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

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Well....

At 04:40 PM 6/11/2002, Peter wrote:
> 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"/>

...

> The last step is to look at using more keys.  I'd still like to know if I
> can use a key or some combination of keys to eliminate the //dataset
> reference in the for-each...(anyone)???

The expression

(//dataset/*/*[generate-id() =
               generate-id(key('dataids', concat(local-name(..),@dataId)))])

would seem on the face of it to find the same node as

key('dataids', concat(local-name(..),@dataId))[1]

since the node inside any dataset that has the same generated ID as the 
first node in the set returned by the key expression, is that node (or 
nothing).

To introduce the constraint that your node must be two levels under a 
dataset, you could do

key('dataids', concat(local-name(..),@dataId))[1][../parent::dataset]

... But there's only ever one such node; sorting it has no effect.

What am I missing? (I admit I'm looking at this thread late.)

Cheers,
Wendell


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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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