Subject: RE: Sorting in descending order on the sum of a calculation
From: "delay" <delay@xxxxxxxxx>
Date: Thu, 29 Nov 2001 11:34:57 -0600
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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
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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