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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: producing xhtml using xsl
 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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list 
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