RE: [xsl] sorting and grouping

[Mukul Gandhi <mukul_gandhi@xxxxxxxxx>]

Hi Mick,
Please find below the Muenchian Grouping solution to
the problem described by you -

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

<xsl:output method="text"/>
	
<xsl:key name="by-jan" match="person[birthmonth =
'January']" use="birthday"/>
	
<xsl:template match="/persons">
  <xsl:for-each select="person[birthmonth =
'January']">
    <xsl:sort select="birthday" data-type="number"/>
    <xsl:if test="generate-id(.) =
generate-id(key('by-jan', birthday)[1])">
       <xsl:value-of
select="birthday"/><xsl:text>&#xa;</xsl:text>
       <xsl:for-each select="key('by-jan', birthday)">
	 <xsl:sort select="name"/>
	 <xsl:value-of
select="name"/><xsl:text>&#xa;</xsl:text>
       </xsl:for-each>
    </xsl:if>
  </xsl:for-each>
</xsl:template>
	
</xsl:stylesheet>

Regards,
Mukul


--- "m.vanrootseler" <m.vanrootseler@xxxxxxxxx> wrote:
> Thanks for the link, Michael. I was already afraid
> I'd have to use Muenchian
> grouping, something that I've never done before. 
> 
> Mick
> 
> 
> -----Oorspronkelijk bericht-----
> Van: Michael Kay [mailto:mhk@xxxxxxxxx] 
> Verzonden: vrijdag 2 juli 2004 14:57
> Aan: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Onderwerp: RE: [xsl] sorting and grouping
> 
> You'll find the answer at
> http://www.jenitennison.com/xslt/grouping
> 
> Michael Kay 
> 
> > -----Original Message-----
> > From: m.vanrootseler
> [mailto:m.vanrootseler@xxxxxxxxx] 
> > Sent: 02 July 2004 13:40
> > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > Subject: [xsl] sorting and grouping
> > 
> > I've got a sorting problem. My XML is as follows:
> > 
> > <person>
> > 	<name>Kermit</name>
> > 	<birthday>3</birthday>
> > 	<birthmonth>January</birthmonth>
> > </person>
> > etc. 
> > 
> > XSLT:
> > 
> > <xsl:for-each
> select="person[birthmonth='January']">
> >     <xsl:sort select="birthday"
> data-type="number"/>
> >     <xsl:sort select="name"/>
> >     <xsl:value-of select="birthday"/>
> >     <xsl:text> - </xsl:text>
> >     <xsl:value-of select="name"/>
> >     <br/>
> > </xsl:for-each>
> > 
> > With the above code, each birthday number is
> repeated. What I 
> > would like is
> > to have the birthday number appear only once
> followed by the 
> > names of people
> > whose birthday that is. I suspect it can be done
> by testing 
> > if the birthday
> > value is the same as the preceding sibling, but I
> can't get 
> > it right. Does
> > anyone have any idea how to solve this? 
> > 
> > Mick



		
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