xsl-list
[Top] [All Lists]

Re: Finding the position of node in foreign node list

2005-03-03 13:53:09
Tempore 21:18:31, die 03/03/2005 AD, hinc in xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com scripsit Jan Rademan <jrademan(_at_)iafrica(_dot_)com>:

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)

--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--



<Prev in Thread] Current Thread [Next in Thread>