how to grep string from hourly basis files

dear all,

pls help on this script..

i have many files which will be created every mins in particular directory.

i want to grep a particular string from only for unique hour files.

from the below code i want to grep a string from only 9th hour files .
Ex files:

-rw-r--r--   1 root     root      580988 Aug 25 09:46 IE1_CV_25_08_2010_09_39_50_006382.cdr
-rw-r--r--   1 root     root      594808 Aug 25 09:51 IE1_CV_25_08_2010_09_44_51_006383.cdr
-rw-r--r--   1 root     root      594468 Aug 25 09:56 IE1_CV_25_08_2010_09_49_52_006384.cdr
-rw-r--r--   1 root     root      638385 Aug 25 10:01 IE1_CV_25_08_2010_09_54_53_006385.cdr
-rw-r--r--   1 root     root      668880 Aug 25 10:05 IE1_CV_25_08_2010_09_59_54_006386.cdr
-rw-r--r--   1 root     root      691399 Aug 25 10:10 IE1_CV_25_08_2010_10_04_55_006387.cdr
-rw-r--r--   1 root     root      662416 Aug 25 10:15 IE1_CV_25_08_2010_10_09_56_006388.cdr

thanks

grep "string"  *_2010_09_*

hi Girija,

I want to grep with respect to system timing .. not with files timing ..

could u help me?

thanks

 ls -ltr | awk '{print $8 " " $9}' | grep "09" | awk '{print $2}' | xargs  grep "string"

string -> represents string to grep

---------- Post updated at 06:55 PM ---------- Previous update was at 06:53 PM ----------

Try the above command and see whatever u excepted is achieved

grep "09:"

or

awk '$8 ~ /^09:/'

# ls -ltr | sed -n '/09:.*/p'