Hi Mike,
Thanks for your observations.
The following solution is likely more efficient than Ken's (because,
it calculates 'max' only once):
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; version="2.0">
<xsl:output method="text" />
<xsl:template match="/Sample">
<xsl:variable name="max-count" select="max(*/count(Car))" />
<xsl:for-each select="*">
<xsl:if test="count(Car) = $max-count">
<xsl:value-of select="local-name()" /><xsl:text> </xsl:text>
</xsl:if>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
On 11/4/07, Michael Kay <mike(_at_)saxonica(_dot_)com> wrote:
Yes, I read the requirement "return the node", not the example "return Honda
and Mitsubishi".
The solution *[count(*) = max(current()/*/count(*))] is easy to write, but
it's very dependent on optimization. Saxon will move the condition
max(current()/*/count(*)) out of the loop if it's written this way, but not
if it's written *[count(*) = max(../*/count(*))]. Even if the max() is
calculated outside the loop, you're visiting each node twice and calculating
the "key" (count(*) twice for each node. Hence the slight preference for the
sorting approach.
The most efficient solution is probably a recursive function, but that's not
the easiest to write. It really calls out for a higher-order function along
the lines of saxon:highest().
Michael Kay
http://www.saxonica.com/
--
Regards,
Mukul Gandhi
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