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[xsl] selecting node with longest data string

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Hi All,

I'm using MSXML3 sp1.

I've got an XML that looks like :

<?xml version="1.0" encoding="ISO8859-1"?>
<items>
	<item>dd</item>
	<item>eeeee</item>
	<item>aaa</item>
	<item>ffffffffff</item>
	<item>c</item>
	<item>bbbb</item>
</items>

I want to get the item which has the longest data string as its value.
I managed to do it with the following xsl, but it seems very crude. Could
someone please tell me a more efficient way of doing this.


<?xml version="1.0" ?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.
0" >
<xsl:output method='html' />

<xsl:template match="/">
	<xsl:variable name="SortedItems" >
		<xsl:call-template name="SortItems" />
	</xsl:variable>
	
	<xsl:variable name="FinalValue">
		<xsl:value-of select="substring-before($SortedItems,'|')" />
	</xsl:variable>
	
	longest = <xsl:value-of select="string-length($FinalValue)" />
	
</xsl:template>


<xsl:template name="SortItems">
	<xsl:for-each select="items/item">
		<xsl:sort select="string-length(.)" data-type="number"
order="descending" />
		<xsl:value-of select="." />|
	</xsl:for-each>
</xsl:template>

</xsl:stylesheet>

Thanks in advance,
Mayura

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