Techno_Dex wrote:
> I'm not sure I agree with you on the <xsl:copy> issue as when using the 
> <xsl:copy-of> the element is copied as is from the inport node list to the 
> output node list and if it were a serializer issue then this would have been 
> converted over to contain both the start and end tags as well which is not 
> the case.  I am using the simple XslTransform class to perform the 
> conversion using the .NET 3.5 framework which by default will use the MSXML 
> 6 parsing engine which doesn't add an end element tag unless an empty value 
> has been added to the node which is what I suspect the xsl:copy function is 
> doing.  The attributes on this function don't appear to to contain any 
> formating instructions as well to indicate not adding an empty value to the 
> element. Is there a way to do a deep copy on the current node() since there 
> are no children in order to obtain the desired results?  I'm assuming I 
> might have to use a xsl:foreach to obtain this functionality which is not 
> very generic.  Am I missing something on the XslTransform perhaps which will 
> convert the empty value elements to empty element instead?
> 
>             XslTransform xtXslTransform;
>             xtXslTransform = new XslTransform();
>             xtXslTransform.Load(asXSLTPath);
>             xtXslTransform.Transform(asXMLInputPath, asXMLOutputPath);

Note that XslTransform is obsolete since .NET 2.0, you should use 
XslCompiledTransform instead, unless you have a stylesheet originally 
run with XslTransform and you have compatibility issues with 
XslCompiledTransform.
And MSXML 6 is a completely different software, it has nothing to do 
with either XslTransform or XslCompiledTransform.
I think with XslCompiledTransform the issue with the Record elements 
goes away.


-- 

	Martin Honnen --- MVP XML
	http://JavaScript.FAQTs.com/