Question

在PHP中扩展DOMElement时，不会调用子类的构造函数。在文档中没有任何东西像我预期的行为那样跳出来，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，如果我自己实例化它很好......

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

为什么当我使用DOMDocument注册节点类时，它是否会调用__construct，即使它以其他方式为我提供了正确的继承？

更新对于真正好奇的人或知道C的人

============================================ =========================== 的调查...

这是取自 PHP src on github

的DOM扩展源

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);

如果DOMDocument实例化它们，那么你似乎没有从扩展DOMNode或它内置的扩展类（DOMElement，DOMText）中获得真正的继承。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。实施例

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性的，还是一个错误，应该发生真正的继承？

Answer 1

到目前为止，我已经在某些情况下找到了解决方法。 DOMNode保存在内存中，因此只要你可以将它一起放入文档中（你的构造函数被调用），那么无论你用它做什么或者之后如何访问它都没关系......

这是一个示例，我们可以使用文档

来获取构造以进行调用

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";

THE CATCH：

只有在创建所有元素时才有效。如果使用$ document-＆gt; loadHTML（$ html）加载HTML标记，则您的扩展构造函数不会被调用。到目前为止，我只能想到解决这个问题的方法，比如加载标记然后将每个节点循环到实例副本并插入它们。绝对可能，但很慢......

没有调用扩展PHP DOMElement

1 个答案: