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Home/ Questions/Q 227747
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T19:37:28+00:00 2026-05-11T19:37:28+00:00

Given a source doc like so: <A> <B> <C>item1</C> </B> <B> <C>item2</C> </B> <B>

  • 0

Given a source doc like so:

<A>
  <B>
    <C>item1</C>
  </B>
  <B>
    <C>item2</C>
  </B>
  <B>
    <C>item3</C>
  </B>
</A>

What XSLT can I use to product something like this:

<A>
  <B>
    <C>item1</C>
    <C>item2</C>
    <C>item3</C>
  </B>
</A>

I have tried a sheet with…

  <xsl:template match="B">
    <xsl:choose>
      <xsl:when test="count(preceding-sibling::B)=0">
        <B>
          <xsl:apply-templates select="./C"/>
          <xsl:apply-templates select="following-sibling::B"/>
        </B>
      </xsl:when>

      <xsl:otherwise>
          <xsl:apply-templates select="./C"/>
      </xsl:otherwise>

    </xsl:choose>
  </xsl:template>

but I am getting…

<A>
  <B>
    <C>item1</C>
    <C>item2</C>
    <C>item3</C>
  </B>
  <C>item2</C>
  <C>item3</C>
</A>

A second question: I have a hard time debugging XSLT. hints?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-11T19:37:28+00:00Added an answer on May 11, 2026 at 7:37 pm

    Simplest approach:

    <xsl:template match="/A">
      <A>
        <B>
          <xsl:copy-of select=".//C" />
        </B>
      </A>
    </xsl:template>
    

    To answer the question why you see the output you see with your XSLT:

    I aussume that you have a

    <xsl:apply-templates select="B" />
    

    in place. This means:

    • The <xsl:template match="B"> is called three times, once for each <B>.
    • For the first <B>, it does what you intend, the other times it branches right into the <xsl:otherwise>, copying the <C>s via the <xsl:template match="C"> you probably have. This is where your extra <C>s come from.
    • To fix it (not that I endorse your approach), you should change it like…

    …this:

    <xsl:apply-templates select="B[1]" />
    
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