RE: [xsl] count all the following nodes of a perticular pare

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Subject: RE: count all the following nodes of a perticular parent
From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>
Date: Thu, 28 Jun 2001 12:38:45 +0100

Or

count(following-sibling::node()/descendent-or-self::node() |
      following-sibling::node/descendent-or-self::node()/@*)

(depending on what you want).

Cheers,
Wendell

At 01:22 PM 6/28/01, Mike wrote:

> i want to count all the following nodes from the current node
> of the same
> parent
> can anybody give me a way to do this?

count(following-sibling::node())

Mike Kay
Software AG

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Current Thread

count all the following nodes of a perticular parent
- Jayaranga Subasinghe - Thu, 28 Jun 2001 06:00:29 -0400 (EDT)
  - Jarno Elovirta - Thu, 28 Jun 2001 06:14:48 -0400 (EDT)
  - Michael Kay - Thu, 28 Jun 2001 08:24:31 -0400 (EDT)
    - Wendell Piez - Thu, 28 Jun 2001 15:31:34 -0400 (EDT) <=
  - <Possible follow-ups>
  - Ben Robb - Thu, 28 Jun 2001 06:12:57 -0400 (EDT)
  - Jayaranga Subasinghe - Fri, 29 Jun 2001 00:14:54 -0400 (EDT)
    - Jarno Elovirta - Fri, 29 Jun 2001 00:27:58 -0400 (EDT)
    - Michael Kay - Fri, 29 Jun 2001 05:09:07 -0400 (EDT)

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