Subject: Re: [xsl] cannot transform to <html xmlns="http://www.w3.org/1999/xhtml"> From: Ryan Neil Gillespie <scoobie@xxxxxxxx> Date: Sun, 21 Jul 2002 18:17:46 -0400 (EDT) |
I think you want this instead for your stylesheet. Try it, but I'm new to XSL. ------------------------- <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:template match="/"> <html> <xsl:attribute name="xmlns">http://www.w3.org/1999/xhtml</xsl:attribute> </html> </xsl:template> </xsl:stylesheet> ---------------------- _ryan On Sun, 21 Jul 2002, Phillip Rhodes wrote: > Hi everyone, > I created an xsl file that will generate an xhtml file. With the xhtml > file, I run it through an xhtml to fo converter (another xsl > file http://www.antenna.co.jp/XML/downfree/Xhtml2fo.xsl). > My problem is that the xhtml to fo conversion does not work unless the html > element appears as: > <html xmlns="http://www.w3.org/1999/xhtml"> > </html> > > Unfortunately, my xhtml file's html element(after transformation) is: > <html xmlns:html="http://www.w3.org/1999/xhtml"> > </html> > > > I have the following stylesheet that does my transformation to xhtml: > > <?xml version="1.0" encoding="UTF-8"?> > <xsl:stylesheet > xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > xmlns:html="http://www.w3.org/1999/xhtml" > version="1.0"> > <xsl:template match="/"> > <html> > </html> > </xsl:template> > </xsl:stylesheet> > > It generates the following: > <?xml version="1.0" encoding="UTF-8"?> > <html xmlns:html="http://www.w3.org/1999/xhtml"> > </html> > > I want it to generate: > <?xml version="1.0" encoding="UTF-8"?> > <html xmlns="http://www.w3.org/1999/xhtml"> > </html> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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