Subject: RE: producing xhtml using xsl
From: "Clapham, Paul" <pclapham@xxxxxxxxxxxxx>
Date: Fri, 20 Apr 2001 09:16:36 -0700
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You use the doctype-public and doctype-system attributes in your
<xsl:output> element instead of trying to force the DOCTYPE. This also
means you don't have to mess about by saying omit-xml-declaration="yes" and
then trying to sneak in an XML declaration.
PC2
-----Original Message-----
From: Andrew Welch [mailto:andrew@xxxxxxxxxxxxxxxxxxxxxxx]
Sent: April 20, 2001 08:55
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: producing xhtml using xsl
Hi,
This is probably a stock question, please bear with me...
Im trying to produce XHTML output using the following stylesheet:
=========
<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes"/>
<xsl:template match="/">
<![CDATA[
<?xml version="1.0" encoding="UTF-8" ?>
<! DOCTYPE html
PUBLIC "-//W3C//DTD XHTML 1.0 Transitional //EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
]]>
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
....
==========
This fails, as the parser converts < in the cdata section to <. How do I
go about outputting the correct xhtml header from a stylesheet??
thanks
andrew
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