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Re: [xsl] xsl:element will not create an output element, in any context

2007-05-31 06:39:16
Ok, thanks David and Abel for your latest emails; I'll have to wait
a
few hours until I'm back home from work to test things in the
browser
now.

You can just put the example in the filesystem and open it in Firefox.



One last question for David, before I test things when back at home.
What you've written below produces 'ul' and 'li' elements in no
namespace, according to what you said earlier. I'm assuming this
means
they will NOT be processed as XHTML elements, and hence will not
produce the result I want (which is XHTML ul and li elements)?

For a browser it is more important that the DOCTYPE is correct. You
can set it in the <xsl:output /> instruction. However, it will not
effect the result when you use a doctype in an inlined stylesheet
(another difference from server side and client side rendering using
<?xml-stylesheet ?>)

In addition of being conformant XHTML, the elements may not have a
prefix and must be in the XHTML namespace. The previous example
changed such a way that it produces conformant XHTML when run from a
commandline processor, may look like this (place it in your
filesystem and point your firefox browser to it):

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="embedded.xslt"?>
<xsl:stylesheet
   xmlns="http://www.w3.org/1999/xhtml";
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
   version="1.0" >

    <xsl:output method="html"
      doctype-public="-//W3C//DTD XHTML 1.0 Strict//EN"
      doctype-system="http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"/>

    <xsl:template match="/">
        <html xml:lang="en" lang="en">
            <head><title>hello world</title></head>
            <body>
                <h1>Hello world!</h1>
            </body>
        </html>
    </xsl:template>
</xsl:stylesheet>


Cheers,
-- Abel Braaksma


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