使用浏览器转换XML(谷歌浏览器或IE7)时,是否可以通过URL将参数传递给XSLT样式表？
例:
data.xml中
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="sample.xsl"?>
<root>
<document type="resume">
<author>John Doe</author>
</document>
<document type="novella">
<author>Jane Doe</author>
</document>
</root>
sample.xsl
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet
version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:output method="html" />
<xsl:template match="/">
<xsl:param name="doctype" />
<html>
<head>
<title>List of <xsl:value-of select="$doctype" /></title>
</head>
<body>
<xsl:for-each select="//document[@type = $doctype]">
<p><xsl:value-of select="author" /></p>
</xsl:for-each>
</body>
</html>
</<xsl:stylesheet>