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Just as the title says.

My original plan was to just run

grep "GET /files/filename.lol" access.log | wc -l

The problem is there are multiple access per IP per timestamp. What I'd like to do is group the lines by IP (which is the first entry in every line) and count those groups.

Ideas? I'm relatively new to bash, but an experienced programmer.

Quote linuxdynasty:

grep "GET /files/filename.lol" access.log | sort | uniq -c
uniq -c will give you the count of each occurance... Hope this helps

What I ended up doing was running

grep "/files/filename.lol" access.log | awk -F' ' '{print $1}' > out.txt
which gave me a list of the IPs, and I planned on writing a quick python script to find the unique IPs. But after reading your posting I combined it with sort and uniq -c and wc -l, and that gave me exactly what I wanted.

Thank you very much friend, I can't express how grateful I am. You saved me a bit of time!

Thanks friend
[Modified by: Novi on May 29, 2008 04:58 AM]

awk has regex matching built-in natively, so grep "/files/filename.lol" access.log | awk -F' ' '{print $1}' > out.txt

might more efficiently be written as
awk -F' ' '/\/files\/filename\.lol/{print $1}' access.log > out.txt

Of course, it's also possible to implement the functionality of the `sort` and `uniq` programs in awk, as it's a complete programming language, but that would not actually benefit performance and/or complexity.
[Modified by: Johannes Truschnigg on June 15, 2008 09:17 PM]

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