使用 XSLT 合并 XML 节点 [英] Merge XML nodes using XSLT

问题描述

我需要将我的 XML 转换为另一种数据结构.我收到如下 XML:

I need to transform my XML into another datastructure. I recieve the XML like below:

<results>
  <resultset>
    <result>
      <name>BMW Cars</name>
      <code>BMW Pkw</code>
      <model.model>730d Saloon</model.model>
      <model.name>KM21</model.name>
    </result>
    <result>
      <name>BMW Cars</name>
      <code>BMW Pkw</code>
      <model.model>120i 3 doors</model.model>
      <model.name>UA51</model.name>
    </result>
    <result>
      <name>BMW Cars</name>
      <code>BMW Pkw</code>
      <model.model>Z4 sDrive23i</model.model>
      <model.name>LM31</model.name>
    </result>
    <result>
      <name>Audi</name>
      <code>AUDI</code>
      <model.model>A4 SAL.3.0 Q SPT TIP 5SPD</model.model>
      <model.name>8E2SFZ04</model.name>
    </result>
    <result>
      <name>Audi</name>
      <code>AUDI</code>
      <model.model>A6 SAL. 2.5TDI SPORT MAN.6SP.</model.model>
      <model.name>4B2BBC04</model.name>
    </result>
    <result>
      <name>AUdi</name>
      <code>AUDI</code>
      <model.model>A8 4.2 QUATTRO 6-SPD TIP</model.model>
      <model.name>4E201L04</model.name>
    </result>
  </resultset>
</results>

我需要它是这样的:

<results>
  <resultset>
    <result>
      <name>BMW Cars</name>
      <code>BMW Pkw</code>
      <model.model>730d Saloon</model.model>
      <model.name>KM21</model.name>
      <model.model>120i 3 doors</model.model>
      <model.name>UA51</model.name>
      <model.model>Z4 sDrive23i</model.model>
      <model.name>LM31</model.name>
    </result>
    <result>
      <name>Audi</name>
      <code>AUDI</code>
      <model.model>A4 SAL.3.0 Q SPT TIP 5SPD</model.model>
      <model.name>8E2SFZ04</model.name>
      <model.model>A6 SAL. 2.5TDI SPORT MAN.6SP.</model.model>
      <model.name>4B2BBC04</model.name>
      <model.model>A8 4.2 QUATTRO 6-SPD TIP</model.model>
      <model.name>4E201L04</model.name>
    </result>
  </resultset>
</results>

我花了很多时间来解决这个问题，但到目前为止还没有运气.有谁知道如何解决这个问题?

I've spent a lot of time to solve this, but no luck so far. Does anyone know how to solve this problem?

推荐答案

<?xml version="1.0" encoding="utf-8"?>
<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:key name="groupName" match="//results/resultset/result" use="concat(name, code)" />

<xsl:template match="/">

  <results>
  <resultset>
  <xsl:for-each select="//results/resultset/result[generate-id() = generate-id( key('groupName', concat(name, code))   [1] ) ]" >


      <xsl:call-template name="group">
        <xsl:with-param name="k1" select="name" />
        <xsl:with-param name="k2" select="code" />
      </xsl:call-template>

  </xsl:for-each>

  </resultset>
  </results>
</xsl:template> 

<xsl:template name="group">
<xsl:param name="k1" /> 
<xsl:param name="k2" /> 

    <result>
      <xsl:copy-of select="name" />       
      <xsl:copy-of select="code" />       

      <xsl:for-each select="//results/resultset/result[name = $k1][code = $k2]">

        <xsl:copy-of select="model.model" />       
        <xsl:copy-of select="model.name" />       

      </xsl:for-each>
    </result>

</xsl:template> 
</xsl:stylesheet>

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