Subject: RE: [xsl] Grouping and Sorting on value inside group From: "Hunsberger, Peter" <Peter.Hunsberger@xxxxxxxxxx> Date: Wed, 12 Jun 2002 10:07:37 -0500 |
> 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. As I said in my reply to Wendell, I think this is why the generate-id works: it groups the sets of elements by the matching dataid? > > So won't this work just as well: > <xsl:for-each select="key('dataset','all')/*/*"> > By itself, that returns everything ungrouped. If I try to add grouping logic to it I get nothing, I'm not sure why, but I'll spend some more time trying to figure it out... > The sort looks really cunning and probably is doing all the hard work here > ;) Hah, I wish, the group was returning the proper number of rows before I every added the sort. It just wasn't sorting them... I'm afraid to admit this, but the more I work on this, the more confused I get... XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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