Shell Programming and Scripting

Hello Everyone,
I'm trying to create a bash script that will count the number of files containing a certain keyword. The twist is that the keyword is found in a unix File and is part of a list. For example, the file DELETE may look like this:

John
Mike
Dana

The script will look for files containing the first keyword and then tell the user how many files contain that word. The script will then move on to the next key word and do the same thing. The filenames themselves are irrelevant.

I currently have the following code:

Code:

for i in `cat DELETE2`
do
  echo "The word $i is found in `grep -il "$i" `find .`` files"
done

Any suggestions?

Just subtract 1 from the number of files found with matches and that should give you the correct number.

Have you tried out, what you have shown ?

What is the error or result you got ?

A more better solution, step by step is:

Code:

for i in `cat DELETE2`
do
  for j in `find .`
  do
     grep -il "$i" $j 1>>/dev/null
     if [[ $? == 0 ]]
     then
        echo "The word $i is found in $j file"
     fi
  done
done

If you want the files count (and not the exact list), you can do :

Code:

while read word
do
   count=$(grep -il "$word" $(find .) | wc -l)
   echo "The word '$word' is found in $count file(s)"
done < DELETE2

Jean-Pierre.

Everyone, thank you for your help! Aigles, I'm using a modified version of your code. Your original code gave me an incorrect output (an added file), but I was able to modify it a bit.

Thanks again!