Easiest (even in 2.0) is to sort elements according to the number of
children and take the last:
<xsl:for-each select="Sample/*">
<xsl:sort select="count(child::*)" data-type="number"/>
<xsl:if test="position()=last()">
<xsl:value-of select="name()"/>
</
</
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Avaneesh Ramprasad [mailto:avaneesh(_at_)rocketmail(_dot_)com]
Sent: 03 November 2007 19:05
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Find the node with maximum elements
Hello,
I have a requirement to write a xsl transformation to find
the node which has the maximum number of elements
Below is a sample xml file
<Sample>
<Toyota>
<Car>Camry</Car>
<Car>Corrola</Car>
</Toyota>
<Honda>
<Car>Accord></Car>
<Car>Civic</Car>
<Car>Pilot</Car>
</Honda>
<Mitsubishi>
<Car>Lancer</Car>
<Car>Lancer</Car>
<Car>Lancer</Car>
</Mitsubishi>
<Hyundai>
<Car>Sonata</Car>
<Car>Accent</Car>
</Hyundai>
</Sample>
The xsl should return Honda and Mitsubishi
Would appreciate your help.
Thanks
Avaneesh
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