Re: [xsl] Find the node with maximum elements

["G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>]

At 2007-11-04 19:08 +0530, Mukul Gandhi wrote:
> This solution is nice. But if the OP wants to do something else with
> the result (not just printing the result (sequence), as this solution
> demonstrates), then I think xsl:for-each loop will be necessary.

Excellent point, Mukul.  Thank you.

Then, my original post stands ... in that solution there is a loop, 
but whereas you looped over all car manufacturers and you do a check 
in side the loop every time, this will only loop for those that have 
the maximum number of cars.

<xsl:template match="/">
  <xsl:for-each select="/Sample/*[count(Car)=max(/Sample/*/count(Car))]">
    <xsl:value-of select="name(.)"/>
    <xsl:text>
</xsl:text>
  </xsl:for-each>
</xsl:template>

And I'm assuming that since the expression 
"max(/Sample/*/count(Car))" is based on an absolute XPath address, 
there is an opportunity for processor optimization because the value 
will never change for the entire execution of the stylesheet.

. . . . . . . . . . Ken

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