Subject: RE: [xsl] Sorting in descending order on the sum of a calculation From: "delay" <delay@xxxxxxxxx> Date: Thu, 29 Nov 2001 11:34:57 -0600 |
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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