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(_dot_)heath(_at_)gmail(_dot_)com>
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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