This is the mail archive of the
xsl-list@mulberrytech.com
mailing list .
Re: Copy XHTML element contents without base tag
- To: xsl-list at lists dot mulberrytech dot com
- Subject: Re: [xsl] Copy XHTML element contents without base tag
- From: cutlass <cutlass at secure0 dot com>
- Date: Fri, 09 Mar 2001 13:09:38 +0000
- References: <487B2B5FD092D411977400D0B73EB0A2B361@titan.neoworks.co.uk>
- Reply-To: xsl-list at lists dot mulberrytech dot com
look at faq on copying
http://www.dpawson.co.uk/xsl/N1930.html
Nick Vincent wrote:
> Hi all,
>
> This is probably painfully easy, but I can't think of a solution:
>
> I need to include a raw XHTML fragment in my XSL:
>
> <document>
> <xfragment>
> This is a fragment of XHTML
> <table><tr><td>
> [snip dull HTML]
> </td></tr></table>
> </xfragment>
> </document>
>
> Now, as far as I can see there are two ways of processing this in XSLT:
>
> <xsl:template match="xfragment">
> <xsl:copy-of select="." />
> </xsl:template>
>
> Unfortunately this ends up including the <xfragment> tags in the output,
> which I can't really have.
>
> The other way I can see is
>
> <xsl:template match="xfragment">
> <xsl:copy-of select="./*" />
> </xsl:template>
>
> Unfortunately this removes any text elements which are direct children of
> the xfragment tags (e.g. the phrase 'This is a fragment of XHTML' in the
> above example).
>
> Does anyone know any better ways of making this copy?
>
> Thanks,
>
>
> Nick Vincent
> NeoWorks
>
>
> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
>
>
>
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list