Using grep to get the next WORD after a match in each line

Question

I want to get the "GET" queries from my server logs.

For example, this is the server log

1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:32:27] code 404, message File not fo$
1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:32:27] "GET /hello HTTP/1.1" 404 -   
1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:41:57] code 404, message File not fo$
1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:41:57] "GET /ss HTTP/1.1" 404 -

When I try with simple grep or awk,

Adi:~ adi$ awk '/GET/, /HTTP/' serverlogs.txt

it gives out

1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:32:27] "GET /hello HTTP/1.1" 404 -
1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:41:57] "GET /ss HTTP/1.1" 404 -

I just want to display : hello and ss

Is there any way this could be done?

Answer 1

Assuming you have gnu grep, you can use perl-style regex to do a positive lookbehind:

grep -oP '(?<=GET\s/)\w+' file

If you don't have gnu grep, then I'd advise just using sed:

sed -n '/^.*GET[[:space:]]\{1,\}\/\([-_[:alnum:]]\{1,\}\).*$/s//\1/p' file

If you happen to have gnu sed, that can be greatly simplified:

sed -n '/^.*GET\s\+\/\(\w\+\).*$/s//\1/p' file

The bottom line here is, you certainly don't need pipes to accomplish this. grep or sed alone will suffice.