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Re: [xsl] how to exclude namespaces from copy-of


Subject: Re: [xsl] how to exclude namespaces from copy-of
From: cert21 <cert21@xxxxxxx>
Date: Thu, 05 Nov 2009 10:52:39 -0500

Thank you. That makes sense. I took your advice and created this template
   <xsl:template match="*" mode="fragment">
       <xsl:element name="{local-name()}">
           <xsl:value-of select="." />
           <xsl:apply-templates select="*" mode="fragment" />
       </xsl:element>
   </xsl:template>

Now from another template that processes the atom feed item I call it like this:
<xsl:when test="@type='xhtml' or @type = 'application/xhtml+xml'">
<fragment>
<xsl:apply-templates mode="fragment" select="xhtml:div" /> </fragment>
</xsl:when>
<xsl:when test="@type='html' or @type = 'text/html'">
<xsl:value-of select="." />
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="normalize-space()" />
</xsl:otherwise>


IT works well now!


Michael Kay wrote:
Although xsl:copy-of in XSLT 2.0 can exclude unused namespaces
(copy-namespaces="no"), it can't remove declarations of namespaces that are
actually in use. You don't actually want to make a copy of the data, you
want to transform it by changing the names of the elements. Remember that
the difference between <rss/> and <rss xmlns="xyz"/> is not just a namespace
declaration - the two elements have different names. So you need to do a
"renaming identity transform" along the lines

<xsl:template match="*">
  <xsl:element name="{local-name()}">
    <xsl:apply-templates select="*"/>
  </xsl:element>
</xsl:template>

Regards,

Michael Kay
http://www.saxonica.com/
http://twitter.com/michaelhkay


-----Original Message-----
From: cert21 [mailto:cert21@xxxxxxx] Sent: 05 November 2009 15:13
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] how to exclude namespaces from copy-of


Hello!

I am trying to parse the atom feed using the xsl template.

The item in the feed contains this element: (example) <div xmlns="http://www.w3.org/1999/xhtml" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:thr="http://purl.org/syndication/thread/1.0">
<p>
October and November are busy months for Amazon Web Services!</p> </div>


I want to copy the div with all its child elements, so I use copy-of The problem is that I don't want the xmlns="http://www.w3.org/1999/xhtml"
or xmlns:dc or any other extra data in the div attribute


I just want the result to be
<div>
<p>
October and November are busy months for Amazon Web Services!</p> </div>


How can I do this?

Thank you.


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