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: "Jarno Elovirta" <jarno@xxxxxxxxxxxxxx>
Date: Thu, 28 Jun 2001 13:12:59 +0300

> i want to count all the following nodes from the current node of the same
> parent

count(following-sibling::node())

Sini

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list

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)

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count all the following nodes, Jayaranga Subasinghe	Thread	RE: count all the following n, Michael Kay
RE: count all the following n, Ben Robb	Date	Re: Default attribute value t, Trevor Nash
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