Subject: RE: [xsl] transforming xml to xhtml From: "Michael Kay" <michael.h.kay@xxxxxxxxxxxx> Date: Tue, 10 Dec 2002 14:38:01 -0000 |
Specifying the DTD in <xsl:output> merely declares your intent to produce valid XHTML. Whether or not the XHTML is actually valid depends entirely on your stylesheet. If your code is producing invalid XHTML, then you need to work out what's wrong with the XHTML (from the error messages produced when it's validated) and correct the relevant parts of your stylesheet. Michael Kay Software AG home: Michael.H.Kay@xxxxxxxxxxxx work: Michael.Kay@xxxxxxxxxxxxxx > -----Original Message----- > From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of True Name > Sent: 10 December 2002 13:02 > To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx > Subject: [xsl] transforming xml to xhtml > > > hello, i would like to use xsl transforming xml file to xhtml file > > xsl like: > > <xsl:stylesheet version="1.0" > xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> > > > <xsl:output method="xml" doctype-public="-//W3C//DTD XHTML > 1.0 Strict//EN" > > doctype-system="http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd" > indent="yes" > omit-xml-declaration="yes" /> > > .... > > but every time i cannot get a valid xhtml file, would you > please tell me how > to transform xml to xhtml?? > > thanks in advance > > > _________________________________________________________________ > MSN 8 with e-mail virus protection service: 2 months FREE* > http://join.msn.com/?page=features/virus > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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