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# 1  


I am trying to grep a number in more then 1000 files which are .gz.
Below is the code :-

for i in `ls 20090618yz*_07.znf.gz`
a=`gzcat 20090618yz*_07.znf.gz | grep 9814843011`
if [ -n "$a" ]
echo $i 

My objective is to echo the file name also with the data.

Please help!!!!
Thanks in advance....

Last edited by vidyadhar85; 06-24-2009 at 12:24 PM.. Reason: CODE TAGS ADDED
# 2  
Please use code tags! do you have zgrep? or gzgrep?
for i  in 20090618yz*_07.znf.gz
  [[  zgrep -q 9814843011 $i ]] && echo "$i"

This works with my zgrep. It will also run a lot faster without all of the `` and extra processes your were creating.
# 3  
Do you have awk:

gzcat filename | nawk '/pattern/ {printf "%s %s\n",$0,FILENAME}'

# 4  
No the awk didn`t worked.
It gave me the same output as before, no filename in it.
Below is the output string without filename.

# # #20090618# # # # # # # # # #9#919814843011# #91# # # #3496#-100# # # # # # # #D@27#1# -

Also i dont have zgrep installed.
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