Subject: Re: [xsl] Select nodes with equal position From: David Carlisle <davidc@xxxxxxxxx> Date: Sun, 3 Apr 2005 13:29:47 +0100 |
Thank you David. This also works - <xsl:value-of select="../../colspec[count(current()/preceding-sibling::entry)+1]/@colname"/> (got idea from your answer) Seems more programmer friendly to me(not a XPath expert like you) The system most likely already knows position() but counting preceding siblings can be an expensive operation, although actually here, for table columns, we are presumably talking about 4 or 5 siblings not 4 or 5 thousand, so efficiency may not be a concern. There is an alternative of course which doesn't require counting at all, instead of iterating through one set of nodes and then indexing into the other set using this number, you can iterate over both sets in parallel passing each node into a template as a parameter and iterating by selecting the next following sibling of _each_ node. 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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