Subject: Re: arbitrary sorting (Part IV) From: David Carlisle <davidc@xxxxxxxxx> Date: Thu, 20 Jul 2000 15:36:04 GMT |
> Tricky, because there's no XPath facility to determine the position of a > node relative to its siblings: you can only do this using <xsl:number/>, and > you can't invoke XSLT-level operations while computing a sort key. but you could presumably use count(preceding-sibling::*)+1 David XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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