Compare all files in directory, list filenames only

MOH

This should be very simple, but I'm making a hash of it. Probably cos my Unix scripting knowledge is minimal

I've got some software for which I have my own amended versions of some of the files. When a new release of the software comes out, I want to find out if I have local versions of any of the base files have changed, as I may need to incorporate those changes.

I'm installing the new version in parallel anyway, so the simplest thing to do would be a diff of the old base files against the new base files to find what files have changed, and then a diff of the those against my files.

So something like:

diff -ru oldbase newbase > basediff.txt
for filename in `cat basediff.txt`
do 
diff myfiles/filename newbase/filename >> mydiff.txt
done

However, I need the first diff to only list the filenames, instead of 'files oldbase/filename and newbase/filename differ' and 'only in newbase: filename' In fact, I don't care about new base files either, only changed ones.

I've been through the diff man page but can't find a way of just listing the filenames. An alternative would probably be a regexp when reading the basediff.txt file to extract the filename, but no sure how to do that in a script.

Tillotson

That is going to get very complicated very fast. I'd recommend using some sort of version control such as git.

This tutorial is probably enough to get you started

MOH

Tillotson wrote: »

That is going to get very complicated very fast. I'd recommend using some sort of version control such as git.

This tutorial is probably enough to get you started

Yeah, I know. Thanks, I'll have a look at that - it's a bit messy in that I'm on a shared server with no SSH access. I might be able to do something with SVN.

In the meantime, solved my script issue using cut to extract the filename:

diff -qr oldbase newbase > basediff.txt
grep 'differ' basediff.txt | cut -d ' ' -f 4 > grep1.txt
for filename in `cat grep1.txt`
do .....