UNIX for Dummies Questions & Answers

Hi,

My requirement is to search for a paritcular string from a group of .gz files and to print the lines containing that string and the name of the files in which that string is present.

Daily 500 odd .gz files will be generated in a directory(directory name will be in the form of daymonthyear i.e 230709) in /data3/home path.
Means the mentioned path has many such directories with .gz files.

I used below command, it's printing the lines containing that string but not filename.


find /data3/home/ -type f -name 'TRCansLog*.gz' -exec gzcat *.gz {} \; | nawk '/xfvc010423465107/ {print $0"\t" FILENAME}'

Pls help me

If possible pls tell how to print the directory name in which the requried file and string present.


Thanks in advance

Code:

find /data3/home/ -type f -name 'TRCansLog*.gz' |
while read filename 
do   
   gzcat $filename | nawk -v fname=$filename '/xfvc010423465107/ {print $0 "\t" fname}'
done

Thank you very much JIM,

It's working.

Pls do one more favor to me, I want to run above one as script, pls tell me what changes to be done inorder to give pattern at the begining instead of giving at nawk command. i.e ./pattern_filename_script xfvc010423465107

Thanks in advance.

You do something like the following:

Code:

SEARCH="$1"
find /data3/home/ -type f -name 'TRCansLog*.gz' |
while read filename 
do   
   gzcat $filename | nawk -v fname=$filename '/'$SEARCH'/ {print $0 "\t" fname}'
done

Thanks a lot
Its working