PHP mysqli包装器:通过引用传递__call()和call_user_func_array() [英] PHP mysqli wrapper: passing by reference with __call() and call_user_func_array()

问题描述

我一直是stackoverflow的狂热粉丝，这是第一次海报.我很想看看是否有人可以帮助我.让我用一些代码深入研究，然后我将解释我的问题.我有以下包装器类:

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 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
}

我的问题是，当我在mysqli_stmt_wrapper类上调用bind_result()时，我的变量似乎没有通过引用传递，并且什么也没有返回.为了说明这一点，如果我运行这段代码，我只会得到NULL:

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();

我还从PHP中得到一个很好的错误，告诉我:PHP Warning: Parameter 1 to mysqli_stmt::bind_result() expected to be a reference, value given in test.php on line 48

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

我试图重载bind_param()函数，但是我不知道如何通过引用获取可变数量的参数. func_get_args()似乎也无法提供帮助.

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.

如果像$stmt->bind_result(&$cfg_key, &$cfg_value)那样通过引用传递变量，它应该可以工作，但这是不建议使用的行为，并且会引发更多错误.

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.

有人对此有想法吗?非常感谢您的宝贵时间.

Does anyone have some ideas around this? Thanks so much for your time.

推荐答案

如果从mysqli_stmt类扩展，将绕过参考问题. (没有干净的解决方案)

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);
  }
}

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