Re: grouping, sorting, splitting

Is that something I'd need XSLT2.0 for or am I misunderstanding the
RTF thing from XSLT1.0?

On 4/18/05, Michael Kay <mike(_at_)saxonica(_dot_)com> wrote:
> You should do this as a pipelined (or multi-phase) transformation. Put the
> results of the first grouping/sorting operation in a variable (a temporary
> tree) and then apply another transformation to put the elements into groups
> of three.
>
> Michael Kay
> http://www.saxonica.com/
>
>
> > -----Original Message-----
> > From: beowulf [mailto:carisenda(_at_)gmail(_dot_)com]
> > Sent: 18 April 2005 11:12
> > To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
> > Subject: [xsl] grouping, sorting, splitting
> >
> > Hi,
> >
> > I have XML if the form:
> >
> > <page>
> > <entry date="2005-04-15">
> >  <title>foo</title>
> > </entry>
> > <entry date="2005-04-15">
> >  <title>bar</title>
> > </entry>
> > <entry date="2005-02-05">
> >  <title>baz</title>
> > </entry>
> > ...
> > </page>
> >
> > Which I am trying to group by date, sort by tiltle and then split into
> > sets of 3, 3 being the number of columns in the HTML TABLE element I
> > am trying to produce as an end result.
> >
> > I've got the grouping and sorting:
> > <xsl:for-each select="entry[key('days', @date) and count(.|key('days',
> > @date)[1])= 1]">
> > <xsl:sort select="title"/>
> >
> > and I've even got the first item in each group of three from that
> > grouped and sorted set:
> > <xsl:for-each select="key('days', @date)[position() mod 3 = 1]">
> >
> > But I just can't seem to make the final leap to displaying the
> > following siblings of the above, making the 3 cell rows. Should I be
> > doing this some other way or can someone help me where I am?
> >
> > Many thanks,
> > Stephen
> >
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