Hi Lee,

The problem is (as you may or may not have found out) that it's not allowed to use a variable or parameter in the select attribute of a sort instruction, so the way to get around it is to use a choose statement and test for all the possible values:

<?xml version="1.0" encoding="utf-8" ?>
<xsl:stylesheet
    version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:umbraco.library="urn:umbraco.library"
    exclude-result-prefixes="umbraco.library"
>

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

    <xsl:param name="currentPage" />


    <xsl:variable name="sortby" select="umbraco.library:RequestQueryString('sortby')" />
    <xsl:variable name="nodes" select="$currentPage/node" />


    <xsl:template match="/">

        <xsl:choose>

            <!-- Sort by name? -->
            <xsl:when test="$sortby = 'name'">
                <xsl:apply-templates select="$nodes">
                    <xsl:sort select="@nodeName" data-type="text" order="ascending" />
                </xsl:apply-templates>
            </xsl:when>

            <!-- Sort by id? -->
            <xsl:when test="$sortby = 'id'">
                <xsl:apply-templates select="$nodes">
                    <xsl:sort select="@id" data-type="number" order="ascending" />
                </xsl:apply-templates>        
            </xsl:when>

            <!-- More ways to sort... -->

            <!-- Just take the Umbraco Content order then... -->
            <xsl:otherwise>
                <xsl:apply-templates select="$nodes" />
            </xsl:otherwise>
        </xsl:choose>

    </xsl:template>

    <!-- Template for how a single node should be output -->
    <xsl:template match="node">
        <p>
            Name: <xsl:value-of select="@nodeName" />, Id: <xsl:value-of select="@id" />
        </p>
    </xsl:template>


    <!-- No output for hidden nodes -->
    <xsl:template match="node[data[@alias = 'umbracoNaviHide'] = 1]" />


</xsl:stylesheet>

/Chriztian

PS: I hope you get to like XSLT some day, 'coz I loooove beer :-)

Thanks for the help chaps, very, very appreciated :)

@ Chris - I actually found that page, but couldn't quite get my head round the Xpath in the sort select=""?  I understand what she is saying and tried this

   <xsl:sort select="./data [@alias = 'distance']" data-type="number" />
        <xsl:sort select="./data [@alias='eventDate'][$resultsort = 'date']" data-type="text" order="ascending" />

But it didn't do anything :S   Do I just add the condition after the XPath?  And is this right?  I want to sort on this

./data [@alias='eventDate']

So added the following condition afterwards, which based on that ladies code should say if the variable $resultsort is equal to 'date' then apply the sort?

[$resultsort = 'date']

Which gives me

./data [@alias='eventDate'][$resultsort = 'date']

@ Chriztian - Thanks for the code, I'll have a good look at it in a minute and see if I can use it if I can't get the XSLt working based on the link Chris gave.  The problem I have is the XSLT is quite complex already and I'm worried about adding all that extra code :(   But if thats the only way to do it, then so be it ;)   If it works, yet another person I'll owe beer too :)

IT'S ALIVE I TELL YA, ALLLIIIIIIIIIVVEEEE

Thanks for the help again everyone, I used idea from the link Chris put up and think I have it working now

http://www.biglist.com/lists/xsl-list/archives/200206/msg00612.html

Pretty fiddly, will double check in the morning when I have had a break from this screen...lol  

Thanks again :)

Yeah - I had a "Weee" moment too - the template with the choose an' all in my example can actually be written llike this, using Jeni's trick from the link above:

   <xsl:template match="/">

        <xsl:apply-templates select="$nodes">
            <xsl:sort select="@nodeName[$sortby = 'name']" data-type="text" order="ascending" />
            <xsl:sort select="@id[$sortby = 'id']" data-type="number" order="ascending" />
        </xsl:apply-templates>

    </xsl:template>

/Chriztian

Hey Chriztian, thats 99% the same way I have done it on the page - But my sorts are for a number or by a date and the date is a DocType property

<xsl:for-each select="$searchResults">
          <xsl:sort select="umbraco.library:FormatDateTime(./data [@alias = 'eventDate'] [string($resultsort) = 'date'], 'yyyyMMdd')" order="ascending" />
          <xsl:sort select="./data [@alias = 'distance'] [string($resultsort) = 'distance']" data-type="number" />
    .....
    .....
</xsl:for-each>

Hope this thread helps someone else

I'm having a tough time with this.  I have a property named sortOrder which is a numeric type.  When I put this inside my for-each loop,

 <xsl:sort select="./data [@alias = 'sortOrder'] [string($resultsort) = 'sortOder']" data-type="number" />

But when I put this line in, I get this error:

Error occured

System.Xml.Xsl.XslLoadException: The variable or parameter 'resultsort' is either not defined or it is out of scope. An error occurred at D:\www\beta.bayarts.net\html\xslt\634018299623240000_temp.xslt(25,3).
at System.Xml.Xsl.XslCompiledTransform.LoadInternal(Object stylesheet, XsltSettings settings, XmlResolver stylesheetResolver)
at System.Xml.Xsl.XslCompiledTransform.Load(XmlReader stylesheet, XsltSettings settings, XmlResolver stylesheetResolver)
at umbraco.presentation.webservices.codeEditorSave.SaveXslt(String fileName, String oldName, String fileContents, Boolean ignoreDebugging)

Any ideas?  I thought ResultSort is a variable created by the sort.

Thanks!

--Kent

Hi Kent, can you post your entire XSLT here? resultsort would need to be defined in the your XSLT manually. With the rest of your file, I should be able to help out. 

Off the bat, if you are simply tring to sort by that property it should look like this:

<xsl:sort select="data [@alias = 'sortOrder']" data-type="number" />

Thanks,
Nik

I'm trying to do the same task, sort based on a column if that coumn is specifed in the Url as the 'sort' querystring, but it's not working. The basic sort on 'lastName' works if I remove:

[string($sortby)='lastName']

 Any ideas?

...
<xsl:variable name="sortby"select="umbraco.library:RequestQueryString('sort')"/>

<xsl:template match="/">

<!--The fun starts here -->
<table border="1" cellspacing="0" cellpadding="2">
<tr class="gridHeader">
  <td><b><a href="?sort=number">#</a></b></td>
  <td><b><a href="?sort=lastName">Player</a></b></td>
  <td><b><a href="?sort=position">Position</a></b></td>
...
</tr>
<xsl:for-each select="$currentPage/node [string(data [@alias='hide'])!='1']">
<xsl:sort select="data [@alias='lastName'][string($sortby)='lastName']" data-type="text" order="ascending" />
...

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