Hi Yael

Could you please paste all the XSLT code? I think it will be much easier to provide you with a good answer if we can see all of the context. I'm not sure the above approach of using a a for-each loop is the correct choise in your case - but let's see some code so we can guide you :)

/Jan

Hi Jan,

Here is the code. Hope this would help :)

<?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" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets" 
  xmlns:random="urn:random"
  exclude-result-prefixes="msxml random umbraco.library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets ">

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

    
    
<xsl:param name="currentPage"/>

<!-- Input the documenttype you want here -->
<xsl:param name="MaxNoChars" select="100" />
<xsl:variable name="numberOfItems" select="2"/> 
<msxml:script implements-prefix="random" language="C#">
<msxml:assembly name="umbraco"/>
<msxml:using namespace="umbraco"/>
<![CDATA[
public int GetRandom(int minValue, int maxValue)
{
return umbraco.library.GetRandom().Next(minValue, maxValue);
}
]]>
</msxml:script>

<xsl:template match="/">

<!-- The fun starts here -->
  <xsl:for-each select="umbraco.library:GetXmlNodeById(1273)/*/* [@isDoc and string(umbracoNaviHide) != '1' and string(chooseSubCategory) = '1748']">
     <xsl:sort select="random:GetRandom(1, count(umbraco.library:GetXmlNodeById(1273)/*/* [@isDoc and string(umbracoNaviHide) != '1']))" order="ascending" />
    <xsl:if test="position() &lt;= $numberOfItems">
      
      <div class="question_box article_box">
      <img src="/images/question_box_top.png" alt="question_box_top" />
      <div class="question_box_bg_repeat">
        <a href="{umbraco.library:NiceUrl(@id)}">
        <img class="item_img" src="{./thumb}" alt="" />
        </a>
        <a href="{umbraco.library:NiceUrl(@id)}"><h3><xsl:value-of select="title" disable-output-escaping="yes"/></h3></a>
        <p><a href="{umbraco.library:NiceUrl(@id)}"><xsl:value-of select="umbraco.library:TruncateString(umbraco.library:StripHtml(text), $MaxNoChars, '...')"
          disable-output-escaping="yes" /></a></p>
       </div>
      <a href="{umbraco.library:NiceUrl(@id)}"><img src="/images/box_bottom.png" alt="" /></a>
    </div>

      
  </xsl:if> 
</xsl:for-each>


</xsl:template>

</xsl:stylesheet>

This code is only for the articles section.

 1273 referes to the parent articles node.

 1748 is a random sub-category. 

Thank you!

Yael

Hi all,

Does anybody know of a way to do this in XSLT? Isn't there a way to refere to a property "same as current page"?

Would really appreciate any help, even if you just tell me that it's not possible in XSLT and I should turn to Razor.

Thanks in advance,

Yael

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