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RE: [xsl] Generating xmlns attribute in root element

Subject: RE: [xsl] Generating xmlns attribute in root element
From: Graham Hannington <Ghannington@xxxxxxx>
Date: Wed, 12 Nov 2003 11:13:32 -0000
I've just figured this out... spurred on by the embarrassment of having to
ask it in public ;-): 

> Can someone please post a small XSLT stylesheet that outputs a result tree
containing an xmlns attribute on the root element, similar to this?

I was "trying too hard" on my previous attempts.  All it took was an
exclude-result-prefixes, and a simple xmlns attribute in the template. Doh!

<?xml version='1.0'?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:html="http://www.w3.org/1999/xhtml";
exclude-result-prefixes="html">
<xsl:output method="xml" version="1.0" indent="yes"
omit-xml-declaration="yes"/>

<xsl:template match="/">
<html xmlns="http://www.w3.org/1999/xhtml";>
<head/>
<body>
<ul>
<li>Any XHTML elements</li>
</ul>
</body>
</html>
</xsl:template>

</xsl:stylesheet>

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
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RE: [xsl] Generating xmlns attribut, Michael Kay Thread RE: [xsl] Generating xmlns attribut, Graham Hannington
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