CODEWITHSUNDEEP
xsltgetdistinct
AlbionKyle
AlbionKyle

Reputation: 53

XSLT select unique node

this is driving me crazy, for xslt newbie like me.

Input:

<root>
    <a><name>kyle</name></a>
    <b><name>stan</name></b>
    <b><name>wendy</name></b>
    <b><name>cece</name></b>
</root>

Expected output:

<root>
        <a><name>kyle</name></a>
        <b><name>stan</name></b>
</root>

I was asked to return first unique node under 'root', how do I do that?

Either xslt 1.0 or 2.0 is fine.

Thank you so much!!!!

Upvotes: 4

Views: 859

Answers (2)

FailedDev
FailedDev

Reputation: 26920

XSLT 2.0 solution : 

<?xml version="2.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

  <xsl:output method="xml" indent="yes"/>

  <xsl:template match="/">
  <root>
    <xsl:for-each-group select="root/*" group-by="local-name()">
      <xsl:copy-of select="."/>
    </xsl:for-each-group>
    </root>
  </xsl:template>
</xsl:stylesheet>

Output:

<?xml version="1.0" encoding="UTF-8"?>
<root>
   <a>
      <name>kyle</name>
  </a>
   <b>
      <name>stan</name>
  </b>
</root>

Upvotes: 1

Daniel Haley
Daniel Haley

Reputation: 52848

You can match any element that has a preceding sibling with the same name and not output anything.

Example XSLT:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="node()|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="/*/*[preceding-sibling::*[name() = current()/name()]]"/> 

</xsl:stylesheet>

Output (using Saxon 9 HE):

<root>
   <a>
      <name>kyle</name>
   </a>
   <b>
      <name>stan</name>
   </b>
</root>

Upvotes: 1

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