Question

如何从标签中获取所有属性？

<a href="URL1">TEXT 1</a>
<a href="URL2" rel="rel2">TEXT 2</a>
<a href="URL3" rel="me" target="_blank">TEXT 3</a>

如何使用PHP获得此结果？有没有办法用PHP DOMDocument做到这一点？

Array
(
    [0] => Array
        (
            [href] => URL1
            [text] => TEXT 1
        )

    [1] => Array
        (
            [href] => URL2
            [text] => TEXT 2
            [rel] => rel2
        )

    [2] => Array
        (
            [href] => URL3
            [text] => TEXT 3
            [rel] => me
            [target] => _blank
        )

)

Answer 1

可以使用DOMNode::attributes DOMNamedNodeMap检索节点的属性，但它可以像数组一样使用：

$html = '<a href="URL1">TEXT 1</a><a href="URL2" rel="rel2">TEXT 2</a><a href="URL3" rel="me" target="_blank">TEXT 3</a>';

$output = array();

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

$anchors = $dom->getElementsByTagName('a');
foreach($anchors as $a)
{
    $attrList = array();
    if($a->hasAttributes())
    {
        foreach ($a->attributes as $attribute)
        {
            $attrList[$attribute->nodeName] = $attribute->nodeValue
            $attrList['text'] = $a->nodeValue;
        }
    }
    $output[] = $attrList;
}

print_r($output);

<强>输出

Array
(
    [0] => Array
        (
            [href] => URL1
            [text] => TEXT 1
        )

    [1] => Array
        (
            [href] => URL2
            [text] => TEXT 2
            [rel] => rel2
        )

    [2] => Array
        (
            [href] => URL3
            [text] => TEXT 3
            [rel] => me
            [target] => _blank
        )

)

Answer 2

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

然后使用DomDocument方法（http://www.php.net/manual/en/class.domdocument.php）

进行解析

这样的事情应该有效

$a = $dom->getElementsByTagName('a');
if ($a->hasAttributes()) {
  foreach ($a->attributes as $attribute) {
    $name = $attribute->nodeName;
    $value = $attribute->nodeValue;
    echo "Attribute '$name' => '$value'<br />";
  }
}

标签中的所有属性

2 个答案: