使用XSLT程序将具有逗号分隔值的XML元素分组

Question

我们是xslt编程的新手，你能帮助我们学习xslt程序吗。我们需要基于"；id"；标记，并用逗号连接另一个xml标记。

输入xml文件：

<?xml version="1.0" encoding="UTF-8"?>
<root>
<row>
<id>123</id>
<functional_manager__c.users>1234567</functional_manager__c.users>
</row>
<row>
<id>123</id>
<functional_manager__c.users>1200000</functional_manager__c.users>
</row>
<row>
<id>111</id>
<functional_manager__c.users>11111111</functional_manager__c.users>
</row>
<row>
<id>111</id>
<functional_manager__c.users>2222222</functional_manager__c.users>
</row>
<row>
<id>123</id>
<editor__v.users>1234567</editor__v.users>
</row>
<row>
<id>123</id>
<editor__v.users>1200000</editor__v.users>
</row>
<row>
<id>111</id>
<learning_partner__c.users>11111111</learning_partner__c.users>
</row>
<row>
<id>111</id>
<learning_partner__c.users>2222222</learning_partner__c.users>
</row>
</root>

所需输出：

<?xml version="1.0" encoding="UTF-8"?>
<root>
<row>
<id>123</id>
<functional_manager__c.users>1234567,1200000</functional_manager__c.users>
</row>
<row>
<id>111</id>
<functional_manager__c.users>11111111,2222222</functional_manager__c.users>
</row>
<row>
<id>123</id>
<editor__v.users>1234567,1200000</editor__v.users>
</row>
<row>
<id>111</id>
<learning_partner__c.users>11111111,2222222</learning_partner__c.users>
</row>
</root>

我们尝试过的代码：

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0" exclude-result-prefixes="xsl wd xsd this env"
xmlns:wd="urn:com.workday/bsvc"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:env="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:this="urn:this-stylesheet">
<xsl:output indent="yes" method="xml"/>
<xsl:template match="/">
<Sharingsettings>
<xsl:for-each-group select="/root/row" group-by="id">
<row>
<ID>
<xsl:value-of select="id"/>
</ID>
<functional_manager__c.users>
<xsl:value-of select="//current-group()//functional_manager__c.users">
</xsl:value-of>
</functional_manager__c.users>
</row>
</xsl:for-each-group>
</Sharingsettings>
</xsl:template>
</xsl:stylesheet>

我们正在尝试使用XSLT程序，但它不能正确地提供所需的输出。

提前感谢

Answer 1

使用XSLT3可以使用

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="3.0"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="#all"
expand-text="yes">
<xsl:output method="xml" indent="yes"/>
<xsl:mode on-no-match="shallow-copy"/>

<xsl:template match="root">
<xsl:copy>
<xsl:for-each-group select="row" group-adjacent="id">
<xsl:copy>
<xsl:apply-templates/>
</xsl:copy>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>

<xsl:template match="row/*[not(self::id)]">
<xsl:copy>
<xsl:value-of select="current-group()/*[node-name() = node-name(current())]" separator=","/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>