Hi,

I'm attempting to create a menu structure that I can manipulate using jquery. I'm ver nearly there, but just need (I guess) an if statement to test whether the current node has children, if it does, tag it with a class "has sub".

My code so far:

<?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:Examine="urn:Examine" 
  exclude-result-prefixes="msxml umbraco.library Examine ">

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

<xsl:param name="currentPage"/>

<xsl:variable name="maxLevelForMenu" select="3"/>

<xsl:template match="/">

<xsl:call-template name="drawNodes">  
<xsl:with-param name="parent" select="$currentPage/ancestor-or-self::* [@isDoc and @level=1]"/>  
</xsl:call-template>
</xsl:template>

<xsl:template name="drawNodes">
<xsl:param name="parent"/> 
<xsl:if test="umbraco.library:IsProtected($parent/@id, $parent/@path) = 0 or (umbraco.library:IsProtected($parent/@id, $parent/@path) = 1 and umbraco.library:IsLoggedOn() = 1)">
    <ul>

  <xsl:for-each select="$parent/* [@isDoc and string(showInMenu) = '1' and @level &lt;= $maxLevelForMenu]"> 
<li>  
                <xsl:if test="$currentPage/@id = current()/@id">
                  <xsl:attribute name="class">active</xsl:attribute>
            </xsl:if>
<a href="{umbraco.library:NiceUrl(@id)}">
<xsl:value-of select="@nodeName"/></a>  
<xsl:if test="count(./* [@level &lt;= $maxLevelForMenu]) &gt; 0">   
  <xsl:call-template name="drawNodes">    
<xsl:with-param name="parent" select="."/>    
</xsl:call-template>  
</xsl:if> 
</li>
</xsl:for-each>
</ul>
</xsl:if>
</xsl:template>
</xsl:stylesheet>

Any help would be much appreciated!

Thanks,

Tom