如何确定在哪个类中启动了对变量的引用(并且当前存在)?
例子:
<?php
class MyClass {
public $array = array(
"this",
"is",
"an",
"array"
);
}
$class = new MyClass();
$arrayReference = &$class->array;
GetClassForVariable($arrayReference); //Should return "MyClass"
?>
我最好的选择是某种Reflection,但我还没有找到任何似乎适合于此的功能。
编辑:
一个更适合我想要的示例如下:
<?php
class API_Module {
public $module;
public $name;
private $methods = array();
public function __construct($module, $name) {
$this->module = $module;
$this->name = $name;
$this->methods["login"] = new API_Method($this, "login", "Login");
}
public function GetMethod($method) {
return $this->methods[$method];
}
public function GetURL() {
return $this->module; //Should return "session"
}
}
class API_Method {
public $method;
public $name;
private $parentReference;
private $variables = array();
public function __construct(&$parentReference, $method, $name) {
$this->parentReference = $parentReference;
$this->method = $method;
$this->name = $name;
$this->variables["myvar"] = new API_Variable($this, "myvar");
}
public function GetURL() {
return $this->GetParentURL() . "/" . $this->method; //Should return "session/login"
}
public function GetVariable($variableName) {
return $this->variables[$variableName];
}
private function GetParentURL() {
// Need to reference the class parent here
return $this->parentReference->GetURL();
}
}
class API_Variable {
public $name;
private $parentReference;
public function __construct(&$parentReference, $name) {
$this->parentReference = $parentReference;
$this->name = $name;
}
public function GetURL() {
return $this->GetParentURL() . "/" . $this->name; //Should return "session/login/myvar"
}
private function GetParentURL() {
// Need to reference the class parent here
return $this->parentReference->GetURL();
}
}
$sessionModule = new API_Module("session", "Session");
var_dump($sessionModule->GetMethod("login")->GetVariable("myvar")->GetURL()); //Should return "session/login/myvar"
?>
现在,这很好用,但我希望能够在不使用$parentReference
每个子变量的情况下做到这一点。这可能是不可能的,但我很想知道它是否是。