PHP mysqli wrapper: passing by reference with __call() and call_user_func_array()
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开发者_如何学编程 I'll explain my problem. I have the following wrapper classes:
class mysqli_wrapper
{
private static $mysqli_obj;
function __construct() // Recycles the mysqli object
{
if (!isset(self::$mysqli_obj))
{
self::$mysqli_obj = new mysqli(MYSQL_SERVER, MYSQL_USER, MYSQL_PASS, MYSQL_DBNAME);
}
}
function __call($method, $args)
{
return call_user_func_array(array(self::$mysqli_obj, $method), $args);
}
function __get($para)
{
return self::$mysqli_obj->$para;
}
function prepare($query) // Overloaded, returns statement wrapper
{
return new mysqli_stmt_wrapper(self::$mysqli_obj, $query);
}
}
class mysqli_stmt_wrapper
{
private $stmt_obj;
function __construct($link, $query)
{
$this->stmt_obj = mysqli_prepare($link, $query);
}
function __call($method, $args)
{
return call_user_func_array(array($this->stmt_obj, $method), $args);
}
function __get($para)
{
return $this->stmt_obj->$para;
}
// Other methods will be added here
}
My problem is that when I call bind_result()
on the mysqli_stmt_wrapper
class, my variables don't seem to be passed by reference and nothing gets returned. To illustrate, if I run this section of code, I only get NULL's:
$mysqli = new mysqli_wrapper;
$stmt = $mysqli->prepare("SELECT cfg_key, cfg_value FROM config");
$stmt->execute();
$stmt->bind_result($cfg_key, $cfg_value);
while ($stmt->fetch())
{
var_dump($cfg_key);
var_dump($cfg_value);
}
$stmt->close();
I also get a nice error from PHP which tells me: PHP Warning: Parameter 1 to mysqli_stmt::bind_result() expected to be a reference, value given in test.php on line 48
I've tried to overload the bind_param()
function, but I can't figure out how to get a variable number of arguments by reference. func_get_args()
doesn't seem to be able to help either.
If I pass the variables by reference as in $stmt->bind_result(&$cfg_key, &$cfg_value)
it should work, but this is deprecated behaviour and throws more errors.
Does anyone have some ideas around this? Thanks so much for your time.
If you'll extend from the mysqli_stmt class you'll bypass the reference problem. (which has no clean solution)
class mysqli_stmt_wrapper extends mysqli_stmt {
public function __construct($link, $query) {
parent::__construct($link, $query);
}
}
class mysqli_wrapper extends mysqli {
public function prepare($query) {
return new mysqli_stmt_wrapper($this, $query);
}
}
With a little help from the guys from the #php irc channel I came up with the following solution:
// We have to explicitly declare all parameters as references, otherwise it does not seem possible to pass them on without
// losing the reference property.
public function bind_result (&$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);
}
I'm guessing this is because the original function signature specifies that it expects references, whereas your __call
cannot do so. Therefore, try not using __call
but explicitly adding the bind_result
with the same function signature as the original.
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