RE: [xsl] Sorting in descending order on the sum of a calculation

Thanks for the example Joshua...

It didn't work for me but it did help me to understand some other features
of xsl.  Your example makes a lot of sense to me.  A programming friend
helped me with this solution and I figured I would post it to help others
with similar troubles.  Thanks for the example you provided.


<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:template match="/">
  <xsl:for-each select="//referers/referer">
   <xsl:sort
select="format-number(sum(//referers//referer[@page=current()/@page]/@hits),
'0000000')" order="descending"/>
   <xsl:sort select="@page" order="ascending"/>
   <xsl:if test="not(@page=preceding::referer/@page)">
     <xsl:value-of
select="format-number(sum(//referers//referer[@page=current()/@page]/@hits),
'######00')"/> -- <xsl:value-of select="@page"/><br/>
   </xsl:if>
   </xsl:for-each>
 </xsl:template>
</xsl:stylesheet>

Also I apologize for the double post yesterday.  For some reason I was
getting it returned as undeliverable mail on both posts I made.  Of course
they both ended up showing in the list:-)

Thanks,
-Delay


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