If i have the following code -- how would i sort by the element "name"

Dim elJobItemList As Xml.XmlNode

XSLT has a sort function built in that allows you to sort on an element.  But that means that you will need to transform your XML with XSLT and parse it that way

<xsl:sort select="*[name() = 'name']" order="ascending"/>

do you have a code example of how to do this

The following site gives you a good example on how to write an XSLT using the sort function. 

http://www.developer.com/xml/article.php/1560361

If you have any questions on what the author is doing in the XSLT example he gives I'll be happy to answer.

 To apply the transformation in .net 2.0 you would use XSLCompiledTransform and we can help you out with that if you need it.

i think i got my xsl the way it needs to be but im having trouble getting it to transform in my code

Dim xmlDocJobItemList As New Xml.XmlDocument()

Dim myXslTransform As XslCompiledTransform = New XslCompiledTransform

Dim settings As New XmlReaderSettings()

settings.IgnoreWhitespace = True

Dim reader As XmlReader = XmlReader.Create(myXPathDocument.CreateNavigator.Name, settings)

What you have to do is load your XslCompiledTransform object with the stylesheet, then set your xmldocument in the transform along with a object to output the xml to.  In this case here we are loading an xmlwriter with a StringWriter object.  This will allow us to output the transfomred XML to a string.

Dim myXslTransform As XslCompiledTransform = new XslCompiledTransform
Dim tempStringWriter As New StringWriter
Dim tempXMLWriter As XMLWriter = New XMLTextWriter(tempStringWriter)

myXslTransform.Load("xsltztest.xslt")
myXSLTransform(myxmlstring, tempstring)

textbox1.text = tempStringWriter.ToString()

I found a link to something similar right after I posted but thanks

how do i turn it back into xml?

I am still having some trouble getting the xml to output something which im guessing is the xsl is not properly put together

xml/xsl segment below -- im trying to get the name, listid, and ParentRef/FullName and sort by name -- it returns nothing -- can you see the error? 

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">


<xsl:param name="sortBy" select="'Name'"/>
<xsl:param name="strXPath" select="//ItemServiceRet"/>
<xsl:template match="/">
  <xsl:apply-templates select="$strXPath">
    <xsl:sort select="*[name()=$sortBy]" order="ascending"/>
  </xsl:apply-templates>
</xsl:template>
<xsl:template match="Person">
 <xsl:value-of select="Name"/>
    <xsl:value-of select="ParentRef/Fullname"/><
 
</xsl:template>
</xsl:stylesheet>

<ListID>2A40000-1171382505</ListID>

<TimeModified>2007-07-03T10:19:50-05:00</TimeModified>

<Name>01.820 Other</Name>

<IsActive>true</IsActive>

<ListID>2A30000-1171382402</ListID>

</ParentRef>

<SalesOrPurchase>

<Price>0.00</Price>

<ListID>B10001-1124894555</ListID>

</AccountRef>

</ItemServiceRet>

okay few things I see here.

 First off in your XSLT you have a lonely "<" right after your xsl:select statement for /parentref/fullname.

 Once you remove that you will get an output.  However it wont be selecting what you want.

Your sorting in one template and then selecting values in a 2nd template but not applying that template.

I've created another sample XSL for you to work from.   Take alook at this and see if this helps out.  It will print out the 1st <name> element and any of the <parentRef/Fullname> elements.

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:param name="sortBy" select="'Name'"/>
<xsl:param name="strXPath" select="//ItemServiceRet"/>
<xsl:template match="/">
  <xsl:for-each select="$strXPath">
    <xsl:sort select="*[name()=$sortBy]" order="ascending"/>
<xsl:value-of select="Name"/>
    <xsl:value-of select="ParentRef/FullName"/>
    </xsl:for-each>
</xsl:template>
</xsl:stylesheet>