[英]PHP Query XML with xPath
我有一個如下的XML結構:
<Tickets>
<EventsPoints>
<Event ID="23">
<PerformanceName>U2</PerformanceName>
<EventDate>25/05/2012</EventDate>
<EventPrice>75.00</EventPrice>
</Event>
<Event ID="27">
<PerformanceName>Jedward</PerformanceName>
<EventDate>28/05/2012</EventDate>
<EventPrice>20.00</EventPrice>
</Event>
<Event ID="27">
<PerformanceName>Rolling Stones</PerformanceName>
<EventDate>03/12/2012</EventDate>
<EventPrice>80.00</EventPrice>
</Event>
</EventsPoints>
</Tickets>
基本上,我想在XML中搜索某個性能名稱,例如“ U2”,然后返回整個XML塊(即性能名稱,事件日期和價格-全部以格式化的XML格式保存,並保存在單獨的xml文件中)
這是我的PHP代碼,但似乎無法正確提取數據:
$srcDom = new DOMDocument;
$srcDom->load('/var/www/html/xml/searchfile.xml');
$xPath = new DOMXPath($srcDom);
foreach ($srcDom->getElementsByTagName('Event') as $event) {
$dstDom = new DOMDocument('1.0', 'utf-8');
$dstDom->appendChild($dstDom->createElement('EventsPricePoints'));
$dstDom->documentElement->appendChild($dstDom->importNode($event, true));
$allEventsForVenue = $xPath->query(
sprintf(
'/Tickets/EventsPoints/Event/PerformanceName[.="U2"]'
)
);
foreach ($allEventsForVenue as $event) {
$dstDom->documentElement->appendChild($dstDom->importNode($event, true));
}
$dstDom->formatOutput = true;
$dstDom->save(sprintf('/var/www/html/xml/searchresults1.xml'));
}
當您需要父元素時,您的XPath將獲得PerformanceName
元素。 更改為
/Tickets/EventsPoints/Event/PerformanceName[.="U2"]/..
要么
/Tickets/EventsPoints/Event[PerformanceName[.="U2"]]
或導入
$event->parentNode
另外,您不需要第一個foreach
。 刪除它,然后移動將$dstDom
寫入到XPath結果的代碼中的代碼。 看到http://codepad.org/zfOZXycZ
另請參閱:
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