I solved the problem partially, I changed my xslt-code to this:

<xsl:variable name="boekentips" select="umbraco.library:GetXmlNodeById($currentPage/boekentips)" />
<xsl:template match="/">
  <a href="{umbraco.library:NiceUrl($boekentips/@id)}"><xsl:value-of select="$boekentips/@nodeName" /></a>
</xsl:template>

And this works, but only for the homepage. If I go to a page under the homepage I get this error:

Error parsing XSLT file: \xslt\GetBoekentips.xslt

Hi Anthony,

I am not an xslt expert either, but have you tried  your initial xslt by using $boekentips/@id instead of boekentips/@id to fill in your "id" variable. So, something like this:

<xsl:param name="currentPage"/>
<xsl:variable name="boekentips" select="umbraco.library:GetXmlNodeById($currentPage/boekentips)"/>
<xsl:variable name="id" select="$boekentips/@id" />    
<xsl:template
 match="/">
  <xsl:element name="a">
    <xsl:attribute name="href">
      <xsl:copy-of select="umbraco.library:NiceUrl($id)"/>
    </xsl:attribute>
  </xsl:element>
</xsl:template>

Hope this helps.

Cheers,

Michael.

Hi Anthony

Do you have the content picker on each page of your site or just the homepage?

If you've got it on every page try this:

<xsl:if test="$currentPage/boekentips!=''">   
    <xsl:variable name="boekentips" select="umbraco.library:GetXmlNodeById($currentPage/boekentips)"/>
    <a href="{umbraco.library:NiceUrl($currentPage/boekentips)}"><xsl:value-of select="$boekentips/@nodeName" /></a>
</xsl:if>

/Kim A

Hi Michael,

I already solved the problem partially, my xslt-code looks like this now:

<xsl:variable name="boekentips" select="umbraco.library:GetXmlNodeById($currentPage/boekentips)" />
    <xsl:template match="/">
  <a href="{umbraco.library:NiceUrl($boekentips/@id)}"><xsl:value-of select="$boekentips/@nodeName" /></a>
</xsl:template>

The only problem I still have is that this only works on my homepage, the boekentips property is defined on the homepage document type.

But if I go to a page that is a child of the homepage I get the error:

Error parsing XSLT file: \xslt\GetBoekentips.xslt

Thanks for your help (again),

Anthony

@Kim: I only got the boekentips property defined on my homepage document type.

Thanks for your help,

Anthony

Hi Anthony,

Just to make sure I understand what you are trying to do:

1/ You have a "boekentips" property defined on your homepage document type only

2/ You want to display the "boekentips" link only on the homepage and you get an error on sub-pages
OR
2 bis/ You want to display the boekentips link on all pages but it only works on the homepage.

Another question: Do you get anything else then "Error parsing XSLT file: \xslt\GetBoekentips.xslt". Maybe some indication on where it has problems?

Cheers,

Michael.

Hi Michael,

it's 2 bis :)

I define the boekentips property on my homepage document type, but it should show on all child pages under my homepage.

I have adapted a little the code snippet of Kim by using an XPath axis, like this:

<xsl:template match="/">
  <xsl:if test="ancestor-or-self::boekentips!=''">   
    <xsl:variable name="boekentips" select="umbraco.library:GetXmlNodeById($currentPage/boekentips)"/>
    <a href="{umbraco.library:NiceUrl($currentPage/boekentips)}"><xsl:value-of select="$boekentips/@nodeName" /></a>
</xsl:if>

</xsl:template>

but it doesn't work

Hi Anthony,

It is in fact possible to identify the document type name by using the name() method. So maybe you might try something like this:

<xsl:variable name="homepage" select="$currentPage/ancestor-or-self::* [name()='YOUR_DOCTYPE_ALIAS']"
<xsl:template match="/">
  <xsl:if test="$homepage/boekentips!=''">   
    <xsl:variable name="boekentips" select="umbraco.library:GetXmlNodeById($homepage/boekentips)"/>
    <a href="{umbraco.library:NiceUrl($homepage/boekentips)}"><xsl:value-of select="$boekentips/@nodeName" /></a>
</xsl:if>
</xsl:template>

Hope this helps/works :-)

Cheers,

Michael.

Hi Kim,

I tried your code example and it works, thanks a lot!

Anthony

I'm glad to hear that Anthony. You are very welcome.

/Kim A

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