I've created a menu in umbraco using XSLT. The menu is using the usual ul and li elements and I'm displaying only the first level of the menu. The aim is to create a menu that expands to show the sub menu when I click a parent node (in the top level).
I am after the xslt I would need to expose the sub menu when clicked.
I think I would need to make use of ancestor-or-self to detect the current menu and parent menu and display them and also the $currentPage variable.
I have the following xslt:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:stylesheet [ <!ENTITY nbsp " "> ]>
<xsl:stylesheet
version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxml="urn:schemas-microsoft-com:xslt"
xmlns:umbraco.library="urn:umbraco.
library" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets" xmlns:tagsLib="urn:tagsLib" xmlns:urlLib="urn:urlLib"
exclude-result-prefixes="msxml umbraco.
library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets tagsLib urlLib ">
<xsl:output method="xml" omit-xml-declaration="yes"/>
<xsl:param name="currentPage"/>
<xsl:template match="/">
<div id="kb-categories">
<h3>Categories</h3>
<xsl:call-template name="drawNodes">
<xsl:with-param name="parent" select="$currentPage/ancestor-or-self::node [@level=1]"/>
</xsl:call-template>
</div>
</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 $parent/@level = 1">
<ul class="kb-menuLevel1" >
<xsl:for-each select="$parent/node [string(./data [@alias='showInMenu']) = 1]">
<li>
<a href="/kb{umbraco.library:NiceUrl(@id)}">
<xsl:value-of select="@nodeName"/>
</a>
<xsl:variable name="level" select="@level" />
<xsl:if test="(count(./node [string(./data [@alias='showInMenu']) = '1']) > 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:if test="(umbraco.library:IsProtected($parent/@id, $parent/@path) = 0 or (umbraco.library:IsProtected($parent/@id, $parent/@path) = 1)) and $parent/@level > 1">
<ul class="kb-menuLevel{@level}" style="display: none;">
<xsl:for-each select="$parent/node [string(./data [@alias='showInMenu']) = 1]">
<li>
<a href="/kb{umbraco.library:NiceUrl(@id)}">
<xsl:value-of select="@nodeName"/>
</a>
<xsl:variable name="level" select="@level" />
<xsl:if test="(count(./node [string(./data [@alias='showInMenu']) = '1']) > 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>
I suspect this could be improved using apply-templates, but I'm not yet up to speed with that (this being only the second day of my learning xslt).
My menu:
Item 1
Item 2
Item 3
Item 4
when I click on Item 2 I want to see it's child menu too:
Item 1
Item 2
-- Item 2.1
-- Item 2.2
Item 3
Item 4
and so on down the nested menu.
