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RE: xsl:copy
- From: "Michael Kay" <michael dot h dot kay at ntlworld dot com>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Tue, 19 Feb 2002 09:30:28 -0000
- Subject: RE: [xsl] xsl:copy
- Reply-to: xsl-list at lists dot mulberrytech dot com
If you need an exact copy (of an element and all its descendants), then use
xsl:copy-of. However, if you need to make any changes anywhere in the
structure, you'll need to do a recursive descent: it's useful here to use
the "identity template rule" which you'll find the W3C XSLT spec.
Michael Kay
Software AG
home: Michael.H.Kay@ntlworld.com
work: Michael.Kay@softwareag.com
> -----Original Message-----
> From: owner-xsl-list@lists.mulberrytech.com
> [mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of Garrick
> Besterwitch
> Sent: 19 February 2002 06:09
> To: xsl-list@lists.mulberrytech.com
> Subject: [xsl] xsl:copy
>
>
> Hi,
> I need to generate a copy of an xml file using xsl ...I tried using
> xsl:copy..but this gives me only the value and not the node
> names ....Is
> there a workaround to accomplish this.
>
> Thanks and cheers
> Garry
>
>
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>
>
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