Question

说我的XML看起来像这样：

<record>
  <row name="title">this item</row>
  <row name="url">this url</row>
</record>

现在我正在做这样的事情：

$xml = new DOMDocument();
$xml->load('xmlfile.xml');

echo $xml->getElementByTagName('row')->item(0)->attributes->getNamedItem('title')->nodeValue;

但这只是给了我：

注意：尝试获取非对象ID的属性

有人知道如何获取“name”属性值为“title”的节点值吗？

Answer 1

尝试：

$xml = new DOMDocument();
$xml->loadXml('
<record>
  <row name="title">this item</row>
  <row name="url">this url</row>
</record>
');

$xpath = new DomXpath($xml);

// traverse all results
foreach ($xpath->query('//row[@name="title"]') as $rowNode) {
    echo $rowNode->nodeValue; // will be 'this item'
}

// Or access the first result directly
$rowNode = $xpath->query('//row[@name="title"][1]')->item(0);
if ($rowNode instanceof DomElement) {
    echo $rowNode->nodeValue;
}

Answer 2

foreach ($xml->getElementsByTagName('row') as $element)
{
if ($element->getAttribute('name') == "title")
{
 echo $element->nodeValue;
}
}

Answer 3

$xpath = new DOMXPath( $xml );
$val = $xpath->query( '//row[@name="title"]' )->item(0)->nodeValue;

php domdocument获取属性值所在的节点值

3 个答案: