--- "Martin Lormes" <martin dot lormes at gmx dot net> wrote:
----- Original Message -----
From: "Dimitre Novatchev" dnovatchev at yahoo dot com
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Thursday, September 26, 2002 5:11 PM
Subject: [xsl] Re: Selecting a random node from source-tree
I was thinking of an XPath expression like this:
document('funnies.xml')/funnies/quote[random()]
Can I avoid using extension elements?
Yes, use the "randNext" template from FXSL.
More information is contained in the article:
"Casting the Dice with FXSL: Random Number Generation Functions in
XSLT"
http://fxsl.sourceforge.net
/articles/Random/Casting%20the%20Dice%20with%20FXSL-htm.htm
Hope this helped.
Thanks, Dimitre. Awsome article!
Unfortunately I always get the same random number (which makes
perfect
sense, since I can't dynamically generate a starting number :-( )
So I must add this as a condition to my original question: the
stylesheet is
static. No way of changing that.
Thanks for your help.
Martin Lormes
That's easy -- have the minutes or seconds part of the current time (or
whatever else you find appropriate) passed as a parameter to the
stylesheet and use that as the seed (or use a subsequence of the
original sequence, starting with the element having index equal to this
parameter.
=====
Cheers,
Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL
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