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RE: Can't Obtain Results
- From: "Lindy Quick" <lindyq at iwon dot com>
- To: xsl-list at lists dot mulberrytech dot com
- Date: Fri, 3 May 2002 09:58:42 -0400 (EDT)
- Subject: RE: [xsl] Can't Obtain Results
- Reply-to: xsl-list at lists dot mulberrytech dot com
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>
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