how to execute a shell built-in if find found certain files

Question

In general, if we want to execute a command if find identifies files corresponding to certain specs, we can use -exec. But in my case I want to exit the shell/script if find found a match and -exec can't do that. [Apparently find's -exec cannot call bash functions either.)

For instance in a script to remove a tree if and only if it contains only empty directories, I'd like to first check for presence of any non-directory and abort with an error message at the first such unexpected entry.

So far whenever I need to do something like this, I put the output to a variable (possibly using -quit with find) and testing for it being empty or not.

So my script looks like:

#! /bin/sh
INTRUDER="$(find "$@" ! -type d -print -quit)"
if [ -n "$INTRUDER" ] ; then
    echo "Found non-dir $INTRUDER; leaving the arguments untouched"
    exit 1
fi
rm -R "$@"

Is there a different/better approach than this? By "better" I mean, smaller to code, more portable (if there is any portability issue with the above), etc. By "different" I mean by avoiding the extra variable and testing for its length.

Answer 1

This one worked for me:

find "$@" ! -type d -exec kill -9 $$ \; -quit && rm -R "$@"

• If find exits normally (nothing is found) rm -R "$@" will be executed.
• If find finds something the current shell/script is killed ($$ stores the pid). The rm-part will never be executed in this case.