Can I make `find` return non-0 when no matching files are found?

find /tmp -name something | grep .

The return status will be 0 when something is found, and non-zero otherwise.

EDIT: Changed from egrep '.*' to the much simpler grep ., since the result is the same.

Exit 0 is easy with find, exit >0 is harder because that usually only happens with an error. However we can make it happen:

if find -type f -exec false {} +
then
  echo 'nothing found'
else
  echo 'something found'
fi

Simplest solution that doesn't print, but exits 0 when results found

find /tmp -name something | grep -q "."

Having just found this question whilst trying to find my way to solve a problem with Puppet (change permissions on folders under a directory but not on the directory itself), this seems to work:

test -n "$(find /tmp -name something)"

My specific use case is this:

test -n "$(find /home -mindepth 1 -maxdepth 1 -perm -711)"

Which will exit code 1 if the find command finds no files with the required permissions.

It's not possible. Find returns 0 if it exits successfully, even if it didn't find a file (which is a correct result not indicating an error when the file indeed doesn't exist).

To quote the find manpage

EXIT STATUS

find exits with status 0 if all files are processed successfully, greater than 0 if errors occur. This is deliberately a very broad description, but if the return value is non-zero, you should not rely on the correctness of the results of find.

Depending on what you want to achieve you could try to let find -print the filename and test against it's output:

#!/bin/bash
MYVAR=`find . -name "something" -print`
if [ -z "$MYVAR" ]; then
    echo "Notfound"
else
   echo $MYVAR
fi

I feel that this is the most concise and direct method:

test `find /tmp/ -name something -print -quit 2>/dev/null`

Here's a little script I called test.py. It improves upon other methods posted in that it will return an error code if one is set, and it additionally set one if find didn't list any files:

from subprocess import Popen
import sys

p = Popen(['find'] + sys.argv)
out, err = p.communicate()
if p.returncode:
    sys.exit(p.returncode)
if not out:
    sys.exit(1)

Here's the command-line output:

$ python test.py . -maxdepth 1 -name notthere
$ echo $?
1
$ find . -maxdepth 1 -name notthere
$ echo $?
0
$ find . -failedarg
find: unknown predicate `-failedarg'
$ echo $?
1

Then, for a result where find had errors but found files:

$ ls -lh
$ d---------  2 jeff users   6 Feb  6 11:49 noentry
$ find .
.
./noentry
find: `./noentry': Permission denied
$ echo $?
1
$ find . | egrep '.*'
.
./noentry
find: `./noentry': Permission denied
$ echo $?
0
python ../test.py 
../test.py
$ echo $?
1

Then, if you want the list of files you can make use of -print 0 passed to find and split the out variable on nulls, or you can just add a print statement for it.

It is not only find that returns the exit status codes as zero when it successful. In unix what ever the command you execute, if its succeeds then it returns the exit status as zero.