Try this

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

<xsl:param name="currentPage"/>
<xsl:variable name="HomeNode" select="$currentPage/parent::node" />
<xsl:template match="/">

<xsl:if test="$HomeNode != '' ">

  <tr>
       <td class="TableCellNavLeft">    
       <a href="{umbraco.library:NiceUrl($HomeNode/@id)}">
          Home</a>
      </td>
  </tr>

</xsl:if>

</xsl:template>
</xsl:stylesheet>

Explaination here http://our.umbraco.org/forum/developers/xslt/3102-Using-niceurl-Value-was-either-too-large-or-too-small-for-an-Int32-

Rich

Hi Rich,

Thanks for your feedback. The error is gone. That's super. But I also have no output. For some reason the link is not generated. 

Does anyone know why ?

Tim

Are you trying to find the home node or the parent node? 

Also which version of Umbraco are you using.

Rich

Hi Rich,

I'm trying to create a link pointing to the home node (level 0), to be used on pages below the home node...  

I'm currently using 4.5.2.

Tim

Hey Kim,

Thanks for your feedback. Your provided solution is perfect ... I was trying to solve a problem by using XSLT but the solution was simple plain old html...

Thanks Kim

Tim

No problem Tim.

In XSLT, you can grab the top most node in your site with this piece of code:

<xsl:variable name="HomeNode" select="$currentPage/ancestor-or-self::*[@level='1']" />
<xsl:template match="/">

<xsl:if test="$HomeNode != '' ">

  <tr>
       <td class="TableCellNavLeft">    
       <a href="{umbraco.library:NiceUrl($HomeNode/@id)}">
          Home</a>
      </td>
  </tr>

</xsl:if>

</xsl:template>

Well, at least that would work (with the new XML schema) if yout top most node is located on level 1 which is the the most used way to structure a site in Umbraco :)

/Kim Andersen

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