RE: [xsl] XSL distinct group by date

["Mindy McCutchan" <karma@xxxxxxxxxxxxxx>]

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] 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/