Question

我有这个示例代码，它将提取每个标记的值。除此之外，获取该标签的类名..

<?php
$doc = new DOMDocument;
$doc->loadxml( <<< eox
<tr class="calendar_row" data-eventid="42023">
    <td class="date"/>
    <td class="time">All Day</td>
    <td class="currency">CAD</td>
    <td class="impact">
        <span title="Non-Economic" class="holiday"/>
    </td>
    <td class="event">
        <span>Bank Holiday</span>
    </td>
    <td class="detail">
        <a class="calendar_detail level1" data-level="1" title="Open Detail"/>
    </td>
    <td class="actual"/>
    <td class="forecast"/>
    <td class="previous"/>
    <td class="graph"/>
</tr>
eox
);
$xpath = new DOMXPath($doc);

foreach( $xpath->query('//tr[@data-eventid="42023"]/td[@class]') as $n ) {
    echo $n->nodeName.'-'.$n->nodeValue."<br />";
}
?>

使用上面的代码片段，我想要的就是获取这些值，即使某些标签格式不正确（我正在废弃网络资源）。我如何在DOMDocument XPath查询中执行此操作。我遇到了麻烦，因为要获取的值是：

td-
td-All Day
td-CAD
td- 
td-Bank Holiday 
td- 
td-
td-
td-
td-

而不是：

date-
time-All Day
currency-CAD
impact- 
event-Bank Holiday 
detail- 
actual-
forecast-
previous-
graph-

Answer 1

而不是$n->nodeName你应该这样做$n->getAttribute('class')。

演示：http://codepad.viper-7.com/ktpnv2

Answer 2

echo $n->getAttribute("class") . '-' . $n->nodeValue . "<br />";

DOMDocument XPath查询以获取类名和标记值

2 个答案: