UNIX for Dummies Questions & Answers

Greetings,

I wasn't able to find this exact issue through a forum search, so I hope it is not a repost.

I have a directory of files that all end in ".log", and I would like to grep out something from the most recently created file in the directory without knowing the exact file name (aka I want to just grab the last file, not look at it's name).

So far, I have tried this:

Code:

ls -ltr | tail -1 | grep "Something" *.log

If I do just a ls -ltr | grep -1, it shows the last file like I want, but the grep ends up grepping on all the files in the directory (and there are a crap ton of files), so I guess I am not understanding exactly how the pipe thing works.

This gives me what I want:

Code:

grep "Something" *.log | tail -1

But it is searching every file not just the last one, so it takes forever.

Thanks in advance!

Moderator's Comments:

Code tags for code, please.

You don't need to use -l unless you actually need the long-style listing.

You want to feed a filename into grep, which doesn't take filenames on standard input. The xargs tool can be used to turn a stream into arguments though.

Here is how xargs works in short:

Code:

# These two commands are equivalent.
cat a b c
echo a b c | xargs cat

So you can feed the file from head into grep like this:

Code:

ls -tr | head -n 1 | xargs grep "search string"

That worked! Thanks for the help