Subject: Re: [xsl] grouping, sorting, splitting From: beowulf <carisenda@xxxxxxxxx> Date: Mon, 18 Apr 2005 13:29:50 +0100 |
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@xxxxxxxxxxxx> 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@xxxxxxxxx] > > Sent: 18 April 2005 11:12 > > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > > 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 > > -- "Nothing spoils fun like learning how it builds character." Calvin, Calvin & Hobbes
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