Subject: RE: question with using Muenchian/xsl:key (Re: sort/group/count p robl em) From: Xiaocun Xu <XXu@xxxxxxxxxxxxxxxxxx> Date: Sat, 11 Nov 2000 19:22:23 -0500 |
Thanks for the suggestion, but this did not seems to work. When I changed to use="concat(generate-id(..),@itemid)", the group by @itemid was not built. Xiaocun -----Original Message----- From: David Carlisle To: xsl-list@xxxxxxxxxxxxxxxx Sent: 11/11/00 3:30 PM Subject: Re: question with using Muenchian/xsl:key (Re: sort/group/count probl em) > > key('items-by-itemid', @itemid) returns all items with the same @itemid in > the entire XML document. I just want all items with the same @itemid in > each of the itemlist element, how can I do that? don't you just want to replace <xsl:key name="items-by-itemid" match="item" use="@itemid"/> by something like <xsl:key name="items-by-itemid" match="item" use="concat(generate-id(..),@itemid)" /> so that your key values are all specific to a given itemlist. Not that I've tried it.... David _____________________________________________________________________ This message has been checked for all known viruses by Star Internet delivered through the MessageLabs Virus Control Centre. For further information visit http://www.star.net.uk/stats.asp XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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