迭代具有相同值的子节点 - XSLT

Question

想知道是否可以使用 XSLT 实现以下目标：我需要遍历 XML 树的子节点，并将字段 auditid 标记中具有相似值的节点组合在一起。要求是为每个具有相同 auditid 的“记录”加入“附件”节点。

示例输入XML：

<?xml version='1.0' encoding='UTF-8'?>
<Root>
  <total_count>53</total_count>
  <records>
    <auditid>00000001</auditid>
    <completationdate>2021-08-04</completationdate>
    <organization>testcompany1</organization>
    <streetaddress>1 peel street, melbourne</streetaddress>
    <suburb>South Melbourne</suburb>
  </records>
  <records>
    <auditid>00000002</auditid>
    <completationdate>2021-07-07</completationdate>
    <organization>testcompany2</organization>
    <streetaddress>1 test road, West Footscray</streetaddress>
    <suburb>West Footscray</suburb>
    <attachments>
      <label>71270-2_A</label>
      <url>http://test/53X/00000002/71270-2_A.pdf</url>
    </attachments>
  </records>
  <records>
    <auditid>00000002</auditid>
    <completationdate>2021-07-07</completationdate>
    <organization>testcompany2</organization>
    <streetaddress>1 test road, West Footscray</streetaddress>
    <suburb>West Footscray</suburb>
    <attachments>
      <label>71270-2_B</label>
      <url>http://test/53X/00000002-2_B.pdf</url>
    </attachments>
  </records>
  <records>
    <auditid>00000002</auditid>
    <completationdate>2021-07-07</completationdate>
    <organization>testcompany2</organization>
    <streetaddress>1 test road, West Footscray</streetaddress>
    <suburb>West Footscray</suburb>
    <attachments>
      <label>71270-2_C</label>
      <url>http://test/53X/00000002/71270-2_C.pdf</url>
    </attachments>
  </records>
  <records>
    <auditid>00000003</auditid>
    <completationdate>2021-07-08</completationdate>
    <organization>testcompany3</organization>
    <streetaddress>1 test road, Hampton Park</streetaddress>
    <suburb>Hampton Park</suburb>
    <attachments>
      <label>60797-25_A</label>
      <url>http://test/53V/00000003/60797-25_A.pdf</url>
    </attachments>
  </records>
  <records>
    <auditid>00000003</auditid>
    <completationdate>2021-07-08</completationdate>
    <organization>testcompany3</organization>
    <streetaddress>1 test road, Hampton Park</streetaddress>
    <suburb>Hampton Park</suburb>
    <attachments>
      <label>60797-25_B</label>
      <url>http://test/53V/00000003/60797-25_B.pdf</url>
    </attachments>
  </records>
</Root>

我的 XSLT：到目前为止，我已经能够生成每个具有相同值的 auditid 的计数。但无法弄清楚如何迭代这些。

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:key name="dogo" match="Root/records" use="auditid" />
<xsl:template match="/Root">
<root>
    <total_count>
        <xsl:value-of select="total_count"/>
    </total_count>
    <xsl:for-each select="records[count(key('dogo', auditid)) = 1]">
    <xsl:variable name="count">
        <xsl:value-of select="count(key('dogo', auditid))"/>
    </xsl:variable>
        <records>
            <count>
                <xsl:value-of select="$count"/>
            </count>
            <xsl:copy-of select="auditid"/>
            <xsl:copy-of select="completationdate"/>
            <xsl:copy-of select="organization"/>
            <xsl:copy-of select="streetaddress"/>
            <xsl:copy-of select="suburb"/>
            <xsl:copy-of select="attachments"/>
        </records>
    </xsl:for-each>
</root>
</xsl:template>
</xsl:stylesheet>

预期输出：

<?xml version='1.0' encoding='UTF-8'?>
<Root>
  <total_count>53</total_count>
  <records>
    <auditid>00000001</auditid>
    <completationdate>2021-07-20</completationdate>
    <organization>testcompany1</organization>
    <streetaddress>1 peel street, melbourne</streetaddress>
    <suburb>South Melbourne</suburb>
  </records>
  <records>
    <auditid>00000002</auditid>
    <completationdate>2021-07-07</completationdate>
    <organization>testcompany2</organization>
    <streetaddress>1 test road, West Footscray</streetaddress>
    <suburb>West Footscray</suburb>
    <attachments>
      <label>71270-2_A</label>
      <url>http://test/53X/00000002/71270-2_A.pdf</url>
    </attachments>
    <attachments>
      <label>71270-2_B</label>
      <url>http://test/53X/00000002-2_B.pdf</url>
    </attachments>
    <attachments>
      <label>71270-2_C</label>
      <url>http://test/53X/00000002/71270-2_C.pdf</url>
    </attachments>
  </records>
  <records>
    <auditid>00000003</auditid>
    <completationdate>2021-07-08</completationdate>
    <organization>testcompany3</organization>
    <streetaddress>1 test road, Hampton Park</streetaddress>
    <suburb>Hampton Park</suburb>
    <attachments>
      <label>60797-25_A</label>
      <url>http://test/53V/00000003/60797-25_A.pdf</url>
    </attachments>
    <attachments>
      <label>60797-25_B</label>
      <url>http://test/53V/00000003/60797-25_B.pdf</url>
    </attachments>
  </records>
</Root>

Answer 1

如果您能够使用 XSLT 2.0 那么为什么不利用它的 grouping 功能并简单地做：

<xsl:stylesheet version="2.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<xsl:template match="/Root">
    <xsl:copy>
        <xsl:copy-of select="total_count"/>
        <xsl:for-each-group select="records" group-by="auditid">
            <records>
                <xsl:copy-of select="* except attachments"/>
                <xsl:copy-of select="current-group()/attachments"/>
            </records>
        </xsl:for-each-group>
    </xsl:copy>
</xsl:template>

</xsl:stylesheet>

迭代具有相同值的子节点 - XSLT

Iterate over Child-Nodes with Same Value - XSLT

xml

xslt

integration

xslt-2.0