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(_at_)lists(_dot_)mulberrytech(_dot_)com>
From: "Michael Kay" <mike(_at_)saxonica(_dot_)com>
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/
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