Re: [xsl] Grouping problem

["graham.heath" <graham.heath@xxxxxxxxx>]

Thanks Andrew.

I was trying to do it the other way round (only select pairs that could 
be used) rather than find all and filter out the unacceptable combos.

Cheers

On 08/15/2011 05:37 PM, Andrew Welch wrote:
> Ok I've hacked this together, it gets all possible combos then filters
> them down:
>
> <xsl:variable name="pairs" select="//pair"/>
>
> <xsl:variable name="combos" as="element(table)*">
> 	<xsl:for-each select="$pairs">
> 		<xsl:variable name="outer" select="."/>
> 		<xsl:for-each select="$pairs[not(. is $outer)]">
> 			<table>
> 				<xsl:for-each select="$outer, .">
> 					<xsl:sort select="."/>
> 					<xsl:copy-of select="."/>
> 				</xsl:for-each>
> 			</table>
> 		</xsl:for-each>
> 	</xsl:for-each>
> </xsl:variable>
>
> <xsl:template match="/">
> 	<root>
> 		<xsl:for-each-group
> select="$combos[count(distinct-values(.//player)) = 4]" group-by=".">
> 			<xsl:copy-of select="."/>
> 		</xsl:for-each-group>
> 	</root>
> </xsl:template>
>
>
>
>
> On 15 August 2011 16:00, graham.heath<graham.heath@xxxxxxxxx>  wrote:
> > Okay Andrew,
> >
> > I forgot to write out all the permutations :-)
> > You could say that the example was cut down a tad too far..
> > <table>
> >
> > <pair>
> >       <player>player1</player>
> >       <player>player4</player>
> > </pair>
> > <pair>
> >       <player>player2</player>
> >       <player>player3</player>
> > </pair>
> > </table>
> >
> >
> >
> > In your example output, where should the pair:
> > > <pair>
> > >        <player>player1</player>
> > >        <player>player4</player>
> > > </pair>
> > >
> > > go?
> > >
> > >
> > >
> > > On 15 August 2011 14:38, graham.heath<graham.heath@xxxxxxxxx>    wrote:
> > > > Hello mega brains,
> > > >
> > > > I have an xml grouping problem that has so far defeated me.
> > > >
> > > > Given the xml
> > > > <pairings>
> > > > <pair>
> > > >        <player>player1</player>
> > > >        <player>player2</player>
> > > > </pair>
> > > > <pair>
> > > >        <player>player1</player>
> > > >        <player>player3</player>
> > > > </pair>
> > > > <pair>
> > > >        <player>player1</player>
> > > >        <player>player4</player>
> > > > </pair>
> > > > <pair>
> > > >        <player>player2</player>
> > > >        <player>player3</player>
> > > > </pair>
> > > > <pair>
> > > >        <player>player2</player>
> > > >        <player>player4</player>
> > > > </pair>
> > > > <pair>
> > > >        <player>player3</player>
> > > >        <player>player4</player>
> > > > </pair>
> > > > </pairings>
> > > >
> > > > is it possible to group pairs such that each player occurs only once in
> > > > each
> > > > group?
> > > > <table>
> > > > <pair>
> > > >        <player>player1</player>
> > > >        <player>player2</player>
> > > > </pair>
> > > > <pair>
> > > >        <player>player3</player>
> > > >        <player>player4</player>
> > > > </pair>
> > > > </table>
> > > > <table>
> > > > <pair>
> > > >        <player>player1</player>
> > > >        <player>player3</player>
> > > > </pair>
> > > > <pair>
> > > >        <player>player2</player>
> > > >        <player>player4</player>
> > > > </pair>
> > > > </table>
> > > >
> > > > I was of the opinion that some form of xsl:for-each-group would suffice
> > > > but
> > > > have been unable to devise a group-by expression that works..
> > > >     <xsl:for-each-group select="pairings/pair"
> > > > group-by="count(distinct-values(player))=1">
> > > >       <table>
> > > >         <xsl:apply-templates select="current-group() "/>
> > > >       </table>
> > > >     </xsl:for-each-group>
> > > >
> > > > Many thanks
> > > >
> > > > Hector