php - 伪造闭包的函数名

撇开长话，我有一个这样的场景:

class Foo {

  function doSomething() {
    print "I was just called from " . debug_backtrace()[1]['function'];
  }

  function triggerDoSomething()
  {
    // This outputs "I was just called from triggerDoSomething".
    // This output makes me happy.
    $this->doSomething();
  }

  function __call($method, $args)
  {
    // This way outputs "I was just called from __call"
    // I need it to be "I was just called from " . $method
    $this->doSomething();

    // This way outputs "I was just called from {closure}"
    // Also not what I need.
    $c = function() { $this->doSomething() };
    $c();

    // This causes an infinite loop
    $this->$method = function() { $this->doSomething() };
    $this->$method();
  }
}

在我调用$ foo-> randomFunction()的情况下，我需要将输出读取为“我刚刚从randomFunction中被调用”

是否有办法命名闭包或以其他方式解决此问题？

注意:我无法更改doSomething函数。这是我正在调用的第三方代码的示例，该代码考虑了为执行某项操作而调用该代码的人员的功能名称。

最佳答案

您可以将名称传递给doSomething()，例如

$this->doSomething($method);

或像

$c = function($func_name) { $this->doSomething($func_name); };
$c($method);

在doSomething中，您可以使用该参数。

function doSomething($method) {
    print "I was just called from " . $method;
}

关于php - 伪造闭包的函数名，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/27478110/