How to get value of unique XML element by using XS

问题:

I want the unique list of elements from XML document. If the occurrence of element is more than 1, i want to have the last occurrence in my output:

Please refer the below XML for getting unique list:

<Organization>
    <Fund>
      <id>001</id>
      <name>ABC Ltd</name>
    </Fund>
    <Fund>
      <id>002</id>
      <name>DEF Limited</name>
    </Fund>
    <Fund>
      <id>001</id>
      <name>ABC Ltd.</name>
    </Fund>
    <Fund>
      <id>002</id>
      <name>DEF Corporation</name>
    </Fund>
    <Fund>
      <id>003</id>
      <name>XYZ LLC.</name>
    </Fund>
 </Organization>

The transform should output the below result:

<Organization>
    <Fund>
      <id>001</id>
      <name>ABC Ltd.</name>
    </Fund>
    <Fund>
      <id>002</id>
      <name>DEF Corporation</name>
    </Fund>
    <Fund>
      <id>003</id>
      <name>XYZ LLC.</name>
    </Fund>
 </Organization>

*Please note the change in name tag of Fund with id 001 and 002.

Need the sample code in XSLT1. Thanks in advance.

回答1:

Using muenching grouping:

Create a key for each fund_by_id

copy the last fund from each key by selecting only those funds whose ID matches the last ID in a key group.

<xsl:key name="funds_by_id" match="Fund" use="id"/>
<xsl:template match="Organization">
  <Organization>
      <xsl:copy-of select="Fund[generate-id() = 
                generate-id(key('funds_by_id',id)[last()])]"/>
  </Organization>
</xsl:template>