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(_dot_)H(_dot_)Kay(_at_)ntlworld(_dot_)com
work: Michael(_dot_)Kay(_at_)softwareag(_dot_)com
-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
True Name
Sent: 10 December 2002 13:02
To: XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com
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
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