Hello Tom,
first thanks for trying to help.
I know what you mean, but the problem I have is how do I produce a tree with
distinct values that I can access afterwards. What I do know is first
sorting, so all the same values are together. Than I check every time with
the preceding value, when this values are different I write the node to the
output. But when I now want to use the position function I have the problem
that the nodes that are double are also count. Sending the stylesheet is
difficult because it's more complicated than the examples I have sent, and I
am afraid it would make everything a little bit confusing. The only thing I
actually need is storing a copy of my tree with distinct values that I can
access afterwards. But is seems not as easy as putting it in a variable.
All ideas are welcome !
Kind regards,
Ismaël
-----Original Message-----
From: TSchutzerWeissmann(_at_)uk(_dot_)imshealth(_dot_)com
[mailto:TSchutzerWeissmann(_at_)uk(_dot_)imshealth(_dot_)com]
Sent: vrijdag 22 november 2002 12:33
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] How do can I select distinct nodes and adding a
counter at the sa me time
Hi Ismaël
What I do know is first grouping on the attributes, so I get
every value
only once. Therefore I have used the sorting function. The
problem is I
don't find a method to generate the id. I can't use the count function
because some nodes of the nodeset are not used (only unique values are
needed). Somebody an idea ?
If you're apply-templates to the distinct nodes, or running over them with a
for-each, then you should be able to use position() in either case. What
does your xslt look like?
Regards,
Tom
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