Please, ignore the comment tags, the code is:

         <a>
            <xsl:attribute name="href">
              <xsl:value-of select="umbraco.library:NiceUrl($currentPage/ancestor-or-self::root//@id)" />
            </xsl:attribute>
            <xsl:value-of select="$currentPage/ancestor-or-self::root//@nodeName" />
          </a>

Thanks again,

Hi Sergion,

Change your value to

<xsl:value-of select="umbraco.library:NiceUrl($currentPage/@id)"/> 

This should get you to the first node url "/"

//fuji

@Sergio: Can you please try and Publish (hit Save and Publish) the root node and run your code again? A node that doesn't exist in the XML cache gets the hashtag as it's URL :-)

@Fuji: I think that XSLT snippet gets get nice url of the current page :-) Segio want's the url of the site root node.

Hi,

Thanks for your replies.

@Fuji: I tried that code before posting but as Bo says it gets the Url of the current page when I want the root page

@Bo: I republished the entire site and all the individual pages but no luck so far :( still getting #

Still not working, Any other ideas????

Thanks

Hi Sergio,

I usually create a siteRoot variable just after the currentPage param, so it's easy (and readable) to get to the home page:

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

So to get the URL of that, you would do:

<a href="{umbraco.library:NiceUrl($siteRoot/@id)}">
    <xsl:value-of select="$siteRoot/@nodeName" />
</a>

(Note the use of curly-braces to grab a value inside an attribute)

But - as it turns out, you will *not* get the "/" URL that you're expecting, but rather something like this:

<a href="/myhome.aspx">My Home</a>

- which is why I usually hardcode the URL to "/" whenever I need that particular link...

/Chriztian 

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