.com | sed -n '1p; $p;'
<HTML><HEAD><meta http-equiv="content-type" content="text/html;charset=utf-8">
HTTPCode=301_TotalTime0.270267s

If you specifically mean the first line of the body from the response, now you need to define "first line" a little, and you should really get an HTML-aware parser. I could probably do it in sed, but it's really kind of a bad idea in most cases.

If that's really what you need, let me know, supply some additional details, and we'll work on it.

CodePudding user response：

You can use this curl awk solution:

curl -sw "HTTPCode=%{http_code}_TotalTime%{time_total}s\n" 'http://test.com' |
awk '{p=p $0 ORS} END{print; sub(/\n[^\n] \n$/, "", p); print p}'