Subject: [xsl] Re: Re: Can one _generate_ namespace nodes? From: "Dimitre Novatchev" <dnovatchev@xxxxxxxxx> Date: Mon, 24 Feb 2003 21:18:32 +0100 |
"Joseph Kesselman" <keshlam@xxxxxxxxxx> wrote in message news:OF61D39BC9.77924166-ON85256CD7.00611483-85256CD7.0062E761@xxxxxxxxxxxxx > Yes, generating the <xsl:stylesheet> explicitly as <newxsl:stylesheet> > rather than using <xsl:copy> would result in newxsl: being declared at the > root. > > But then any elements -- or XPaths -- copied from the source stylesheet > would have the reverse problem; they'd still be using the xsl: prefix and > would need to declare *that* prefix. What about the following: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:newxsl="http://dymmy/Transform" > <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:namespace-alias stylesheet-prefix="newxsl" result-prefix="xsl"/> <xsl:template match="/"> <newxsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <newxsl:text>foo</newxsl:text> </newxsl:stylesheet> </xsl:template> </xsl:stylesheet> The "newxsl" prefix will be bound to the XSLT namespace and the serializer will generate the necessary namespace definition. In addition to this, a definition of the XSLT namespace bound to the prefix "xsl" is also present in the newxsl:stylesheet. The result will be: <newxsl:stylesheet xmlns:newxsl="http://www.w3.org/1999/XSL/Transform" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <newxsl:text>foo</newxsl:text> </newxsl:stylesheet> Therefore, any XPath expressions containing the "xsl" prefix will be evaluated correctly. In the same way, any other necessary prefixes can be specified on the newxsl:stylesheet element. Is there an example, which is still problematic? ===== Cheers, Dimitre Novatchev. http://fxsl.sourceforge.net/ -- the home of FXSL XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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