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Editorial Team
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Asked: 2026-06-09T00:49:15+00:00

I am getting duplicate records with the following xml and xsl file. I only

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I am getting duplicate records with the following xml and xsl file. I only want to transform 1 set of the list items. Try not removing anything from the xsl part (just add to it), if possible.

<?xml version="1.0" encoding="utf-8" ?>
<data>
  <listitems name="Select..." CtrId="Id2"/>
  <listitems name="Item A" CtrId="Id2"/>
  <listitems name="Item B" CtrId="Id2"/>
  <listitems name="Select..." CtrId="Id4"/>
  <listitems name="Item A" CtrId="Id4"/>
  <listitems name="Item B" CtrId="Id4"/>  
  <listitems name="Select..." CtrId="Id6"/>
  <listitems name="Item C" CtrId="Id6"/>
  <listitems name="Item D" CtrId="Id6"/>  
</data>

 

  <xsl:template match="data/listitems">
    <html>
      <head>
        <title>Untitled</title>
      </head>
      <body>
        <xsl:value-of select="@name"/>
      </body>
    </html>
  </xsl:template>

 

Result (Incorrect behavior; Duplicate)
Select… Item A Item B Select… Item A Item B

Desired behavior (Only get 1 set)
Select… Item A Item B

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1 Answer

  1. Editorial Team

    Here’s a relatively simple way of doing this:

    <xsl:param name="useId" select="/data/listitems[1]/@CtrId" />

<xsl:template match="/">
  <html>
    <head>
      <title>Untitled</title>
    </head>
    <body>
      <xsl:apply-templates select="data/listitems[@CtrId = $useId]"/>
    </body>
  </html>
</xsl:template>

<xsl:template match="listitems">
  <xsl:value-of select="concat(@name, ' ')" />
</xsl:template>

    Your existing template will actually put in an html element for every listitems element- it seems fairly likely that you probably only want the one.

    The <xsl:param> declaration at the top picks the first CtrId in the file, and uses that. You can change that to a literal value with select="'Id2'" (note the single quotes inside the double quotes), or you can pass a parameter into the stylesheet with the ID you want to pick out.

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