Question

我有这个DOMElement。

我有两个问题：

1）什么意思是省略了对象值？

2）如何从此DOMElement中获取属性？

 object(DOMElement)#554 (18) {
        ["tagName"]=>
        string(5) "input"
        ["schemaTypeInfo"]=>
        NULL
        ["nodeName"]=>
        string(5) "input"
        ["nodeValue"]=>
        string(0) ""
        ["nodeType"]=>
        int(1)
        ["parentNode"]=>
        string(22) "(object value omitted)"
        ["childNodes"]=>
        string(22) "(object value omitted)"
        ["firstChild"]=>
        NULL
        ["lastChild"]=>
        NULL
        ["previousSibling"]=>
        string(22) "(object value omitted)"
        ["nextSibling"]=>
        string(22) "(object value omitted)"
        ["attributes"]=>
        string(22) "(object value omitted)"
        ["ownerDocument"]=>
        string(22) "(object value omitted)"
        ["namespaceURI"]=>
        NULL
        ["prefix"]=>
        string(0) ""
        ["localName"]=>
        string(5) "input"
        ["baseURI"]=>
        NULL
        ["textContent"]=>
        string(0) ""
      }

我让这个类访问该对象。这样做的目的是我可以从输入字段中获取type属性。

<?php

namespace App\Model;

class Field
{
    /**
     * @var \DOMElement
     */
    protected $node;

    public function __construct($node){
        $this->node = $node;
    }

    public function getNode(){
        return $this->node;
    }

    public function getTagName(){

        foreach ($this->node as $value) {
            return $value->tagName;
        }
    }

    public function getAttribute(){


    }
}

Answer 1

我认为(object value omitted)只是一些内部DOM或var_dump()限制，以防止转储太深和/或转储有关对象图的递归信息。

然后，关于获取有关属性的信息：

要获取DOMElement的所有属性，您可以使用attributes属性，该属性在其父类DOMNode上定义并返回DOMNamedNodeMap个DOMAttr个节点：

// $this->node must be a DOMElement, otherwise $attributes will be NULL
$attributes = $this->node->attributes;

// then, to loop through all attributes:
foreach( $attributes as $attribute ) {
  // do something with $attribute (which will be a DOMAttr instance)
}
// or, perhaps like so:
for( $i = 0; $i < $attributes->length; $i++ ) {
  $attribute = $attributes->item( $i );
  // do something with $attribute (which will be a DOMAttr instance)
}

// to get a single attribute from the map:
$typeAttribute = $attributes->getNamedItem( 'type' );
// (which will be a DOMAttr instance or NULL if it doesn't exist)

要从type获取一个名为DOMElement的属性，您可以使用：
- DOMElement::getAttributeNode()，以获取代表DOMAttr属性的type节点，如下所示：
```
$typeAttr = $this->node->getAttributeNode( 'type' );
// (which will be NULL if it doesn't exist)
```
  或
- DOMElement::getAttribute()，获取属性type的字符串值，如下所示：
```
$typeAttrValue = $this->node->getAttribute( 'type' );
// (which will an empty string if it doesn't exist)
```

如何从DOMElement获取属性

1 个答案: