PHP类:从被调用的方法访问调用实例

Question

抱歉这个奇怪的主题,但我不知道如何以另一种方式表达它.

我正在尝试从调用类访问一个方法.就像在这个例子中:

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
    }
}

这两个类都不依赖于继承,我不想将该函数设置为static.

有没有办法实现这一目标？

谢谢你的帮助!

Answer 1

您可以像这样传递对第一个对象的引用:

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

您可以为此使用 debug_backtrace() 。它有点不确定，但出于调试目的它很有用。

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

}