translate menu to new schema

Floffy 26 posts 33 karma points

Jul 23, 2011 @ 21:36

Translate menu to new schema

Tryed to translat xslt 2. and 3. level menu to new schema, but with no luck.

It renders no nodes.

Here is 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"
  exclude-result-prefixes="msxml umbraco.library">

<xsl:output method="xml" omit-xml-declaration="yes" />
<xsl:param name="currentPage"/>
<xsl:variable name="level" select="1"/>
<xsl:template match="/">
  <xsl:call-template name="printListe">
    <xsl:with-param name="node" select="$currentPage/ancestor-or-self::*[@isDoc and @level = 2]"/>  
    <xsl:with-param name="id" select="string('leftMenu')"/>  
  </xsl:call-template>
</xsl:template>
<xsl:template name="printListe">
  <xsl:param name="node"/>
  <xsl:param name="id"/>
<div id="LeftMenu"> 
<ul class="sideMenu">
<xsl:if test="$id != ''">
<xsl:attribute name="id"><xsl:value-of select="$id"/></xsl:attribute>
</xsl:if>
<xsl:for-each select="$node/node [@nodeTypeAlias != 'DateFolder' and @nodeTypeAlias != 'LBKnewsItem']">
<li>
  <xsl:if test="$currentPage/ancestor-or-self::*[@isDoc]/@id = current()/@id">
    <xsl:attribute name="class">here</xsl:attribute>
  </xsl:if>
  <a href="{umbraco.library:NiceUrl(@id)}"><xsl:value-of select="@nodeName"/></a>
  <xsl:if test="$currentPage/ancestor-or-self::*[@isDoc]/@id = current()/@id">
    <xsl:if test="current()/@level = 3">
      <xsl:if test="count(./node) &gt; 0">
        <xsl:call-template name="printListe">
          <xsl:with-param name="node" select="."/>
        </xsl:call-template>
      </xsl:if>
    </xsl:if>
  </xsl:if>
</li>
</xsl:for-each>
</ul>
</div>
</xsl:template>
</xsl:stylesheet>

Any ideas?

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Dennis Aaen 4500 posts 18255 karma points admin hq c-trib

Jul 23, 2011 @ 23:23

Hi Fioffy,

I think you are very close to a solution that works. As I see it, it's just small changes as needed if I'm right.
You might try something like.

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

<xsl:outputmethod="xml"omit-xml-declaration="yes"/>
<xsl:paramname="currentPage"/>
<xsl:variablename="level"select="1"/>
<xsl:templatematch="/">
  <xsl:call-templatename="printListe">
    <xsl:with-paramname="node"select="$currentPage/ancestor-or-self::*[@isDoc and @level = 2]"/>  
    <xsl:with-paramname="id"select="string('leftMenu')"/>  
  </xsl:call-template>
</xsl:template>
<xsl:templatename="printListe">
  <xsl:paramname="node"/>
  <xsl:paramname="id"/>
<divid="LeftMenu"> 
<ulclass="sideMenu">
<xsl:iftest="$id != ''">
<xsl:attributename="id"><xsl:value-ofselect="$id"/></xsl:attribute>
</xsl:if>
<xsl:for-eachselect="$node/*[@isDoc][name() != 'DateFolder' and name() != 'LBKnewsItem']">
<li>
  <xsl:iftest="$currentPage/ancestor-or-self::*[@isDoc]/@id = current()/@id">
    <xsl:attributename="class">here</xsl:attribute>
  </xsl:if>
  <ahref="{umbraco.library:NiceUrl(@id)}"><xsl:value-ofselect="@nodeName"/></a>
  <xsl:iftest="$currentPage/ancestor-or-self::*[@isDoc]/@id = current()/@id">
    <xsl:iftest="current()/@level = 3">
      <xsl:iftest="count(./*[@isDoc]) &gt; 0">
        <xsl:call-templatename="printListe">
          <xsl:with-paramname="node"select="."/>
        </xsl:call-template>
      </xsl:if>
    </xsl:if>
  </xsl:if>
</li>
</xsl:for-each>
</ul>
</div>
</xsl:template>
</xsl:stylesheet>

Hope this can help you.

/Dennis

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