simple shell script question

[wvsa]

Good afternoon all,

I need to create a script that pulls all lines that have entries on 08/Sep/2004, note the following example:

192.168.200.51 - - [08/Sep/2004:11:59:42 -0700] "GET /OA_MEDIA/xxwvus/istore/GCL

Attempted to use awk but could not get it to work.

Your input would be greatly appreciated

[Rick Garland]

How about using grep?

This will pull the lines that match the pattern.

[Sundar_7]

Yeah, grep will do

# grep "[08/Sep/2004:" inputfile

[Sridhar Bhaskarla]

Hi,

Did you want to make the day as the variable? then do something like this

DATE=$(date +%d/%b/%Y)
grep "[${DATE}:" logfile

-Sri

[wvsa]

Gentlemen, note the following:

grep "[08/Sep/2004:" access_log
grep: [] imbalance.

Am I missing something? Trying to print all lines with the occurance of 08/Sep/2004.


Thank you for input.

[Sundar_7 · Accepted solution]

Sorry, you should escape "["

grep "\[08/Sep/2004:" inputfile

[wvsa]

The escape did the trick, THANKYOU!

[RAC_1]

You need--

grep "\[08\/Sep\/2004:" inputfile

Anil