我需要转换以下XML:
<?xml version="1.0" encoding="UTF-8"?>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
我需要的目标XML是这样的:
<params>
<param>
<key>title</key>
<value>Empire Burlesque</value>
</param>
<param>
<key>artist</key>
<value>Bob Dylan<</value>
</param>
<param>
<key>country</key>
<value>USA</value>
</param>
<param>
<key>company</key>
<value>Columbia</value>
</param>
<param>
<key>price</key>
<value>10.90</value>
</param>
<param>
<key>year</key>
<value>1985</value>
</param>
</params>
XSL文件对我来说应该是什么样的?
答案 0 :(得分:1)
<强>予。 XSLT 1.0
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:output omit-xml-declaration="yes"/>
<xsl:template match="/">
<params>
<xsl:apply-templates select="cd"/>
</params>
</xsl:template>
<xsl:template match="cd/*">
<param>
<key><xsl:value-of select="name()"/></key>
<value><xsl:value-of select="."/></value>
</param>
</xsl:template>
</xsl:stylesheet>
<强> II。输出强>
<params>
<param><key>title</key><value>Empire Burlesque</value></param>
<param><key>artist</key><value>Bob Dylan</value></param>
<param><key>country</key><value>USA</value></param>
<param><key>company</key><value>Columbia</value></param>
<param><key>price</key><value>10.90</value></param>
<param><key>year</key><value>1985</value></param>
</params>