Question

I need your help with a short bash script. I have a folder, which contains about 150,000(!) xml-files. I need a script which extracts all those files, which contain a specified line. The script should be work as fast as possible, because the script have to be used very often.

My first approach was the following, using grep:

for f in temp/*
do
   if grep "^.*the line which should be equal.*$" "$f" 
   then 
      echo "use this file"
   else 
      echo "this file does not contain the line"
   fi
done

This approach works, but it takes too much time. Does somebody know an faster approach? If another scripting language is a better choice, it is also ok.

Best regards,
Michael

Answer 1

You can use grep without any bash handlers.

   -l, --files-with-matches
          Suppress normal output; instead print the name of each input file from which output would normally have been printed.  The scanning will stop on the first  match.   (-l  is
          specified by POSIX.) 

So, try this:

grep "the line which should be equal" --files-with-matches  temp/*