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Copying without namespace?

 
 
Andre-John Mas
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      03-10-2008
On Mar 5, 3:41 pm, Joseph Kesselman <(E-Mail Removed)>
wrote:
> Andre-John Mas wrote:
> > The issue is that I end up with an xhtml document that includes
> > namespaces that were only meant to be used in the xsl and not the
> > final document.

>
> Sounds like you're looking for xsl:stylesheet's exclude-result-prefixes
> attribute.


That what I tried doing, currently I have:

exclude-result-prefixes="i18n ota dataList iItemList math"

which excludes non-anonymous namespaces, such as:

xmlns:i18n=""
xmlns:math=""

but how do you exclude an anonymous namespace:

xmlns=""

Reading elsewhere I was given the suggestion of using:

<!-- copies without the template -->
<xsl:template match="*">
<xsl:element name="{name()}" namespace="">
<xsl:copy-of select="@*"/>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>

and changed the original block to:

<xsl:for-each select="$pageMetaData/*">
<xsl:choose>
<xsl:when test="local-name() = 'meta'">
<xsl:apply-templates select="."/>
</xsl:when>
<xsltherwise>
<meta content="{.}" name="{local-name()}"/>
</xsltherwise>
</xsl:choose>
</xsl:for-each>

but I am not sure this is the best approach.

Andre



 
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Joseph Kesselman
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      03-10-2008
Andre-John Mas wrote:
> but how do you exclude an anonymous namespace:
> xmlns=""


The default namespace declaration is generally not generated into the
output unless it is required, meaning that the parent element is in a
different default namespace. I'd have to take another look at your
specific example to see whether that's what's going on, but if it is
then the proper answer is "if you don't want to change default
namespaces, don't output an element that's in a different default
namespace than its parent."

--
Joe Kesselman / Beware the fury of a patient man. -- John Dryden
 
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