Subject: [xsl] Reverse order From: Janning Vygen <vygen@xxxxxxxxxxxx> Date: Tue, 1 May 2001 15:11:33 +0200 |
i am trying to list the parts of a book in reverse order. <book> <part><title>foo</title></part> <part><title>bar</title></part> <part><title>huhu</title></part> </book> <xsl:for-each select="/book/part[position()=last()]/preceding-sibling::part"> <xsl:value-of select="title"/><br/> <xsl:for-each> huhu<br> bar<br> foo<br> I thought the XPath expression in the for-each statement should return a reverse oder of all parts because it goes to the last part and looks back to all preceding-siblings which are parts. But it doesnt work, so i actually know that XPath doesnt 'look' the way i do :-) Is there a way to get a node set in a reverse order? thanks in advance janning -- Planwerk 6 /websolutions Herzogstraße 86 40215 Düsseldorf fon 0211-6015919 fax 0211-6015917 http://www.planwerk6.de XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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Re: [xsl] Accessing nth element, bharat chintapally | Thread | Re: [xsl] Reverse order, David Carlisle |
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