我想获取属性xsi:schemaLocation的内容。它与php中的getElementsByTagName完美配合(以及之后的foreach),但它很难看,对吗?
如何通过简单的Xpath查询获得相同的内容?
这是xml内容的简短示例:
<?xml version="1.0" encoding="utf-8"?>
<gpx xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" version="1.0" creator="blabla" xsi:schemaLocation="http://www.topografix.com/GPX/1/0 http://www.topografix.com/GPX/1/0/gpx.xsd http://www.groundspeak.com/cache/1/0/1 http://www.groundspeak.com/cache/1/0/1/cache.xsd" xmlns="http://www.topografix.com/GPX/1/0">
...
</gpx>
谢谢!
最佳答案
使用SimpleXMLElement class可以轻松获取属性xsi:schemaLocation的值:
<?php
$xml = <<<XML
<?xml version="1.0" encoding="utf-8"?>
<gpx xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" version="1.0" creator="blabla" xsi:schemaLocation="http://www.topografix.com/GPX/1/0 http://www.topografix.com/GPX/1/0/gpx.xsd http://www.groundspeak.com/cache/1/0/1 http://www.groundspeak.com/cache/1/0/1/cache.xsd" xmlns="http://www.topografix.com/GPX/1/0">
</gpx>
XML;
$sxe = new SimpleXMLElement($xml);
$schemaLocation = $sxe->attributes('xsi', true)->schemaLocation;
echo (string) $schemaLocation;