Re: [xsl] Find the node with maximum elements

Subject: Re: [xsl] Find the node with maximum elements
From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>
Date: Sun, 04 Nov 2007 08:46:17 -0500
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(.)"/>

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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