Subject: Re: [xsl] namespace generation in the output. From: Dave Pawson <davep@xxxxxxxxxxxxx> Date: Wed, 01 Jun 2005 16:40:29 +0100 |
On Tue, 2005-05-31 at 20:56 +0100, David Carlisle wrote: > Dave, in your case the namespace seems to be statically known in the > stylesheet so usually you don't need to do this copying trick, which is > only normally needed if you need to generate the namespace uri > dynamically. I may be wrong, but I think I do. I'm generating a stylesheet, whose default output elements need to be in the (as you say, fixed) namespace. > > I can't really tell as the snippet you posted isn't well formed, Sorry. http://cvs.sourceforge.net/viewcvs.py/docbook/xsl/xhtml/html2xhtml.xsl? view=markup look for the xsl:template match="xsl:stylesheet" > but for > example if you know what element xsl:copy is copying as you know which > element is being matched, then you can simply replace xsl:copy with a > literal result element that has the needed namespace decln. > > When you do need to dynamically generate a namespace declaration xslt2 > has an xsl:namespace instruction that works like xsl:attribute, but for > namespaces. Thanks David. (Please note folks, MK is being 'hard teacher' - go find out, DC is being 'nice teacher' this is it :-) I want: <xsl:stylesheet ..... (standard stuff) xmlns="http://example.com"> so I can write, in xslt 2.0, <xsl:template match="...."> <xsl:element name="xsl:stylesheet" xsl:namespace="http://example.com"> (No, I haven't tried it. docbook isn't xslt 2.0 friendly... yet) regards DaveP
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