Subject: Re: [xsl] XPATH - Finding similar/conflicting nodes From: David Carlisle <davidc@xxxxxxxxx> Date: Tue, 13 Jul 2004 18:03:07 +0100 |
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output indent="yes"/> The equal case is fairly easy I think: <xsl:template match="/"> equal: <xsl:copy-of select="/SCHED/Schedule/Day/Squad/event[@id='2'] [@timebeg = ../preceding-sibling::Squad/event[@id='2']/@timebeg] "/> </xsl:template> </xsl:stylesheet> produces $ saxon squad.xml squad.xsl <?xml version="1.0" encoding="utf-8"?> equal: <event id="2" timebeg="960" timend="1020"/> > Now I have an XPATH question. I don't think that you can find the overlapping cases in a single Xpath 1 statement (although I could be proved wrong) You could do it in Xpath2 or using an XSLT loop construct. David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star Internet. 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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