Question

Notice: Undefined property: DOMNodeList::$id in D:\wamp\www\xml\index.php on line 15
id:

Notice: Undefined property: DOMNodeList::$name in D:\wamp\www\xml\index.php on line 16
name:

<?php
$xml = new DOMDocument();

$xml->load('test.xml');

$xpath = new DOMXPath($xml);

$query = '/people/person[id="33333"]';

$entries = $xpath->query($query);

echo 'id:'. $entries->id.'<br/>';
echo 'name:'.$entries->name.'<br/>';
?>

xml文件样本：

<people>
...
    <person>
        <phone>33333</phone>
        <name>Aadgar</name>
        <last_name>Adas</last_name>
    </person>
...
</people>

Answer 1

首先，id节点不存在...

$query = '/people/person[id="33333"]';

我想你想要：

$query = '/people/person[phone="33333"]';

然后，你必须这样做：

$entries = $xpath->query($query);

foreach ($entries as $entry) {
   echo 'name:'. $entry->getElementsByTagName('name')->item(0)->nodeValue.'<br/>';
   echo 'last_name:'.$entry->getElementsByTagName('last_name')->item(0)->nodeValue.'<br/>';
}

您似乎在混淆DOM和SimpleXML语法。

Answer 2

你的xpath错了。 33333是手机不是id。

尝试

$query = '/people/person[phone="33333"]';

另外

他们没有身份

echo 'id:'. $entries->id.'<br/>';
echo 'name:'.$entries->name.'<br/>';

尝试删除id部分

Answer 3

试试这个：

<?php
    $xml = new DOMDocument();

    $xml->load('test.xml');

    $xpath = new DOMXPath($xml);

    $query = '/people/person/phone';

    $entries = $xpath->query($query);

    foreach($entries  as $entry)
    {
       echo $entry;
    }
    ?>

xml + xpath，print元素

3 个答案: