Subject: Re: [xsl] Problem for : increment a variable in a for-each? From: David Carlisle <davidc@xxxxxxxxx> Date: Mon, 29 Nov 2004 15:57:38 GMT |
It is my problem. no matter what I use, I can't find the position inside node which I filter from original source since position() will tell you the position on the source. position() never refers to the position of a node in the original source tree, it always refers to a position in the current node list. So for example if you select a node with select="." it alsways has position()=1 no matter where it is in teh tree as select="." selects one node so that node is always the first (and only) node in the current node list. David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________
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