Re: Grouping problem

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
>
>



--
Andrew Welch
http://andrewjwelch.com