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(_at_)yahoo(_dot_)com>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Monday, August 30, 2004 11:55 AM
Subject: Re: [xsl] 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(_at_)lists(_dot_)mulberrytech(_dot_)com
From: David Carlisle <davidc(_at_)nag(_dot_)co(_dot_)uk>
CC: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] xhtml xslt?
Message-Id:
<200408260905(_dot_)KAA05731(_at_)penguin(_dot_)nag(_dot_)co(_dot_)uk>
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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