Subject: Re: [xsl] Sorting problem From: David Carlisle <davidc@xxxxxxxxx> Date: Wed, 31 Aug 2005 12:49:18 +0100 |
me> No. keys are what you _do_ use in xslt 1 in this case. (in xslt2 you me> could use xsl:for-each-group) Or rather you could use keys here, but in this case, i think you just want to sort don't you? <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" > <xsl:output method="xml" indent="yes"/> <xsl:template match="root"> <root> <xsl:for-each select="item"> <xsl:sort data-type="number" select="count(preceding-sibling::item[@type=current()/@type])"/> <xsl:copy-of select="."/> </xsl:for-each> </root> </xsl:template> </xsl:stylesheet> produces $ saxon item.xml item.xsl <?xml version="1.0" encoding="utf-8"?> <root> <item type="A" subType="1"/> <item type="B" subType="2"/> <item type="D" subType="1"/> <item type="E" subType="2"/> <item type="A" subType="2"/> <item type="B" subType="3"/> <item type="D" subType="2"/> <item type="E" subType="4"/> <item type="A" subType="4"/> <item type="D" subType="3"/> <item type="D" subType="4"/> </root> which appears to be what you wanted. 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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