RE: [xsl] Grouping and Sorting on value inside group

[TSchutzerWeissmann@xxxxxxxxxxxxxxxx]

Peter
I was just thinking about this and realised that Wendell was completely
right, of course. 
Doesn't <xsl:for-each select="(//dataset/*/*[generate-id() = 
	generate-id(key('dataids', concat(local-name(..),@dataId)))])">
filter out most of the nodes you want because it only lets through one node
per value of @dataId?
But if it works...
Tom

> -----Original Message-----
> From: TSchutzerWeissmann@xxxxxxxxxxxxxxxx
> [mailto:TSchutzerWeissmann@xxxxxxxxxxxxxxxx]
> Sent: 12 June 2002 10:01
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: RE: [xsl] Grouping and Sorting on value inside group
> 
> 
> hello Peter
> 
> If you want to avoid using "//dataset" you could have a key like this:
> <xsl:key name="dset" match="dataset" use="'all'"/> 
> and access it using 
> select="key('dataset','all')"
> 
> > >>I finally found the answer.  Your use of [@dataid = 
> > current()/@dataid]
> > >>provided the missing clue:
> > >>
> > >>       <xsl:for-each select="(//dataset/*/*[generate-id() =
> > >>generate-id(key('dataids', concat(local-name(..),@dataId)))])">
> > >>          <xsl:sort select="parent::node()/*[local-name() = 
> > $sortcol and
> > >>@dataId = current()/@dataId]/value"/>
> 
> 
> I agree with Wendell that the whole generate-id() malarkey is 
> redundant, but
> it doesn't do the same thing as key('dataids',
> concat(local-name(..),@dataId))[1] because you will get every 
> node that
> satisfies the predicate and not just the first one. 
> And every node in //dataset/*/* that has a @dataid will be in 
> the 'dataid'
> key and satisify the predicate.
> 
> So won't this work just as well:
> <xsl:for-each select="key('dataset','all')/*/*">
> 	
> The sort looks really cunning and probably is doing all the 
> hard work here
> ;)
> 
> Regards,
> Tom
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list