没有调用扩展PHP DOMElement的构造函数

Question

在PHP中扩展DOMElement时,不会调用子类的构造函数.就预期的行为而言,没有什么能在文档中跳出来,但也许我错过了一些东西.这是一个简单的测试用例....

 class SillyTestClass extends DOMElement{
    public $foo=null;
    public function __construct($name,$value=null,$namespace=null){
        echo "calling custom construct....";
        $this->foo="bar";
        parent::__construct($name,$value,$namespace);
    }
    public function sayHello(){
        echo "Why, hello there!";
    }
}

$doc=new DOMDocument();
$doc->registerNodeClass('DOMElement','SillyTestClass');
$doc->loadHTML("<div><h1>Sample</h1></div>");
//THIS WORKS! CUSTOM CLASS BEING USED
$doc->documentElement->firstChild->sayHello();

//THIS IS STILL NULL:( Never set by construct, no message saying construct was called either  
echo $doc->documentElement->firstChild->foo; 

当然,如果我自己实例化,那很好......

$elm=new SillyTestClass("foo","Hi there");
//WORKS! Outputs "bar";
echo $elm->foo;

为什么当我用DOMDocument注册节点类时,它不会调用__construct偶数虽然它以其他方式给我正确的继承？

更新对于真正好奇的人或知道C的人

================================================== ===================== 调查......

这是从github上的PHP src获取的DOM扩展源代码

如果你要创建一个元素,这就是发生的事件链::

document.c :: dom_document_create_element
    |  //uses libxml to generate a new DOMNode
    | node = xmlNewDocNode(docp, NULL, (xmlChar *) name, (xmlChar *) value);

    // that node is then sent to 
  php_dom.c :: php_dom_create_object
   |
   |  //the node type is used to figure out what extension class to use
   |     switch (obj->type) {...
   |    
   |     //that class is used to instance an object
   |     if (domobj && domobj->document) {
   |          ce = dom_get_doc_classmap(domobj->document, ce);
   |     }
        object_init_ex(return_value, ce);

看来你没有从扩展DOMNode获得真正的继承,或者如果DOMDocument实例化它们,它就是内置的扩展类(DOMElement,DOMText).在这种情况下,首先创建libxml节点,然后在第二个节点上添加类属性.

这是不幸的,似乎不可能绕过它,因为即使将importNode导入文档,它也会实例化一个新节点.例

class extendsDE extends DOMElement{
   public $constructWasCalled=false;
    public function __construct($name){
       parent::__construct($name);
       $this->constructWasCalled=true;
   }
}

 class extendsDD extends DOMDocument{

    public function __construct(){
        parent::__construct();
        $this->registerNodeClass("DOMElement","extendsDE");
    }
    //@override
    public function createElement($name){
        $elm=new extendsDE($name);
        echo "Element construct called when we create=";
        echo $elm->constructWasCalled?"true":"false";
        return $this->importNode($elm);
   }
}

$doc=new extendsDD();
$node=$doc->createElement("div");
echo "<br/>";
echo "But then when we import into document, a new element is created  and construct called= ";
echo $node->constructWasCalled?"true":"false"; 

现在辩论 - 这是开发人员的意图和文档是误导,或者它是一个错误和真正的继承应该发生？

Answer 1

到目前为止，我已经在某些情况下找到了解决这个问题的方法。DOMNodes 保存在内存中，因此只要您可以将其完整地放入文档中（您的构造函数被调用），那么无论您用它做什么或之后如何访问它都会没问题......

这是一个示例，我们可以调用该构造并且仍然可以处理文档

class extendsDE extends DOMElement{
    public $constructWasCalled=false;
    public function __construct($name){
         parent::__construct($name);
          $this->constructWasCalled=true;
   }
  }

   $doc=new DOMDocument();
   $doc->registerNodeClass("DOMElement","extendsDE");
   $doc->loadHTML("<div></div>");

   //append a node we create manually rather than through createElement
   $node=$doc->getElementsByTagName('div')->item(0)->appendChild(new extendsDE("p"));
    $node->nodeValue="Was my construct called?";
     echo "<br/>";
    echo "A new element was appended and construct called= ";
    echo $node->constructWasCalled?"true":"false";
    echo "<br/>";
    echo "Okay but what happens if I retrieve that node some other way..";
    echo "<br/>";
    echo "what if I get that element through selectors. Custom property still set from constructor=";
   echo $doc->getElementsByTagName('p')->item(0)->constructWasCalled?"true":"false";
  echo "<br/>";
  echo "what if I get that element through relationships. Custom property still set from constructor=";
  echo $doc->getElementsByTagName('div')->item(0)->childNodes->item(0)->constructWasCalled?"true":"false";

要点：

仅当您创建所有元素时才有效。如果您使用 $document->loadHTML($html) 加载 HTML 标记，则不会调用您的扩展构造函数。到目前为止，我只能想到解决这个问题的黑客方法，例如加载标记，然后将每个节点循环到实例副本并插入它们。绝对可以，但是慢...