How to use grep to output unique lines of code fro

Question

I have a large log file that contains lines such as:

82.117.22.206 - - [08/Mar/2013:20:36:42 +0000] "GET /key/0/www.mysite.org.uk/ HTTP/1.0" 200 0 "-" "-"

And i want to extract from each line that matches the above pattern only the ip 82.117.22.206 followed by a space and the text www.mysite.org.uk from it. The ip and text can differ. So given the above line the line in the output file would be:

82.117.22.206 www.mysite.org.uk

How can I use grep or other commands in bash to make the output unique so that the output file won't contain two identical lines? Can someone refer me to a good place to start learnning more about this kind of shell scripting?

Answer 1

With perl you can capture the parts

use strict;
use warnings;

if (m/^(\d+\.\d+\.\d+\.\d+)\s+-\s+-\s+\[.+?\]\s+\"GET\s+\/key\/0\/(.+?)\//) {
    print "$1 $2\n";
}

and call this as

perl -n script.pl logfile.txt | sort -u

This extracts the needed fields, sorts and eliminates duplicate lines.

Answer 2

if you figure out the regex to use, you could do something like:

echo "Hello World" | grep "Hell" | sed 's/\(Hell\).*\(World\)/\1 \2/'

only, you'd cat your log, instead of echoing a string.

Answer 3

grep -Po "^[\d.]*|[^/]*(?=/ HTTP)" file|sed 'N;s/\n/ /'