PHP类：您可以通过调用的方法访问调用实例(PHP Classes: get access to t

遗憾对于怪异的问题，但我不知道该如何表达它以另一种方式。

我试图从一个调用的类访问方法。就像这个例子：

class normalClass {
     public function someMethod() { 
          [...]
          //this method shall access the doSomething method from superClass
     }
}

class superClass {
     public function __construct() {
          $inst = new normalClass;
          $inst->someMethod();
     }
     public function doSomething() {
          //this method shall be be accessed by domeMethod form normalClass
    }
}

两个类都没有继承关系，我不想要的功能设置为静态。

有没有什么办法来实现这一目标？

谢谢你的帮助！

Answer 1:

你可以通过像这样的第一对象的引用：

class normalClass {
    protected $superObject;
    public function __construct(superClass $obj) {
        $this->superObject = $obj;
    }

    public function someMethod() { 
        //this method shall access the doSomething method from superClass
        $this->superObject->doSomething();  
    }
}

class superClass {
    public function __construct() {
          //provide normalClass with a reference to ourself
          $inst = new normalClass($this);
          $inst->someMethod();
    }
    public function doSomething() {
          //this method shall be be accessed by domeMethod form normalClass
    }
}

Answer 2:

你可以为这个使用debug_backtrace（）。这是一个有点玄乎，但用于调试的目的是有用的。

class normalClass {
 public function someMethod() { 
      $trace = debug_backtrace();
      $trace[1]->object->doSomething();
 }

}

Answer 3:

您有几种选择。您可以使用聚合像这样

class normalClass
{
  protected $superClass;

  public function __construct( superClass $superClass )
  {
    $this->superClass = $superClass;
  }

  public function someMethod()
  {
    $this->superClass->doSomething();
  }
}

class superClass
{
  public function __construct()
  {
    $inst = new normalClass( $this );
    $inst->someMethod();
  }

  public function doSomething()
  {  //this method shall be be accessed by domeMethod form normalClass
  }
}

或者只是一个直线上升的二传手

class normalClass
{
  protected $superClass;

  public function setSuperClass( superClass $superClass )
  {
    $this->superClass = $superClass;
  }

  public function someMethod()
  {
    if ( !isset( $this->superClass ) )
    {
      throw new Exception( 'you must set a superclass' );
    }
    $this->superClass->doSomething();
  }
}

class superClass
{
  public function __construct()
  {
    $inst = new normalClass();
    $inst->setSuperClass( $this );
    $inst->someMethod();
  }

  public function doSomething()
  {  //this method shall be be accessed by domeMethod form normalClass
  }
}

Answer 4:

根据你的使用情况，您可能希望实例传递到只有功能：

class normalClass {
    public function someMethod($object) { 
        $object->doSomething();
    }
}

如果normalClass::someMethod()可以通过多个不同被称为$object S，这可能是更好的选择（而不是提供的$object到整个normalClass实例）。

但不管，你可能会考虑创建一个接口用于类型提示：

interface ISomethingDoer {
    public function doSomething();
}

class normalClass {
    public function someMethod(ISomethingDoer $object) {
        # Now PHP will generate an error if an $object is passed
        # to this function which does not implement the above interface.
        // ...

class superClass implements ISomethingDoer {
    // ...