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RE: Sorting in descending order on the sum of a calculation
- From: "delay" <delay at pobox dot com>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Thu, 29 Nov 2001 11:34:57 -0600
- Subject: RE: [xsl] Sorting in descending order on the sum of a calculation
- References: <200111290730.CAA09256@biglist.com>
- Reply-to: xsl-list at lists dot mulberrytech dot com
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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