I have the following line in a script that searches files in several directories and shows the search results on the screen.

ls "$path" | xargs cat | tr -s " " | fgrep -i "$search_arg"

But, because I'm also using CAT and TR, the output from the search does not display the name of the file that the results came from. I need to show the file name, either on the same line of ouput, or preceding it.

I can get the filename if I remove CAT and TR, but I need to use TR to strip redundant blank spaces between words that would otherwise cause the search to fail (unless my search string also contained the same number of spaces, which wouldn't be possible to predict).

Anyone know how I can get the file name, or if there's a search command that ignores redundant spaces? We use the Korn shell.

Driver,
Thanks for the response. I should've mentioned that I'm using a circumvention for this problem, which happens to be very similar to what you've shown. But the problem I have with this method is that it's much *MUCH* slower than using XARGS. Since I have thousands of files to search, I'm hoping that there's a solution that still allows me to use XARGS. Again, this is probably something I should've mentioned in my post, but thanks for your help!

I looked at grep ...

You can use -l (ell) to list the filename that the pattern matches...

You can use -w to Select only those lines containing matches that form whole words.


They are incompatible with each other...

Im out of ideas unless you can redo the tr portion... or change how you search for the pattern.

Maybe if you use grep -f and put all possible combos in a file and use -l ...


Just throwing darts in the dark here though.

If you are searching for two words separated by any number of spaces, try...

ls "$path" | xargs grep -i "word *word2"

How about:
ls "$path" | xargs awk '{gsub(" ","",$0);print FILENAME $0}' | fgrep -i "$search_arg"

Perderabo,
The example you gave removes all spaces from the input line, while I need to remove only redundant spaces. But you did give me a good push in the right direction. Here's what I came up with:

ls "$path" | xargs nawk '{print "("FILENAME") "$0}' | tr -s " " | fgrep -i "$search_arg"


Thank you, and everyone who contributed!

Just another way to do it....
-------------------------------------
for list in `ls -1t $path`
do
if grep $search_arg > /dev/null
then
echo "${path}/${list}" >> yourfile.txt
fi
done
-------------------------------------

This will output filenames that contain the string you are looking for into "yourfile.txt"