Re: [xsl] Grouping problem

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

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