I want to both require and prune two different directories in my find command:

find . -path "*/logs/*"

and -prune "./snapshots" at the same time, any idea how to do this?

I can't seem to do both at the same time.

What are you trying to achieve, and how are you
trying to do this? Give's the complete syntax and
a few bits and pieces of what the directory structure
looks like, and what you want to get rid of.



Cheers,
Tink

Like I said, I want to force one directory, and weed out another in the same find command.


I am trying to accomplish something to this effect:

find . -path "*/logs/*" -maxdepth 4 -path "./snapshots/*" -prune

where it forces the path to have "logs" and skips "snapshots"...the problem being, I have a huge amount of data, and ./snapshots contains duplicates snapshots of the data every hour. So I don't want to traverse that directory, since it contains 5 times the amount of the original file tree.

Make sense? (the above doesn't work)

Hmmmm ... if the "*/logs/*" is to force one directory,
why don't you just
find ./logs -maxdepth 4 -a \( -path "./snapshots/*" -prune \)
or something to that effect? Would make it quite clear,
and probably slightly less costly ...


Cheers,
Tink

Because my directory structure looks like this:

./foo/logs/
./foo2/logs/
./foo3/logs/
./snapshot/hourly/foo/logs, etc. etc.
./snapshot/hourly.1/foo/logs, etc. etc.
./snapshot/hourly.4/foo/logs, etc. etc.


I have 582 "foo" directories, inwhich I want to find all log files. Skipping ./snapshots would make the search faster.

Try that one ...



Cheers,
Tink

Similar problem...

I can't seem to combine any other find arguments with -prune:


find ./ -path "*.snapshots*" -prune -o print


But when I add -user nobody, it just matches all users.