如何确定方法是从对象外部还是内部调用 - How to determine if a method is called from outside of an object or inside?

How to determine if a method is called from outside of an object or inside?

如何确定方法是从对象外部还是从对象内部调用？

例如：

class Example{

   public function Foo(){
      $this->Bar();      
   }

   public function Bar(){
      if(this_function_was_called_from_outside_the_object){
         echo 'I see you''re an outsider!' // this method was invoked from outside the object.
      }else{
         echo 'I''m talking to myself again.'; // this method was invoked from another method in this object.
      }
   }
}
$oExample = new Example();
$oExample->Foo(); // I''m talking to myself again.
$oExample->Bar(); // I see you''re an outsider!

在你调用的方法中，抛出并立即捕获一个异常，如下所示：

public function test()
{
    try 
    {
        throw new Exception("My Exception");
    }
    catch(Exception $e)
    {
        //check the stack trace at desired level
        //print_r($e->getTrace());
    }
    //Your code here
}

在 catch 块中，您可以浏览堆栈跟踪并查看谁调用了该方法。在这个 try/catch 块之后，只需输入您的正常代码即可。

使用 php 的 get_called_class（）函数

它将返回类名。如果从类外部调用，将返回 FALSE。