Subject: RE: [xsl] Can't Obtain Results From: "Lindy Quick" <lindyq@xxxxxxxx> Date: Fri, 3 May 2002 09:58:42 -0400 (EDT) |
Jeni, Your Solution worked while I know that I will only have these vaules in the immediate future. Thank you. Also good idea on the email format, but you can see below it doesn't work. I have notified my provider. Michael, You are correct in your second assumption that I am looking for the values that are not in the second set. I am interested in pursuing this farther, and will spend most of today working with Keys and groups. I am however limited to using the XSLT 1.0. While I see the ease of using 2.0, I am stumped how to create that in 1.0 Can anyone help? Lindy MK Wrote: > What's special about f1 and f6? Is it that they are the first and last > elements in the first keyset, or is it that they are the only ones not > present in the second keyset? (Sometimes reverse engineering the > algorithm > from an example of its output isn't easy...) > > Let's assume the latter. In this case you have an > elimination-of-duplicates > problem, which is essentially the same as a grouping problem. The > grouping > key (the thing that's the same between matching elements) is the element > name rather than it's value, so you can't use the simple > *[not(.=preceding::*)] technique; instead you need to use keys. > > Look up Muenchian grouping in your favourite textbook or at > www.jenitennison.com/xslt/grouping, and use it with a grouping key > declared > as > > <xsl:key name="g" match="keyset/*" > use="name()"/> > > Or if you want to use XSLT 2.0, as implemented in Saxon 7.1, you can > write > > <xsl:for-each-group select="(keyset[1] | > keyset[@outlet=$outlet])/*" > group-by = "name()"> > <xsl:copy-of select="."/> > </xsl:for-each-group> ------------------------------------------------ Visit iWon.com - the Internet's largest guaranteed cash giveaway! XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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