如何在比较中使用子节点合并重复节点

Question

<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="Test1.xsl"?>
<root>
  <RemittanceInformation>
    <EntityAssignedNumber>25</EntityAssignedNumber>
    <IndividualRemittance>
      <IndividualName>
        <LastName>Johnson</LastName>
        <FirstName>Steve</FirstName>
        <ExchangeAssignedSubscriberID>6650442525</ExchangeAssignedSubscriberID>
      </IndividualName>
      <ExchangeAssignedQHPID>38408SC221000101</ExchangeAssignedQHPID>
      <ExchangeAssignedPolicyID>26141334</ExchangeAssignedPolicyID>
      <IssuerAssignedPolicyID>39147964</IssuerAssignedPolicyID>
      <IssuerAssignedSubscriberID>101009913000</IssuerAssignedSubscriberID>
    </IndividualRemittance>
    <RemittanceDetail>
      <ExchangePaymentCode>APTC</ExchangePaymentCode>
      <PaymentAmount>214.00</PaymentAmount>
      <PaymentCoverageStartDate>2016-10-01</PaymentCoverageStartDate>
      <PaymentCoverageEndDate>2016-10-31</PaymentCoverageEndDate>
    </RemittanceDetail>
    <RemittanceDetail>
      <ExchangePaymentCode>UF</ExchangePaymentCode>
      <PaymentAmount>-43.04</PaymentAmount>
      <PaymentCoverageStartDate>2016-10-01</PaymentCoverageStartDate>
      <PaymentCoverageEndDate>2016-10-31</PaymentCoverageEndDate>
    </RemittanceDetail>
  </RemittanceInformation>
    <RemittanceInformation>
    <EntityAssignedNumber>26</EntityAssignedNumber>
    <IndividualRemittance>
      <IndividualName>
        <LastName>Johnson</LastName>
        <FirstName>Steve</FirstName>
        <ExchangeAssignedSubscriberID>0000442525</ExchangeAssignedSubscriberID>
      </IndividualName>
      <ExchangeAssignedQHPID>38408SC001000101</ExchangeAssignedQHPID>
      <ExchangeAssignedPolicyID>26141334</ExchangeAssignedPolicyID>
      <IssuerAssignedPolicyID>39147964</IssuerAssignedPolicyID>
      <IssuerAssignedSubscriberID>101009913000</IssuerAssignedSubscriberID>
    </IndividualRemittance>
    <RemittanceDetail>
      <ExchangePaymentCode>APTC</ExchangePaymentCode>
      <PaymentAmount>556.00</PaymentAmount>
      <PaymentCoverageStartDate>2016-10-01</PaymentCoverageStartDate>
      <PaymentCoverageEndDate>2016-10-31</PaymentCoverageEndDate>
    </RemittanceDetail>
    <RemittanceDetail>
      <ExchangePaymentCode>UF</ExchangePaymentCode>
      <PaymentAmount>-30.50</PaymentAmount>
      <PaymentCoverageStartDate>2016-10-01</PaymentCoverageStartDate>
      <PaymentCoverageEndDate>2016-10-31</PaymentCoverageEndDate>
    </RemittanceDetail>
  </RemittanceInformation>
   <RemittanceInformation>
    <EntityAssignedNumber>27</EntityAssignedNumber>
    <IndividualRemittance>
      <IndividualName>
        <LastName>Masterson</LastName>
        <FirstName>Gene</FirstName>
        <MiddleName>E</MiddleName>
        <ExchangeAssignedSubscriberID>0032171620</ExchangeAssignedSubscriberID>
      </IndividualName>
      <ExchangeAssignedQHPID>384111C001000101</ExchangeAssignedQHPID>
      <ExchangeAssignedPolicyID>26523035</ExchangeAssignedPolicyID>
      <IssuerAssignedPolicyID>38976623</IssuerAssignedPolicyID>
      <IssuerAssignedSubscriberID>101009869500</IssuerAssignedSubscriberID>
    </IndividualRemittance>
    <RemittanceDetail>
      <ExchangePaymentCode>APTC</ExchangePaymentCode>
      <PaymentAmount>448.00</PaymentAmount>
      <PaymentCoverageStartDate>2016-10-01</PaymentCoverageStartDate>
      <PaymentCoverageEndDate>2016-10-31</PaymentCoverageEndDate>
    </RemittanceDetail>
    <RemittanceDetail>
      <ExchangePaymentCode>UF</ExchangePaymentCode>
      <PaymentAmount>-30.50</PaymentAmount>
      <PaymentCoverageStartDate>2016-10-01</PaymentCoverageStartDate>
      <PaymentCoverageEndDate>2016-10-31</PaymentCoverageEndDate>
    </RemittanceDetail>
  </RemittanceInformation>
</root>

我在上述格式的XML文件中有大量数据。一个人由一个"汇款信息"代表，但有一些重复。我想合并这些重复项，并将所有重复项中的"汇款详细信息"实例添加到该人的第一个实例中。一个人由"交换分配的保单 ID"号唯一标识。

在上面的示例中，26 号的两个 RemittanceDetail 节点应移动到 25 号的 RemittanceInformation 节点，因为它们都是具有相同 ExchangeAssignedPolicyID 的同一个人。所有后续的汇款信息节点都应向下递增，以弥补缺失的数字。

我在 S/O 上看到了类似的代码，但经过很多小时和很多咖啡后，我无法弄清楚这一点。任何帮助将不胜感激。

Answer 1

如果在 XSLT 2.0 中使用 for-each-group(请参阅 https://www.w3.org/TR/xslt20/#grouping-examples 中的示例(，则只需要

<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
    <xsl:output indent="yes"/>
    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="root">
        <xsl:copy>
            <xsl:for-each-group select="RemittanceInformation" group-by="IndividualRemittance/ExchangeAssignedPolicyID">
              <xsl:copy>
                  <xsl:apply-templates select="EntityAssignedNumber, IndividualRemittance, current-group()/RemittanceDetail"/>
              </xsl:copy>
            </xsl:for-each-group>
        </xsl:copy>
    </xsl:template>
</xsl:transform>

在线 http://xsltransform.net/ejivdHd。

至于EntityAssignedNumber元素的适应，这里有一些变化来实现这一点：

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
    <xsl:output indent="yes"/>
    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="root">
        <xsl:copy>
            <xsl:for-each-group select="RemittanceInformation" group-by="IndividualRemittance/ExchangeAssignedPolicyID">
              <xsl:copy>
                  <xsl:variable name="pos" select="position()"/>
                  <xsl:apply-templates select="EntityAssignedNumber, IndividualRemittance, current-group()/RemittanceDetail">
                      <xsl:with-param name="pos" select="$pos"/>
                  </xsl:apply-templates>
              </xsl:copy>
            </xsl:for-each-group>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="EntityAssignedNumber">
        <xsl:param name="pos"/>
        <xsl:copy>
            <xsl:value-of select="/root/RemittanceInformation[1]/EntityAssignedNumber + $pos - 1"/>
        </xsl:copy>
    </xsl:template>
</xsl:transform>

在线 http://xsltransform.net/ejivdHd/1。