PHP：函数参数必须与动态类名称的对象

Question

所以我在面向对象编程的世界新和我目前面临这个问题（一切都在规范中描述）：

<?php 
    class MyClass { 
     // Nothing important here 
    } 

    class MyAnotherClass { 
     protected $className; 

     public function __construct($className){ 
      $this->className = $className; 
     } 
     public function problematicFunction({$this->className} $object){ 
      // So, here I obligatorily want an $object of 
      // dynamic type/class "$this->className" 
      // but it don't works like this... 
     } 
    } 

    $object = new MyClass; 
    $another_object = new MyAnotherClass('MyClass'); 

    $another_object->problematicFunction($object); 
?> 

灿任何人都帮助我？

感谢，马克西姆（法国：对不起，我的英语）

Answer 1

你需要的是

public function problematicFunction($object) { 
    if ($object instanceof $this->className) { 
     // Do your stuff 
    } else { 
     throw new InvalidArgumentException("YOur error Message"); 
    } 
} 

Answer 2

尝试这样

class MyClass { 
    // Nothing important here 
    public function test(){ 
     echo 'Test MyClass'; 
    } 
} 

class MyAnotherClass { 
    protected $className; 

    public function __construct($className){ 
     $this->className = $className; 
    } 
    public function problematicFunction($object){ 
     if($object instanceof $this->className) 
     { 
      $object->test(); 
     } 
    } 
} 

$object = new MyClass; 
$another_object = new MyAnotherClass('MyClass'); 

$another_object->problematicFunction($object); 

Answer 3

这就是所谓的type hinting，你想要做什么只是不支持。

如果所有这些动态类名都有一些共同点（例如，它们对于某些功能而言是不同的实现），您可能需要定义一个基类（也许是抽象类）或一个接口并将该共同祖先用作类型提示：

<?php 

interface iDatabase{ 
    public function __contruct($url, $username, $password); 
    public function execute($sql, $params); 
    public function close(); 
} 

class MyClass implements iDatabase{ 
    public function __contruct($url, $username, $password){ 
    } 

    public function execute($sql, $params){ 
    } 

    public function close(){ 
    } 
} 

class MyAnotherClass { 
    protected $className; 

    public function __construct($className){ 
     $this->className = $className; 
    } 
    public function problematicFunction(iDatabase $object){ 
    } 
} 

否则，刚刚转会支票problematicFunction()身体内，其他的答案解释。