CODEWITHSUNDEEP
xmlxsltduplicatesduplicate-data
Jamie Carmichael
Jamie Carmichael

Reputation:

Eliminating duplicate same-level elements in XSLT

I'm looking to display a list of same-level node names, without duplicates.

Let's say I have

<a>
    <b>
        <c />
        <d />
        <d />
    </b>
    <b>
        <e />
        <c />
        <f />
    </b>
</a>

I'd want c,d,e,f to be displayed. I've found several solutions to a similar problem, eliminating duplicate siblings from output, but I'm having trouble eliminating duplicate "cousins".

Upvotes: 2

Views: 2217

Answers (2)

Martin v. L&#246;wis
Martin v. L&#246;wis

Reputation: 127447

I would use the XPath preceding-sibling axis, and check for same local name. Untested:

<xsl:template match="c|d|e|f">
    <xsl:if test="local-name(.) != local-name(preceding-sibling::*[1])">
       <xsl:copy-of select="."/>
    </xsl:if>
</xsl:template>

IOW, if an element has the same name as its preceding-sibling, it does not get copied.

Upvotes: 0

Tomalak
Tomalak

Reputation: 338118

One possibility:

<!-- make all element nodes accessible by their nesting level -->
<xsl:key name="kNodesByLevel" match="*" use="count(ancestor-or-self::*)" />

<xsl:template match="/">
  <!-- select all nodes on one particular level -->
  <xsl:variable name="lvl" select="key('kNodesByLevel', 3)" />

  <!-- step through them... -->
  <xsl:for-each select="$lvl">
    <xsl:sort select="name()" /> 
    <xsl:variable name="name" select="name()" />
    <!-- ... and group them by node name -->
    <xsl:if test="generate-id() = generate-id($lvl[name() = $name][1])"> 
      <xsl:copy-of select="." />
    </xsl:if>
  </xsl:for-each>
</xsl:template>

Output for the XML you provided:

<c />
<d />
<e />
<f />

Upvotes: 1

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