translate menu to new schema

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Floffy 26 posts 33 karma points

Jul 23, 2011 @ 21:36

XSLT

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 4499 posts 18254 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>