XSLT转换:选择不同的和附属的组

Question

<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>

我对一些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' traveler '节点分配为子节点，以产生如下内容:

<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，在Transport节点中还包含许多本示例中未显示的子节点和孙子节点。也可以有许多属于一个TravellerSequence的旅行者。也有许多TransportSequence编号。

Answer 1

下面是使用Saxon 9或AltovaXML等XSLT 2.0处理器运行的XSLT 2.0样式表:

<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>

它转换

<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>

到

<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>

[edit]要完成这个答案，如果您想使用XSLT 1.0处理器，那么使用Muenchian分组的解决方案如下:

<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>