Hi Tommy am using this piece of code at the moment..

<xsl:variable name="source" select="1168"/>
<xsl:variable name="maxLevel" select="4"/> 

<xsl:template match="/">

<!-- The fun starts here -->
<ul>
    <xsl:call-template name="drawNodes">  
        <xsl:with-param name="parent" select="umbraco.library:GetXmlNodeById($source)/* [@isDoc and string(umbracoNaviHide) != '1']"/>  
    </xsl:call-template>
</ul>
</xsl:template>

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

But only getting the Sub Node of the Child that is in my cas Sub Link 1 and Sub Link 2

Btw id (1168) is for the Folder 1 node..

Fuji

You go 1 step too deep in your template call/for-each (in your for-each the parent is actually one of your $source children)

So e.g. replace

<xsl:with-param name="parent" select="umbraco.library:GetXmlNodeById($source)/* [@isDoc and string(umbracoNaviHide) != '1']"/>  
    </xsl:call-template>

with

<xsl:with-param name="parent" select="umbraco.library:GetXmlNodeById($source) [@isDoc and string(umbracoNaviHide) != '1']"/>  
    </xsl:call-template>

Hi Fuji

I was just doing a little local example. Where I've the following XSLT code to get all nodes out, based on a specific node.
So maybe you could try something like this

<xsl:template match="/">


<ul>
  <xsl:for-each select="umbraco.library:GetXmlNodeById($source)//* [@isDoc and string(umbracoNaviHide) != '1']">
  <li>
    <a href="{umbraco.library:NiceUrl(@id)}">
      <xsl:value-of select="@nodeName"/>
    </a>
  </li>
</xsl:for-each>
</ul>

</xsl:template>

And the two slash characters represent the descendant axis. So it will get all childs and thir childs, and so on.

Hope it can helps you.

/Dennis

An apologies to Tommy because I hadn't seen that you had proposed the same example, as I did. And you had added that the * could be replaced by decendants::*

I´m sorry

/Dennis

Dennis no worries - you supplying an answer, trying to help out on the forum is important. Hopefully one of the posts will solve the problem.

Thanks Dennis its working...can you tell me how i can get the subPages to get a class so that the links indent?

 <xsl:attribute name="class">
         <xsl:choose>           
           <xsl:when test="$currentPage/@id and @level=2]">
                 <xsl:text>left-menu-slink</xsl:text>        
            </xsl:when>
           
               <xsl:otherwise>
                <xsl:text></xsl:text>
               </xsl:otherwise>                  
             </xsl:choose>                
          <xsl:if test="$currentPage/@id = current()/@id">
               <xsl:text> current</xsl:text>
          </xsl:if>
   </xsl:attribute>    

Hi Fuji.

Did you find a solution on how to add a class to subpages  so that the links will be indented

/Dennis

Hi Dennis....

I tried the above code in my previous post but its not indenting the subpages, the only thing working so far is the active current page.

Any suggestion what i did wrong?

//fuji

Hi Fuji,
I've tried some different things yesterday, but I did not find a solution on how you can get subpage to be indented.

But hope you find a solution.
/Dennis

Hey Dennis,

Yes got it working, if you need some help on this i'll be please to share.

//Fuji

Hi Fuji,

I know it is an old question. But I do not know if you found a solution.
I have sat and tried to solve a similar question tonight and got the inspiration for the solution to your question.

The code below gives you all the subpages for each page. These subpages is indented under the page they belong

<ul>
<xsl:for-each select="umbraco.library:GetXmlNodeById($source)/* [@isDoc and string(umbracoNaviHide) != '1']">
  <li>
    <a href="{umbraco.library:NiceUrl(@id)}">
      <xsl:value-of select="@nodeName"/>
    </a>
  </li>
 <xsl:if test="count(./ancestor-or-self::* [@isDoc][@level=3]/* [@isDoc and string(umbracoNaviHide) != '1']) &gt; 0">
  <ul id="subMenu">
    <xsl:for-each select="./ancestor-or-self::* [@isDoc and @level=3]/* [@isDoc and string(umbracoNaviHide) != '1']">
    <xsl:sort select="@sortOrder" order="ascending"/>
    <li> 
      <a href="{umbraco.library:NiceUrl(current()/@id)}">
      <xsl:if test="$currentPage/ancestor-or-self::*/@id = current()/@id">
        <xsl:attribute name="class">selected</xsl:attribute>
      </xsl:if>
      <xsl:value-of select="current()/@nodeName"/>
      </a>    
    </li>
  </xsl:for-each>
  </ul>
  </xsl:if>
</xsl:for-each>
</ul>

I'm not quite sure if it's the best way to solve this problem at.

/Dennis

Ahh hadn´t seen you found a solution to the issue. This may be another way to solve the same problem on.

/Dennis

Hey Dennis,

Yes got it working.....how about your solution is it working if not let me know ill try out.

//fuji

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