根据多次出现的节点创建记录副本

Question

我需要根据多次出现的字段创建 XML 文件的副本。

输入XML：XML中有两个Employment节点。我需要将它们分开并复制其余节点，以便我有两个 PerPerson 记录，每个记录有一个 EmpEmployment。

<PerPerson>
    <PerPerson>
        <personalInfoNav>
            <PerPersonal/>
        </personalInfoNav>
        <nationalIdNav>
            <PerNationalId/>
        </nationalIdNav>
        <personIdExternal>AA</personIdExternal>
        <personEmpTerminationInfoNav>
            <PersonEmpTerminationInfo/>
        </personEmpTerminationInfoNav>
        <phoneNav>
            <PerPhone/>
        </phoneNav>
        <employmentNav>
            <EmpEmployment>
                <compInfoNav>
                    <EmpCompensation/>
                </compInfoNav>
                <jobInfoNav>
                    <EmpJob/>
                </jobInfoNav>
            </EmpEmployment>
            <EmpEmployment>
                <compInfoNav>
                    <EmpCompensation/>
                </compInfoNav>
                <jobInfoNav>
                    <EmpJob/>
                </jobInfoNav>
            </EmpEmployment>
        </employmentNav>
        <homeAddressNavDEFLT>
            <PerAddressDEFLT/>
        </homeAddressNavDEFLT>
    </PerPerson>
</PerPerson>

我正在尝试使用 XSLT 1.0 执行此操作

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:hci="http://sap.com/it/" exclude-result-prefixes="hci">
    <xsl:strip-space elements="*"/>
    <xsl:output encoding="utf-8" indent="yes" method="xml"/>
    <xsl:template match="/">
        <PerPerson>
            <xsl:for-each select="PerPerson/PerPerson">
                    <xsl:variable name="var_person" select="./*[not(name()='EmpEmployment')]"/>
                    <xsl:for-each select="employmentNav/EmpEmployment">
                        <xsl:variable name="var_empInfo" select="."/>
                            <PerPerson>
                                <xsl:copy-of select="$var_person"/>
                                <xsl:copy-of select="$var_empInfo"/>
                        </PerPerson>
                    </xsl:for-each>
            </xsl:for-each>
        </PerPerson>
    </xsl:template>
</xsl:stylesheet>

它没有按预期工作。我无法在下面创建所需的输出：

<PerPerson>
    <PerPerson>
        <personalInfoNav>
            <PerPersonal/>
        </personalInfoNav>
        <nationalIdNav>
            <PerNationalId/>
        </nationalIdNav>
        <personIdExternal>AA</personIdExternal>
        <personEmpTerminationInfoNav>
            <PersonEmpTerminationInfo/>
        </personEmpTerminationInfoNav>
        <phoneNav>
            <PerPhone/>
        </phoneNav>
        <employmentNav>
            <EmpEmployment>
                <compInfoNav>
                    <EmpCompensation/>
                </compInfoNav>
                <jobInfoNav>
                    <EmpJob/>
                </jobInfoNav>
            </EmpEmployment>
        </employmentNav>
        <homeAddressNavDEFLT>
            <PerAddressDEFLT/>
        </homeAddressNavDEFLT>
    </PerPerson>
        <PerPerson>
        <personalInfoNav>
            <PerPersonal/>
        </personalInfoNav>
        <nationalIdNav>
            <PerNationalId/>
        </nationalIdNav>
        <personIdExternal>AA</personIdExternal>
        <personEmpTerminationInfoNav>
            <PersonEmpTerminationInfo/>
        </personEmpTerminationInfoNav>
        <phoneNav>
            <PerPhone/>
        </phoneNav>
        <employmentNav>
            <EmpEmployment>
                <compInfoNav>
                    <EmpCompensation/>
                </compInfoNav>
                <jobInfoNav>
                    <EmpJob/>
                </jobInfoNav>
            </EmpEmployment>
        </employmentNav>
        <homeAddressNavDEFLT>
            <PerAddressDEFLT/>
        </homeAddressNavDEFLT>
    </PerPerson>
</PerPerson>

Answer 1

你可以这样做：

XSLT 1.0

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<xsl:template match="/PerPerson">
    <PerPerson>
        <xsl:for-each select="PerPerson/employmentNav/EmpEmployment">
            <PerPerson>
                <xsl:copy-of select="../preceding-sibling::*"/>
                <employmentNav>
                    <xsl:copy-of select="."/>
                </employmentNav>
                <xsl:copy-of select="../following-sibling::*"/>
            </PerPerson>
        </xsl:for-each>
    </PerPerson>
</xsl:template>     

</xsl:stylesheet>

Answer 2

在 XSLT 2.0 中，您可以使用 tunnel-parameters。

然后像下面这样使用模板 macht 引擎

<?xml version='1.0' encoding='UTF-8'?>
<xsl:stylesheet 
  version="2.0" 
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
  xmlns:xs="http://www.w3.org/2001/XMLSchema" 
  exclude-result-prefixes="#all">

  <xsl:template match="@*|node()" mode="#all">
    <xsl:copy>
      <xsl:apply-templates select="@*|node()" mode="#current"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="PerPerson/PerPerson">
    <xsl:variable name="person" select="."/>
    <xsl:for-each select="employmentNav/EmpEmployment">
      <xsl:apply-templates select="$person" mode="denormalize">
        <xsl:with-param name="position" as="xs:integer" select="position()" tunnel="yes"/>
      </xsl:apply-templates>
    </xsl:for-each>
  </xsl:template>
  
  <xsl:template match="employmentNav" mode="denormalize">
    <xsl:param name="position" as="xs:integer" tunnel="yes"/>    
    <xsl:copy>
      <xsl:apply-templates select="EmpEmployment[$position]"/>
    </xsl:copy>
  </xsl:template>

</xsl:stylesheet>

根据多次出现的节点创建记录副本

Create copy of record based on node that occurs multiple times

xslt

xslt-1.0