Subject: Re: [xsl] transforming XML to XML with XSLT 1.0 From: David Carlisle <davidc@xxxxxxxxx> Date: Fri, 20 May 2005 15:24:59 +0100 |
> What do I need to add when I get more then 1 <aid> from the database ? some more code. I thought that was probably the case but wasn't prepared to guess from a sample of one what the possible inputs were or what output you wanted in other cases. Your requested output format didn't appear to group the elements with the same aid by anything other than the top level element so it's not clear how you want multiple aid values reporting, similarly it's not clear what you are supposed to do if the aid values are the same but the type or version numbers are different. Basically assuming that aid, type and version are the key that you want to group on, you want to use muenchian grouping with a key of match="Row" use="concat(@AID,':',@GROUP_TYPE,':',@GROUP_VERSION) and then, for the nodes in each group do the for-each that I posted last time. David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________
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