Subject: Re: [xsl] Muenchian Method From: Jeni Tennison <mail@xxxxxxxxxxxxxxxx> Date: Fri, 3 Aug 2001 15:16:37 +0100 |
Hi Doug, > I would like to group an XML file so the HTML would be like this: Your XSLT is almost there. The only problem is that the path you're using in the use attribute of xsl:key (and later, when you try to get unique nodes) isn't correct. Your key is: <xsl:key name="contacts-by-surname" match="member" use="last_name" /> But you the last_name element isn't a direct child of the member element - you should be stepping down through the name element, i.e.: <xsl:key name="contacts-by-surname" match="member" use="name/last_name" /> ^^^^^ Similarly, when you select the unique members, you need to use a path that takes you from the member element to the last_name element, i.e.: member[count(.|key('contacts-by-surname', name/last_name)[1]) = 1] ^^^^^ In the final use of the key, you got the path right, but added quotes around it, so it's trying to find members with the surname "name/last_name", which I doubt anyone has. There, again, you need: key('contacts-by-surname', name/last_name) I hope that helps, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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