Find files in one directory not in another

Question

Are there any Linux/Unix tools which find all the files in one directory not present in another? Basically I'm looking for diff which works on the output of ls.

Short and sweet scripts are also appreciated.

Answer 1

diff does this already:

diff dir1 dir2

Example output:

Only in dir1: some_file.txt
Only in dir1: some_other_file.txt
Only in dir2: third_file.txt

Answer 2

Bash:

diff <(cd dir1; ls) <(cd dir2; ls)

Compare only the filenames - not the contents of the files.

Answer 3

Like people told you here, you can use DIFF in various usage variations. Or you just use dirdiff instead, which is meant for what you're trying! :-)

But if you want to keep some directories in sync then you really should take a look on rsync.

Regards

Answer 4

If you are wanting to do this through all sub directories as well, a good way to do it is:

diff --brief -r dir1/ dir2/

I prefer using brief, but you can leave that out if you want.

Answer 5

for i in $(ls -1 directory1); do if (test -f directory2/$i) then echo $i; fi; done

Answer 6

Dennis Williamson had a good answer, but I needed to do this recursively. GNU findutils 4.7.0 doesn't sort its output, so here's what I used

diff <(cd $dir1; find | sort) <(cd $dir2; find | sort)

To do this only one way, and produce a list of files, I used this:

diff <(cd $dir1; find | sort) <(cd $dir2; find | sort) \
| grep '< ./' | sed "s,< ./,$dir1/,"

For this to work properly, neither $dir1 nor $dir2 should include the trailing slash.

Answer 7

This works..

ls -a1 /dir1 | sort > /tmp/1
ls -a1 /dir2 | sort > /tmp/2
diff /tmp/1 /tmp/2

Answer 8

untested:

find /dir/A -printf "%P" | while read f; do
  if [ ! -e "/dir/B/$f" ]; then
    echo $f
  fi
done