Question

function outer_function() {
  $nid = 3;

    function exists($var) {
    print $nid;
    return $var->nid == $nid;
    }   

    $a_filtered_array = array_filter($an_array, "exists");
}

I’m trying to filter this array using a variable that’s defined in the outer function, but the variable is not defined. This would work in JS. What am I doing wrong here? How would I accomplish this in PHP?

Answer 1

You could write this (correctly) as:

function outer_function() {
  $nid = 3

    $a_filtered_array = array_filter($an_array, function ($var) use ($nid) {
        print $nid;
        return $var->nid == $nid;
    });
}

You can’t just pass a string containing the function’s name. You could also write

function outer_function() {
  $nid = 3
  $exists =  function ($var) use ($nid) {
        print $nid;
        return $var->nid == $nid;
    }
    $a_filtered_array = array_filter($an_array, $exists);
}

See http://www.php.net/manual/en/functions.anonymous.php for in-depth on the syntax and semantics.