PHP mysqli wrapper: passing by reference with __call() and call_user_func_array()

Question

I\'m a long running fan of stackoverflow, first time poster. I\'d love to see if someone can help me with this. Let me dig in with a little code, then I\'ll explain my probl

Answer 1

With a little help from the guys from the #php irc channel I came up with the following solution:

// We have to explicitly declare all parameters as references, otherwise it does not seem possible to pass them on without
// losing the reference property.
public function bind_result (&$v1 = null, &$v2 = null, &$v3 = null, &$v4 = null, &$v5 = null, &$v6 = null, &$v7 = null, &$v8 = null, &$v9 = null, &$v10 = null, &$v11 = null, &$v12 = null, &$v13 = null, &$v14 = null, &$v15 = null, &$v16 = null, &$v17 = null, &$v18 = null, &$v19 = null, &$v20 = null, &$v21 = null, &$v22 = null, &$v23 = null, &$v24 = null, &$v25 = null, &$v26 = null, &$v27 = null, &$v28 = null, &$v29 = null, &$v30 = null, &$v31 = null, &$v32 = null, &$v33 = null, &$v34 = null, &$v35 = null) {
    // debug_backtrace returns arguments by reference, see comments at http://php.net/manual/de/function.func-get-args.php
    $trace = debug_backtrace();
    $args = &$trace[0]['args'];
    return call_user_func_array(array($this->mysqlObj, 'bind_result'), $args);
}