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Editorial Team
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Asked: 2026-06-02T12:20:47+00:00

I’m currently working on an XSL script that transforms the following xml into an

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I’m currently working on an XSL script that transforms the following xml into an ordered list. Please refere to the input and output sample below. I’m using XSLT 2.0 by the way.

**Input:**
<xml>
    <numberedList>1. Number List 1</numberedList>
    <unnumberedList>Child List 1</unnumberedList>
    <unnumberedList>Child List 2</unnumberedList>
    <unnumberedList>Child List 3</unnumberedList>
    <numberedList>2. Number List 2</numberedList>
    <unnumberedList>Child List 1</unnumberedList>
    <numberedList>3. Number List 3</numberedList>
</xml>

**Output:**
<html>
<ol>
    <li>1. Number List 1</li>
    <li>
        <ul>
            <li>Child List 1</li>
            <li>Child List 2</li>
            <li>Child List 3</li>
        </ul>
    </li>
    <li>2. Number List 2</li>
    <li>
        <ul>
            <li>Child List 1</li>
        </ul>
    </li>
    <li>3. Number List 3</li>
</ol>
</html>

Any help would be greatly appreciated. Thanks!

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1 Answer

  1. Editorial Team

    An attempted improvement on Dimitre’s XSLT 2.0 solution:

    <xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="/*">
  <html>
   <ol>
     <xsl:for-each-group select="*" group-adjacent="name()">
       <xsl:choose>
         <xsl:when test="self::numberedList">
           <xsl:apply-templates select="current-group()"/>
         </xsl:when>
         <xsl:otherwise>
           <li><ul><xsl:apply-templates select="current-group()"/></ul></li>
         </xsl:otherwise>
       </xsl:choose>
     </xsl:for-each-group>
   </ol>
  </html>
 </xsl:template>

 <xsl:template match="numberedList|unnumberedList">
  <li><xsl:apply-templates/></li>
 </xsl:template>

</xsl:stylesheet>
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