How would I extract the user agent strings from a log file?

Question

Currently I am running a command like this, to get the most requested content:

grep "17\/Jul\/2011" other_vhosts_access.log | awk '{print $8}' | sort | uniq -c | sort -nr

I want to now see the user agent strings, but the problem is they include several spaces. Here is a typical log file line. The UA is the last section delimited by quotation marks:

example.com:80 [ip] - - [17/Jul/2011:23:59:59 +0100] "GET [url] HTTP/1.1" 200 6449 "[referer]" "Mozilla/5.0 (Windows NT 6.1) AppleWebKit/534.30 (KHTML, like Gecko) Chrome/12.0.742.122 Safari/534.30"

Is there a better tool than awk for this?

Michael Mrozek · Accepted Answer · 2011-07-18 14:25:48Z

24

If that format is consistent and the field is really wrapped in double quotes you can use either awk or cut with " as the field delimiter:

awk -F\" '{print $6}'

or:

cut -d\" -f 6

edited Jul 18, 2011 at 14:25

Michael Mrozek

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answered Jul 18, 2011 at 14:18

Caleb

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RedGrittyBrick · Accepted Answer · 2011-07-18 14:27:29Z

3

perl -ne 'if(/"([^"]+)"$/){$ua{$1}++;} END{for(keys %ua){print "$ua{$_} $_\n"}}' \
  access_log

edited Jul 18, 2011 at 14:27

answered Jul 18, 2011 at 14:21

RedGrittyBrick

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