0 
<Sections>
    <Products>
      <Transport>
        <TransportSequence>1</TransportSequence>
        <Traveller>001</Traveller>
      </Transport>
      <Transport>
        <TransportSequence>2</TransportSequence>
        <Traveller>001</Traveller>
      </Transport>
    </Products>
  </Sections>
  <Sections>
    <Products>
      <Transport>
        <TransportSequence>1</TransportSequence>
        <Traveller>002</Traveller>
      </Transport>
      <Transport>
        <TransportSequence>2</TransportSequence>
        <Traveller>002</Traveller>
      </Transport>
    </Products>
  </Sections>

我对某些 XML 的排序有一个特定的问题。从上面的示例中，我需要更改格式，以便仅在 TransportSequence 上选择 distinct。然后，我需要将任何“Traveller”节点分配为子节点以生成如下内容：

<Sections>
   <Products>
      <Transport>
         <TransportSequence>1</TransportSequence>
         <Travellers>
            <Traveller>001</Traveller>
            <Traveller>002</Traveller>
         </Travellers>
      </Transport>
      <Transport>
         <TransportSequence>2</TransportSequence>
         <Travellers>
            <Traveller>001</Traveller>
            <Traveller>002</Traveller>
         </Travellers>
      </Transport>
   </Products>
</Sections>

另一个问题是，Transport 节点中还包含许多本例中未显示的子节点和孙节点。也可以有许多属于 TravellerSequence 的 travller。还有许多 TransportSequence 号码。

4

1 回答 1

0

这是一个 XSLT 2.0 样式表，可与 Saxon 9 或 AltovaXML 等 XSLT 2.0 处理器一起运行：

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="2.0">

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

<xsl:template match="*[Sections]">
  <xsl:copy>
    <Sections>
      <Products>
        <xsl:for-each-group select="Sections/Products/Transport" group-by="TransportSequence">
          <Transport>
            <TransportSequence><xsl:value-of select="current-grouping-key()"/></TransportSequence>
            <Travellers>
              <xsl:copy-of select="current-group()/Traveller"/>
            </Travellers>
          </Transport>
        </xsl:for-each-group>
      </Products>
    </Sections>
  </xsl:copy>
</xsl:template>

</xsl:stylesheet>

它转变

<Root>
<Sections>
    <Products>
      <Transport>
        <TransportSequence>1</TransportSequence>
        <Traveller>001</Traveller>
      </Transport>
      <Transport>
        <TransportSequence>2</TransportSequence>
        <Traveller>001</Traveller>
      </Transport>
    </Products>
  </Sections>
  <Sections>
    <Products>
      <Transport>
        <TransportSequence>1</TransportSequence>
        <Traveller>002</Traveller>
      </Transport>
      <Transport>
        <TransportSequence>2</TransportSequence>
        <Traveller>002</Traveller>
      </Transport>
    </Products>
  </Sections>
</Root>

进入

<Root>
   <Sections>
      <Products>
         <Transport>
            <TransportSequence>1</TransportSequence>
            <Travellers>
               <Traveller>001</Traveller>
               <Traveller>002</Traveller>
            </Travellers>
         </Transport>
         <Transport>
            <TransportSequence>2</TransportSequence>
            <Travellers>
               <Traveller>001</Traveller>
               <Traveller>002</Traveller>
            </Travellers>
         </Transport>
      </Products>
   </Sections>
</Root>

[编辑]要完成答案，如果您想使用 XSLT 1.0 处理器，使用 Muenchian 分组的解决方案如下所示：

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="1.0">

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

<xsl:key name="by-seq" match="Sections/Products/Transport" use="TransportSequence"/>

<xsl:template match="*[Sections]">
  <xsl:copy>
    <Sections>
      <Products>
        <xsl:for-each select="Sections/Products/Transport[generate-id() = generate-id(key('by-seq', TransportSequence)[1])]">
          <Transport>
            <xsl:copy-of select="TransportSequence"/>
            <Travellers>
              <xsl:copy-of select="key('by-seq', TransportSequence)/Traveller"/>
            </Travellers>
          </Transport>
        </xsl:for-each>
      </Products>
    </Sections>
  </xsl:copy>
</xsl:template>

</xsl:stylesheet>
于 2013-03-26T14:08:42.347 回答