Re: [xsl] position within same nodes

Subject: Re: [xsl] position within same nodes
From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>
Date: Thu, 19 Jul 2001 15:38:29 -0400

Try (count(preceding-sibling::meep)+1), or (count(preceding-sibling::*[name()=name(current())])+1) for the general case.

<xsl:number/> is another way to go. It can get fairly complex (look it up in the spec), but in this case you shouldn't need any attributes to configure the counting: it'll default to what you want. Bind it to a variable if you need to retrieve it through XPath.


At 02:35 PM 7/19/01, you wrote:
What is the XPath to get the position of a given node relative to its
siblings of the same name?  Eg,


I'm looking for an XPath statement that would return 1 for the first meep,
2 for the second, and 3 for the 3rd meep, rather than 2, 3, and 5 (which
is what plain position() returns).

I'm trying to count the number of siblings and subtract the number of
siblings of the same type, then subtract that from position(), but I can't
get either of the first two parts there to work either.  Does anyone have
any suggestions?

--Larry Garfield

Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
Mulberry Technologies, Inc.      
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