Subject: [xsl] namespace-alias question From: omprakash.v@xxxxxxxxxxxxx Date: Mon, 17 Jan 2005 19:57:05 +0530 |
Hi, Iam using XSL to transform a schema into xsl and it is causing many problems. My stylesheet looks like this: Here Iam using x to prefix my stylesheet while using xsl to prefix the stylesheet being outputted as suggested by Steve Muench. <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:x ="http://tempuri" xmlns:xsd="http://www.omprakashv.org" version="1.0"> <x:namespace-alias stylesheet-prefix="x" result-prefix="xsl"/> In order to output the <xsl:stylesheet> element in to the output I have tried many ways like below: <x:text disable-output-escaping="yes"> <xsl:stylesheet version="1.0" xmlns:xsl ="http://www.w3.org/1999/XSL/Transform"? </x:text> <x:text disable-output-escaping="yes"> <xsl:stylesheet version="1.0" xmlns:xsl ="http://www.w3.org/1999/XSL/Transform"> </x:text> While the above compiles, the result of the transformation is empty document using both oracle xsl parser and xalan. The other way suggested by many doesnt work with the xslt processors. <x:template match="xsd:schema"> <xsl:stylesheet version="1.0" xmlns:xsl ="http://www.w3.org/1999/XSL/Transform"? </x:template> Any suggestions would be appreciated. Regards, Omprakash.V This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e-mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit Us at http://www.polaris.co.in
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