Hello,

Currently I have a function that will list out nodes in alphabetical order from A-Z in one column.

I now want to make it into a 3 column layout like so:

a     e     h

b     f       i

c     g      j

Note there are sub nodes under each of these headings.

Here is my current XSLT.

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:stylesheet [ <!ENTITY nbsp "&#x00A0;"> ]>
<xsl:stylesheet 
  version="1.0" 
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
  xmlns:msxml="urn:schemas-microsoft-com:xslt" 
  xmlns:umbraco.library="urn:umbraco.library" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets" xmlns:umbraco.contour="urn:umbraco.contour" xmlns:PS.XSLTsearch="urn:PS.XSLTsearch" 
  exclude-result-prefixes="msxml umbraco.library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets umbraco.contour PS.XSLTsearch ">

<xsl:output method="xml" omit-xml-declaration="yes"/>

<xsl:param name="currentPage"/>
<xsl:variable name="col" select="3" />

<xsl:template match="/">

<!-- The fun starts here -->
<ul class="alphabet">
<xsl:for-each select="$currentPage/* [@isDoc]" >
  <xsl:sort select="@nodeName" />
  <li>
    <h3><xsl:value-of select="@nodeName"/></h3>
    
    <xsl:if test="count(./* [@isDoc]) &gt; 0">
          <ul class="t2n">
            <xsl:for-each select="./* [@isDoc]">
              <li>
                <a href="{umbraco.library:NiceUrl(@id)}">
                  <xsl:value-of select="@nodeName"/>
                </a>
              </li>
            </xsl:for-each>
          </ul>
      </xsl:if>
  </li>
</xsl:for-each>
    
</ul>

</xsl:template>

Any help would be great.

Owen