Question

Warning: array_filter() expects parameter 2 to be a valid callback,
function ’empty’ not found or invalid function name….

Why is empty considered a invalid callback?

$arr = array_filter($arr, 'empty');

This works: if(empty($arr['foo'])) die();

Answer 1

Answer

empty() is not a function but a language construct and array_filter() can only accept a function as its callback.

This is given as a small note on the manual page:

Note: Because this is a language construct and not a function, it
cannot be called using variable functions

Work around

To work around this you can wrap empty in another function for example:

function empty_test($val) {
    return empty($val);
}

And then call it like so:

$arr = array_filter($arr, 'empty_test');