Php DOMDocument XPath查询以获取类名和标记值
我有这个示例代码,它将提取每个标记的值。 除此之外,还要得到该标记的类名Php DOMDocument XPath查询以获取类名和标记值,php,xpath,domdocument,xpathquery,Php,Xpath,Domdocument,Xpathquery,我有这个示例代码,它将提取每个标记的值。 除此之外,还要得到该标记的类名 <?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=
<?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 />";
}
?>
而不是:
date-
time-All Day
currency-CAD
impact-
event-Bank Holiday
detail-
actual-
forecast-
previous-
graph-
您不应该执行
$n->nodeName
而是执行$n->getAttribute('class')
演示:你应该做这个
$n->getAttribute('class')
而不是做$n->nodeName
演示:echo$n->getAttribute(“类”)。-$n->nodeValue。“
”;
echo$n->getAttribute(“类”)。-”$n->nodeValue。“
”;
date-
time-All Day
currency-CAD
impact-
event-Bank Holiday
detail-
actual-
forecast-
previous-
graph-
echo $n->getAttribute("class") . '-' . $n->nodeValue . "<br />";