php - 伪造一个闭包的函数名

Question

长话短说，我有这样一个场景:

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();
  }
}

如果我调用 $foo->randomFunction()，我需要输出为“I was just called from randomFunction”

有没有办法命名一个闭包或以不同的方式解决这个问题？

注意:我无法更改 doSomething 函数。这是我调用的第三方代码示例，它考虑调用它的函数名称以执行某些操作。

Answer 1

你可以将名称传递给doSomething()，比如

$this->doSomething($method);

或者像这样关闭

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

并且在 doSomething 中您可以使用该参数。

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