Re: [xsl] sort and sum

Subject: Re: [xsl] sort and sum
From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx>
Date: Fri, 30 Nov 2001 17:24:06 +0000
Hi Maneshi,

> i want to get the sum for elements after 10 top  elements

The easiest way to do this is with a two-step transformation. First
sort the rows into the order that they should be in:

  <xsl:variable name="sorted-rows-rtf">
    <xsl:for-each select="row">
      <xsl:sort select="col" data-type="number" order="descending" />
      <xsl:copy-of select="." />

Then convert that result tree fragment into a node set using a
node-set() extension function (which one you use depends on what
processor you're using):

  <xsl:variable name="sorted-rows"
                select="exsl:node-set($sorted-rows-rtf)/row" />

And then sum the top ten with the sum() function:

  <xsl:value-of select="sum($sorted-rows[position() &lt;= 10]/col)" />

If you don't want to use an extension function, then the 'pure' way is
either using two stylesheets or a recursive template that steps
through the rows one by one, keeping track of their sum as you go. Let
us know if you need help putting that together.

I think that Dimitre also has a generic template for doing sorting and



Jeni Tennison

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