是否可以确定某个方法是从一个类的内部还是外部调用的？

Question

示例代码：

class MyClass { 
    function echo_msg { 
     echo // now what... 
    } 

    function echo_from_inside { 
     $this->echo_msg() 
    } 
} 

result should be: 

$my_instance = new MyClass(); 
$my_instance->echo_msg(); // I was called from OUTside 
$my_instance->echo_from_inside(); // I was called from INside 

Answer 1

它可能会更容易，而不是检测函数从哪里被调用，以包装一个公用的私人函数。像这样：

class MyClass{ 
    private function myob(){ 
    //do something 
    } 

    public function echo_msg(){ 
    $this->myob(); 
    //do other stuff like set a flag since it was a public call 
    } 

    private function foo(){ //some other internal function 
    //do stuff and call myob 
    $this->myob(); 
    } 
} 

$obj=new MyClass(); 
$obj->echo_msg();//get output 
$obj->myob(); //throws error because method is private 

Answer 2

你可以试试，让你的方法的调用者：

$trace = debug_backtrace(); 
$caller = array_shift($trace); 
echo 'called by '.$caller['function'] 
echo 'called by '.$caller['class'] 

这应该为你工作。

Answer 3

您可以添加一个可选的参数，像这样：

function echo_msg($ins=false) { 
     if($ins){/*called from inside*/}else{/*called from outside*/} 
     echo // now what... 
    } 

，离开去年。如果你是从课堂内部调用它，通过它true，否则什么都不传！