Subject: Re: [xsl] xslt2 - namespace attributes for output from called templates From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Fri, 22 Aug 2003 14:30:14 +0100 |
Hi Tom, > But if I wasn't making html, then I could have this problem. Would > every function, called template etc. have to re-specify the > namespace attribute? If you want a literal result element to be in the namespace 'http://www.w3.org/1999/xhtml' then it has to be associated with that namespace. If you don't want the element to have a prefix, then the 'http://www.w3.org/1999/xhtml' namespace has to be in-scope as the default namespace. Namespaces are in-scope on the element that they're declared, and on all the descendants of that element. So the usual way in which you guarantee that all the elements you generate in the stylesheet are in a particular default namespace is to declare that namespace as the default on the <xsl:stylesheet> element. So rather than: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/"> <html xmlns="http://www.w3.org/1999/xhtml"> ... </html> </xsl:template> ...lots of other templates in which the default namespace is no namespace... </xsl:stylesheet> you should use: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns="http://www.w3.org/1999/xhtml"> <xsl:template match="/"> <html> ... </html> </xsl:template> ...lots of other templates in which the default namespace is the XHTML namespace... </xsl:stylesheet> Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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