Wendell,
I tried your approach and the xpath returned me all of the person elements.
I then realized that the @name param to your xsl:key call needs to wrapped
within a substring:
//person[count(.|key('person-by-initial', substring(@name,1,1))[1]) = 1]
which is, by the definition of your xsl:key, I'm sure what you meant to do.
-Jeff
-----Original Message-----
From: Wendell Piez [mailto:wapiez(_at_)mulberrytech(_dot_)com]
Sent: Wednesday, November 20, 2002 6:22 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] How do you select all unique first-position
characters?
I should think a key using the initial letter of each name would be a big
help:
<xsl:key name="person-by-initial" match="person"
use="substring(@name,1,1)"/>
Then "key('person-by-initial', 'A')" would get you back the elements for
Anthony, Amy and the other Amy, and "key('person-by-initial', 'A')[1]"
would get you just Anthony.
//person[count(.|key('person-by-initial', @name)[1]) = 1] gets you all
person elements that are first with their initial.
In other words, the usual grouping stuff.
Cheers,
Wendell
(At 04:50 PM 11/20/2002, Greg wrote:
At 01:36 PM 11/20/2002, you wrote:
<report-data>
<person name="Anthony" />
<person name="Amy" />
<person name="Amy" />
<person name="Jim" />
<person name="Jason" />
</report-data>
I would expect to get A,J from the above XML
Here's most of it for you. I'm having trouble with the initial
variable selection though. I'm hoping somebody can take it from here, as
I'm stumped now too....
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Wendell Piez
mailto:wapiez(_at_)mulberrytech(_dot_)com
Mulberry Technologies, Inc. http://www.mulberrytech.com
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