Gebruik XSLT Sort en Order statement om top5 artikelen te selecteren

Voorbeeld van XSLT:

<?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"
 exclude-result-prefixes="msxml 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"/>
<xsl:template match="/">

<div id="Top5">
<p>Laatste 5 aantekeningen:</p>
<xsl:for-each select="$currentPage/ancestor-or-self::node//node [(@level &gt; 2) and (@nodeTypeAlias = 'Aantekening')]">
    <xsl:sort select="@updateDate" order="descending"></xsl:sort>
    <xsl:if test="position() &lt; 6">

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

    </xsl:if>
</xsl:for-each>
</div>
</xsl:template>

</xsl:stylesheet>

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