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Re: xhtml xslt?

2004-08-30 11:51:03
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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