XSLT 将 XML 转换为分层格式

Question

我有一个简单形式的包含 partyId 和 parentPartyId 值的 xml。我想把它转换成层次树格式的形式。

我已根据所需的输出创建了架构。我正在http://www.keller.com/xslt/8/ 中尝试轴名称表达式。我没有得到，如何将源转换为所需的格式？

Source和需要的Target如下。

来源：

<OutputCollection xmlns="http://xmlns.test.com/pcbpel/adapter/db/PartyHierarchy">
            <Output>
               <level>1</level>
               <Parent_Party xsi:nil="true"/>
               <Party>Party-1</Party>
            </Output>
            <Output>
               <level>2</level>
               <Parent_Party>Party-1</Parent_Party>
               <Party>Party-1-1</Party>
            </Output>
            <Output>
               <level>2</level>
               <Parent_Party>Party-1</Parent_Party>
               <Party>Party-1-2</Party>
            </Output>
            <Output>
               <level>3</level>
               <Parent_Party>Party-1-2</Parent_Party>
               <Party>Party-1-2-1</Party>
            </Output>
            <Output>
               <level>3</level>
               <Parent_Party>Party-1-2</Parent_Party>
               <Party>Party-1-2-2</Party>
            </Output>
            <Output>
               <level>3</level>
               <Parent_Party>Party-1-2</Parent_Party>
               <Party>Party-1-2-3</Party>
            </Output>
            <Output>
               <level>2</level>
               <Parent_Party>Party-1</Parent_Party>
               <Party>Party-1-3</Party>
            </Output>
            <Output>
               <level>3</level>
               <Parent_Party>Party-1-3</Parent_Party>
               <Party>Party-1-3-1</Party>
            </Output>
            <Output>
               <level>3</level>
               <Parent_Party>Party-1-3</Parent_Party>
               <Party>Party-1-3-2</Party>
            </Output>
         </OutputCollection>

目标：

<OutputCollection xmlns="http://xmlns.test.com/pcbpel/adapter/db/PartyHierarchy">
            <Output>
               <level>1</level>
               <Parent_Party xsi:nil="true"/>
               <Party>Party-1</Party>
               <Children>
                    <level>2</level>
                    <Parent_Party>Party-1</Parent_Party>
                    <Party>Party-1-1</Party>
               </Children>
               <Children>
                    <level>2</level>
                    <Parent_Party>Party-1</Parent_Party>
                    <Party>Party-1-2</Party>
                    <Children>
                        <level>3</level>
                        <Parent_Party>Party-1-2</Parent_Party>
                        <Party>Party-1-2-1</Party>
                    </Children>
                    <Children>
                        <level>3</level>
                        <Parent_Party>Party-1-2</Parent_Party>
                        <Party>Party-1-2-2</Party>
                    </Children>
                    <Children>
                        <level>3</level>
                        <Parent_Party>Party-1-2</Parent_Party>
                        <Party>Party-1-2-3</Party>                  
                    </Children>
               </Children>
               <Children>
                    <level>2</level>
                    <Parent_Party>Party-1</Parent_Party>
                    <Party>Party-1-3</Party>
                    <Children>
                        <level>3</level>
                        <Parent_Party>Party-1-3</Parent_Party>
                        <Party>Party-1-3-1</Party>                  
                    </Children>
                    <Children>
                        <level>3</level>
                        <Parent_Party>Party-1-3</Parent_Party>
                        <Party>Party-1-3-2</Party>          
                    </Children>                 
               </Children>
            </Output>
         </OutputCollection>

Answer 1

对于 XSLT，如果您需要遵循交叉引用，那么您需要定义一个键 <xsl:key name="children" match="ph:Output" use="ph:Parent_Party"/> 并使用它 <xsl:apply-templates select="key('children', ph:Party)"/>：

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:ph="http://xmlns.test.com/pcbpel/adapter/db/PartyHierarchy"
  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  xmlns="http://xmlns.test.com/pcbpel/adapter/db/PartyHierarchy"
  exclude-result-prefixes="ph xsi"
  version="1.0">

<xsl:output indent="yes"/>
<xsl:strip-space elements="*"/>

<xsl:key name="children" match="ph:Output" use="ph:Parent_Party"/>

<xsl:template match="/*">
  <xsl:copy>
    <xsl:apply-templates select="key('children', '')"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="ph:Output[ph:Parent_Party/@xsi:nil = 'true']">
  <xsl:copy>
    <xsl:copy-of select="node()"/>
    <xsl:apply-templates select="key('children', ph:Party)"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="ph:Output">
  <children>
    <xsl:copy-of select="node()"/>
    <xsl:apply-templates select="key('children', ph:Party)"/>
  </children>
</xsl:template>

</xsl:stylesheet>

请注意，您发布的输入示例使用前缀 xsi，但未声明它，因此我必须为该前缀假设一个命名空间。