Re: [xsl] Finding the position of node in foreign node list

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Subject: Re: Finding the position of node in foreign node list
From: "Joris Gillis" <roac@xxxxxxxxxx>
Date: Thu, 03 Mar 2005 21:53:09 +0100

Tempore 21:18:31, die 03/03/2005 AD, hinc in xsl-list@xxxxxxxxxxxxxxxxxxxxxx scripsit Jan Rademan <jrademan@xxxxxxxxxxx>:

Any ideas how I can find out what the position of the current node is in a foreign node set? (As opposed to the position in the current node set)

Hi,

I think you can use 'count()' and 'preceding::' (or 'preceding-sibling::') for that purpose. try: count($B/*[...]/preceding::*)

were '...' would be something like 'current() = .'

regards,
--
Joris Gillis (http://www.ticalc.org/cgi-bin/acct-view.cgi?userid=38041)
Veni, vidi, wiki (http://www.wikipedia.org)

Current Thread
Finding the position of node in foreign node list Jan Rademan - 3 Mar 2005 20:20:05 -0000 David Carlisle - 3 Mar 2005 20:55:34 -0000 Joris Gillis - 3 Mar 2005 20:57:14 -0000 <= Dimitre Novatchev - 4 Mar 2005 02:15:32 -0000

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