derekmw wrote:

> <a>
> 	<b>
> 		<c type="period">
> 			<key>
> 				<d name="date">2006-09-12 00:00:00.0</d>
> 			</key>
> 			...(many other elements, many which have nested elements 6-7 levels
> deep)
> 		</c>
> 		<c type="period">
> 			<key>
> 				<d name="date">2006-09-15 00:00:00.0</d>
> 			</key>
> 			...(many other elements, many which have nested elements 6-7 levels
> deep)
> 		</c>
> 	</b>
> </a>
> 
> From what I read/understood, with xsl, you have to specify all the
> fields you want to copy over to the new xml transformation?  If so,
> that's just not an option as there are so many other elements within
> the 'c' elements I want to sort by.
> 
> What I need to do is find a quick way to sort the 'c' elements by the
> 'd' element value (date).  Any suggestions/direction would be
> appreciated!

With XSLT 1.0 you can put xsl:sort elements as children of the 
xsl:for-each or xsl:apply-templates elements. What you do then is up to 
you thus if you want to sort and then make a deep copy that is certainly 
possible e.g.

   <xsl:template match="b">
    <xsl:copy>
     <xsl:for-each select="c">
       <xsl:sort select="key/d[@name = 'date']"/>
       <xsl:copy-of select="."/>
     </xsl:for-each>
    </xsl:copy>
   </xsl:template>

So the above is a template for 'b' elements that makes a shallow copy of 
the 'b' element and then processes the 'c' children, sorted by the text 
value of the "key/d[@name = 'date']" grandchild element and then simply 
makes a deep copy.


-- 

	Martin Honnen --- MVP XML
	http://msmvps.com/blogs/martin_honnen/