Question

遗漏了一个长篇故事，我有这样的场景：

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();
  }
}

如果我调用$ foo-＆gt; randomFunction（），我需要将输出读作“我刚从randomFunction调用”

有没有办法命名一个闭包或以不同的方式处理这个问题？

注意：我无法更改doSomething功能。这是我正在调用的第三方代码的示例，它考虑了为了做某事而调用它的函数名称。

Answer 1

您可以将名称传递给doSomething()，如

$this->doSomething($method);

或像

这样的闭包

$c = function($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;
}

Answer 2

我在doSomething（）中不改变任何内容的唯一想法是使用eval()

function __call($method, $args)
{
    eval("
         function {$method}(\$self) {
              \$self->doSomething();
         }

         {$method}(\$this);
    ");
}

伪造一个闭包的函数名

2 个答案: