[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: XSL distinct group by date
Hi, The grouping tutorial helped a bit, but I'm not sure how to match by just part of the content. This is the XSL I have so far: <xsl:key name="documents-by-date" match="newsitem" use="date" /> <xsl:template match="news"> <xsl:apply-templates select="newsitem[generate-id(.) = generate-id(key('documents-by-date', date)[1])]" /> </xsl:template> <xsl:template match="newsitem"> <xsl:value-of select="substring(date, string-length(date)-3, 4)" /> </xsl:template> I have it so it outputs just the year portion of the date( ex: 5/6/2004), but with the dates 5/6/2004, 5/7/2004, 4/8/2005, I want only 2004 & 2005 to show up once. Do I need to alter my xml to store the year seperately to group by it only? This is current structure: <news> <newsitem newsid="1"> <title>news release1</title> <date>5/6/2004</date> <description></description> <fulltext></fulltext> </newsitem> </news> Thanks! I'm really learning a lot. Mindy > >-----Original Message----- > >Date: Sun, 24 Apr 2005 08:56:55 +0100 > >To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> > >From: "Michael Kay" <mike@xxxxxxxxxxxx> > >Subject: RE: XSL distinct group by date > > > >Are you familiar with > > > >http://www.jenitennison.com/xslt/grouping > > > >which gives the standard XSLT 1.0 approaches to grouping problems? > > > >Michael Kay > >http://www.saxonica.com/
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