Subject: Re: position() of the parent From: David Carlisle <davidc@xxxxxxxxx> Date: Thu, 20 Jan 2000 11:05:30 GMT |
> Worked great! But just for my curiosity, is there any way to find the > position() of the parent? the question is not well formed. Nodes do not, of themselves, have a position, they only have a position in a node list. So for example if you select the parent with .. or parent:: the position() will be 1, as it will be the first node in a list of one, but I assume this isn't what you meant. If you mean what is 1 more than the number of preceding siblings of the parent, you can do select="1+count(../preceding-sibling::*)" David XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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