select a random member in xslt

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[email protected] 378 posts 1915 karma points MVP 4x c-trib

Mar 19, 2010 @ 09:09

0

Select a random member in XSLT

I want to display a random member on my homepage. The only way I can find to select a member is by the function umbraco.library:GetMember('1060'), but I don't want to always show member 1060

Is there a way to select a random member in XSLT?

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Dan 1250 posts 3747 karma points admin c-trib

Mar 19, 2010 @ 10:07

Hi Jeffrey,

I can't solve your exact problem right now, but here's some code I use to select a random node of site content on a page - you can probably modify it quite easily to pick members rather than content nodes:

<?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 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:variable name="source" select="/macro/source"/>

<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="/">
 <xsl:for-each select="umbraco.library:GetXmlNodeById($source)/node [string(data [@alias='umbracoNaviHide']) != '1']">
       <xsl:sort select="random:GetRandom(1, count($currentPage/node))" order="ascending" />
           <xsl:if test="position() &lt; 2">
               <p>...Output content...</p>
           </xsl:if>
       </xsl:for-each>
 </xsl:template>
</xsl:stylesheet>

Hope it helps...

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Jan Skovgaard 11258 posts 23500 karma points MVP 7x admin c-trib

Mar 19, 2010 @ 10:11

0

Maybe this Wiki-Entry http://our.umbraco.org/wiki/reference/code-snippets/getmembersbygroupname can be usefull in combination with the random code Dan provided?

/Jan

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Lee Kelleher 3945 posts 15163 karma points MVP 10x admin c-trib

Mar 19, 2010 @ 10:39

Hi Jeffrey,

Here's an alternative suggestion that makes use of the SQL for XSLT package. Install the package, follow the instructions, then try the following SQL query:

SELECT
    TOP 1
    CAST(xml AS XML)
FROM
    cmsContentXml AS x
    INNER JOIN cmsMember AS m ON x.nodeID = m.nodeID
ORDER BY
    NEWID()
FOR XML
    PATH('')
;

This will return a single XML node of the member data - selected randomly, (of course).

An example XSLT would be:

<xsl:template match="/">

    <xsl:variable name="sql">
        <xsl:text disable-output-escaping="yes"><![CDATA[
SELECT
    TOP 1
    CAST(xml AS XML)
FROM
    cmsContentXml AS x
    INNER JOIN cmsMember AS m ON x.nodeID = m.nodeID
ORDER BY
    NEWID()
FOR XML
    PATH('')
;
]]></xsl:text>
    </xsl:variable>

    <xsl:variable name="randomMember" select="jesper.sql:SQLXml($sql)"/>

    <xsl:if test="count($randomMember/node) &gt; 0">
        <xsl:value-of select="$randomMember/node/@nodeName" />
    </xsl:if>

</xsl:template>

I haven't tested this myself, but it should work... might need some playing around with it!

Good luck, Lee.

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Lee Kelleher 3945 posts 15163 karma points MVP 10x admin c-trib

Mar 19, 2010 @ 10:42

0

Quick update, (because we can't edit posts).

Get rid of the "FOR XML PATH('')" bit ... the SQL for XSLT extension does that automatically for you.

It will return the XML in a root element called <x> ... so change the XSLT example to use "$randomMember/x/node".

Cheers, Lee.

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[email protected] 378 posts 1915 karma points MVP 4x c-trib

Mar 22, 2010 @ 10:34

0

Hi Lee,

it works great!Thanks!

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Select a random member in XSLT