我希望学习XSLT,但我更喜欢通过实例来学习。 我想执行一个简单的模式到模式转换。 如何在一次转换中执行此转换(我的当前解决方案使用了两个步骤,并且丢失了客户的原始顺序)?
来源:
<?xml version="1.0" encoding="UTF-8"?>
<sampleroot>
<badcustomer>
<name>Donald</name>
<address>Hong Kong</address>
<age>72</age>
</badcustomer>
<goodcustomer>
<name>Jim</name>
<address>Wales</address>
<age>22</age>
</goodcustomer>
<goodcustomer>
<name>Albert</name>
<address>France</address>
<age>51</age>
</goodcustomer>
</sampleroot>
收件人:
<?xml version="1.0" encoding="UTF-8"?>
<records>
<record id="customer">
<name>Donald</name>
<address>Hong Kong</address>
<age>72</age>
<customertype>bad</customertype>
</record>
<record id="customer">
<name>Jim</name>
<address>Wales</address>
<age>22</age>
<customertype>good</customertype>
</record>
<record id="customer">
<name>Albert</name>
<address>France</address>
<age>51</age>
<customertype>good</customertype>
</record>
</records>
我已经用一种不好的方法解决了这个问题(我失去了客户的顺序,我认为我必须多次解析文件):
<?xml version='1.0'?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/sampleroot">
<records>
<xsl:for-each select="goodcustomer">
<record id="customer">
<name><xsl:value-of select="name" /></name>
<address><xsl:value-of select="address" /></address>
<age><xsl:value-of select="age" /></age>
<customertype>good</customertype>
</record>
</xsl:for-each>
<xsl:for-each select="badcustomer">
<record id="customer">
<name><xsl:value-of select="name" /></name>
<address><xsl:value-of select="address" /></address>
<age><xsl:value-of select="age" /></age>
<customertype>bad</customertype>
</record>
</xsl:for-each>
</records>
</xsl:template>
</xsl:stylesheet>
请问是否有人能帮我用正确的XSLT结构,只需使用一次解析(仅一个for-each)?谢谢,Chris。