Question

我有以下html

$html =  '<body><div style="font-color:#000">Hello</div>
<span style="what">My name is rasid</span><div>new to you 
</div><div style="rashid">New here</div></body>';

$dom = new DOMDocument();
$dom->loadHTML($html);

$elements = $dom->getElementsByTagName('body');

我试过了

foreach($elements as $child)
 {
        echo $child->nodeName;
 }

输出

body

但是我需要遍历身体下面的所有标签而不是身体。我怎么能这样做。

我也在上面的例子中尝试过替换

$elements = $dom->getElementsByTagName('body');

带

$elements = $dom->getElementsByTagName('body')->item(0);

但它给出了错误。任何解决方案??

Answer 1

试试这个

    $elements = $dom->getElementsByTagName('*');
    $i = 1; //counter to output from 3rd one, since foreach loop below will output" html  body div span div div"

    foreach($elements as $child)
    {

    if ($i > 2) echo $child->nodeName."<br>";   //output "div span div div"
    ++$i;
    }

Answer 2

使用此代码

 $elements = $dom->getElementsByTagName('*');


        foreach($elements as $child)
        {

              echo $child->nodeName;

        }

Answer 3

如果您只想要body元素的子节点，可以使用：

$body = $dom->getElementsByTagName( 'body' )->item( 0 );
foreach( $body->childNodes as $node )
{
    echo $node->nodeName . PHP_EOL;
}

如果您想要body元素的所有降序节点，可以使用DOMXPath：

$xpath = new DOMXPath( $dom );
$bodyDescendants = $xpath->query( '//body//node()' );
foreach( $bodyDescendants as $node )
{
    echo $node->nodeName . PHP_EOL;
}

如何遍历PHP DOMDocument中标记下的所有Childs

3 个答案: