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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: xhtml xslt?
I had the same problem, I use this:
<xsl:template match="*">
<xsl:element name="{name(.)}">
<xsl:for-each select="@*">
<xsl:attribute name="{name(.)}">
<xsl:value-of select="."/>
</xsl:attribute>
</xsl:for-each>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
-Jeff Moss
----- Original Message -----
From: "Nathan Shaw" <n8_shaw@xxxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Monday, August 30, 2004 11:55 AM
Subject: Re: xhtml xslt?
> OK, this works, but it is copying the namespace along
> with the elements. My output from this transformation
> should be XHTML, so I need to drop the namespace so
> that I do not end up with xhtml:strong, etc...
>
> How can I do a copy but drop the namespace?
>
> Thanks,
>
> --Nathan
>
>
>
> Date: Thu, 26 Aug 2004 10:05:34 +0100
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> From: David Carlisle <davidc@xxxxxxxxx>
> CC: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: xhtml xslt?
> Message-Id: <200408260905.KAA05731@xxxxxxxxxxxxxxxxx>
>
> >I started to create an XSLT to accomplish this:
>
> you just want to copy the html you don't need a
> template for each
> element, the whole point of namespaces id to make this
> easy, you can
> grab them all at once.
>
> <xsl:template match="h:*">
> <xsl:copy>
> <xsl:copy-of select="@*"/>
> <xsl;apply-templates/>
> </xsl:copy>
> </xsl:template>
>
> David
>
>
>
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