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gzcat/grep

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# 1  
Old 06-24-2009
gzcat/grep
Hi!!!

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


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

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 11:24 AM.. Reason: CODE TAGS ADDED
# 2  
Old 06-24-2009
Please use code tags! do you have zgrep? or gzgrep?
Code:
for i  in 20090618yz*_07.znf.gz
do
  [[  zgrep -q 9814843011 $i ]] && echo "$i"
done

This works with my zgrep. It will also run a lot faster without all of the `` and extra processes your were creating.
# 3  
Old 06-24-2009
Do you have awk:

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

# 4  
Old 06-24-2009
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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