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Subject: Re: [xsl] Specifying the XHTML XMLNS From: David Carlisle <davidc@xxxxxxxxx> Date: Wed, 28 Sep 2005 16:03:27 +0100 |
> Of course not. I have posted the "original" stylesheet that didn't
> include the xmlns declaration. If I make the change you suggest, the
> XSLT becomes (the same but with an extra xmlns declaration in
> xsl:stylesheet tag):
well not really, I meant that the result you posted was not generated by
your stylesheet _after_ you had added the xmlns declaration.
You said that it had xmlns on this link element:
<link rel="stylesheet" href="./css/login.css" type="text/css" xmlns=""></link>
But in this message when you post the entire result, note that link
does not have xmlns.
Note that the script element generated in the stylesheet also doesn't
have xmlns="" either.
Presumably the script that does have this has been copied from the
source. If you have a source element script-in-no-namespace and want to
generate script-in-html this (to xslt) is just like saying you have an
element <a> and want to generate <b>. You can do it but it is _not_ a
copy so you don't want
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
You probably want a default template that renames elements, giving thenm
the same name but the xhtml namespace, which would be
<xsl:template match="node()">
<xsl:element name="{local-name()}">
<xsl:copy-of select="@*"/>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
which will make elements in the xhtml namespace if that is the current
default.
David
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