Shell Programming and Scripting

i have a ksh script :

#!/bin/ksh
TZ=`date +%Z`+24 ;a=`date +%Y-%m-%d`

b=`date +"%H:%M:%S"`
cd /ednadtu3/u01/pipe/logs
for i in Archiver1.log
do

cat $i | grep $a | grep $b >> /ednadtu3/u01/pipe/naveed/Insert_Date.txt
done
------------------------------------------------------------------

here i want to grep certain given values of hour in file Archiver1.log.
but this script greps current time of system. i want to grep 20th hour in file Archiver1.log. how to do that.

actually i want to insert the own values in below format

b=`date +"%H:%M:%S"`

------------------------------

What does the data look like?

something like this what your looking for (obviously cleaned up to only match the date area of the log versus the content portion.

Code:

grep -E "20:[0-9][0-9]:[0-9][0-9]" ...

the data looks like

2008-02-11 20:23:34
2008-02-11 20:53:20
2008-02-11 20:25:44


2008-02-11 21:40:34

2008-02-11 23:59:00

----------------------------------------------------------------------

i want to grep only 20th hour. thanks ur suggestion helped me. now i m able to grep . thanks dude

So this (very cut and paste, you could easily shorten)

Code:

grep -E "^[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] 20:[0-9][0-9]:[0-9][0-9]" ...

Since you have the -E flag

Code:

grep -E "^[0-9]{4}-[0-9]{2}-[0-9]{2} 20:[0-9]{2}:[0-9]{2}" ...