我是XML/XSL的新手.我希望能够在规则字符串中传递var并返回正确的数据.
现在我有这个PHP:
<?php
$params = array('id' => $_GET['id']);
$xslDoc = new DOMDocument();
$xslDoc->load("test.xsl");
$xmlDoc = new DOMDocument();
$xmlDoc->load("test.xml");
$xsltProcessor = new XSLTProcessor();
$xsltProcessor->registerPHPFunctions();
$xsltProcessor->importStyleSheet($xslDoc);
foreach ($params as $key => $val)
$xsltProcessor->setParameter('', $key, $val);
echo $xsltProcessor->transformToXML($xmlDoc);
?>
我的xml文件如下所示:
<Profiles>
<Profile>
<id>1</id>
<name>john doe</name>
<dob>188677800</dob>
</Profile>
<Profile>
<id>2</id>
<name>mark antony</name>
<dob>79900200</dob>
</Profile>
<Profile>
<id>3</id>
<name>neo anderson</name>
<dob>240431400</dob>
</Profile>
<Profile>
<id>4</id>
<name>mark twain</name>
<dob>340431400</dob>
</Profile>
<Profile>
<id>5</id>
<name>frank hardy</name>
<dob>390431400</dob>
</Profile>
</Profiles>
我的xsl看起来像这样
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:param name="id" />
<xsl:template match="*">
<html><body>
<h2>Profile</h2>
<table cellspacing="1" cellpadding="5" border="1">
<caption>User Profiles</caption>
<tr><th>ID</th><th>Name</th><th>Date of Birth</th></tr>
<xsl:for-each select="/Profiles/Profile[id='$id']">
<tr>
<td><xsl:value-of select="id"/></td>
<td><xsl:value-of select="php:function('ucwords', string(name))"/></td>
<td><xsl:value-of select="php:function('date', 'jS M, Y', number(dob))"/></td>
</tr>
</xsl:for-each>
</table>
</body></html>
</xsl:template>
</xsl:stylesheet>
当我像这样测试网址时:
http://foo.com/sanbox/index.php?id=2
我只得到:
Profile User Profiles ID Name Date of Birth.
该错误是因为您没有包含正确的命名空间.
在你的xsl:stylesheet声明中包含xmlns:php ="http://php.net/xsl"
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:php="http://php.net/xsl"
>