What's the fastest way using lsof to find a single open file?

Question

I'm trying to test a single file if it is open using lsof. Is there a faster way than this?

$result = exec('lsof | grep filename | wc -l');

if($result > 0) {
    //file is open
}

I'm thinking their must be a way to just test for one file if you know the filename. All I really need is a true or false.

Answer 1

I found a much better way. With lsof (tested on version 4.63), you can directly query a specific file:

if lsof filename > /dev/null; then
    # file is open
fi

Answer 2

Don't. The answer might change by the time you try to do anything with the result. Instead, try to do whatever you intended to do with the file and handle the "in use" error or "sharing violation".