Re: [xsl] position within same nodes

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

Larry,

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.

Cheers,
Wendell

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,

<blah>
 <argh/>
 <meep/>
 <meep/>
 <stuff/>
 <meep/>
</blah>

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
lgarfiel@xxxxxxxxxxxxxxxxxxx

======================================================================
Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
Mulberry Technologies, Inc.                http://www.mulberrytech.com
17 West Jefferson Street                    Direct Phone: 301/315-9635
Suite 207                                          Phone: 301/315-9631
Rockville, MD  20850                                 Fax: 301/315-8285
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Current Thread
position within same nodes Laurence O Garfield - Thu, 19 Jul 2001 14:39:41 -0400 (EDT) Wendell Piez - Thu, 19 Jul 2001 15:39:10 -0400 (EDT) <= Laurence O Garfield - Thu, 19 Jul 2001 16:04:28 -0400 (EDT) Thomas B. Passin - Thu, 19 Jul 2001 15:57:43 -0400 (EDT) David Carlisle - Thu, 19 Jul 2001 16:47:14 -0400 (EDT) Michael Kay - Fri, 20 Jul 2001 02:33:43 -0400 (EDT)

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