Hi everybody. I've got a problem.I need to sort a file by teh 4th column (it's a date). I attach two rows of the file:
127.0.0.1 - - [01/Nov/2006:15:03:11 +0100] "GET /_oracle_http_server_webcache_static_.html HTTP/1.1" 200 99 "95689802227,0"
127.0.0.1 - - [02/Aug/2006:15:41:49 +0200] "GET /_oracle_http_server_webcache_static_.html HTTP/1.1" 200 99 "91313042432,0"

In theory, the 02/Aug/2006 must be before that 01/Nov/2006

Thank you
Isaac.

Your best bet is to go cycle thru the file changing each date to a numeric, year-month-day format and then sort it. i.e.

02/Aug/2006 -> 20060802
01/Nov/2006 -> 20061101

and then sort numerically on that field.
If it were me, I'd use awk and parse the 4th field using the / as a delimiter. If you need it, I might be able to help.

Sorry, but I see no easy way to do it.

This little trcik may help you...

#!/bin/ksh
cut -c 16-26 inputfile > cutsort
sort -t "/" -k 3,3 -M -k 2,2 -k 1,1 cutsort > sorted
for i in `cat sorted`
do
grep $i inputfile >> outfile
done

Make sure you clear outfile when u have to run twice.

-Devaraj Takhellambam

Regards,
Devaraj Takhellambam