Hi Group,

See the following log file of apache (HTTPD).
I want a script which can calculate the busiest date. This file is long (very long) with the lines below:



60.231.97.192 - - [03/Jan/2005:11:43:27 +1100] "GET"
60.231.97.192 - - [03/Jan/2005:11:43:27 +1100] "GET"
60.231.97.192 - - [03/Jan/2005:11:43:27 +1100] "GET"
60.231.97.196 - - [04/Jan/2005:11:43:27 +1100] "GET"
60.231.97.192 - - [05/Feb/2005:11:43:27 +1100] "GET"
60.231.97.195 - - [05/Feb/2005:11:43:27 +1100] "GET"
60.231.97.194 - - [04/Mar/2005:11:43:27 +1100] "GET"
60.231.97.193 - - [04/Mar/2005:11:43:27 +1100] "GET"
60.231.97.192 - - [06/Feb/2005:11:43:27 +1100] "GET"
60.231.97.192 - - [06/Feb/2005:11:43:27 +1100] "GET"
60.231.97.191 - - [06/Mar/2005:11:43:27 +1100] "GET"
60.231.97.192 - - [06/Mar/2005:11:43:27 +1100] "GET"


I can do it with grep command as below, which works perfectly:

# less access.log |grep -c 03/Jan/2005
03

but with this I need to put the pattern manualli i.e. 03/Jan/2005.

Is there a good way to write this script which will automatically give me a busyiest day of the year. the file has more then 4 yrs access record so its only possible with a good script.

Any thought are welcome.

Frog.

I think this is definitely the wrong forum (suggestions & feedback) for this post. Plus, don't post the same thread into multiple forums.

In general, I wouldn't try to use a shell script to do this. However, in perl, it is only a few lines of code to parse an Apache log file. There are other equally suitable languages for the task. An alternative might be to use a ready-made log file analyzer.

--- rod.

This has been posted twice, I just gave a solution here:
http://www.linuxquestions.org/questi...d.php?t=453461

// closed dupe.

--jeremy