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RE: [xsl] xsl:copy


Subject: RE: [xsl] xsl:copy
From: "Michael Kay" <michael.h.kay@xxxxxxxxxxxx>
Date: Tue, 19 Feb 2002 09:30:28 -0000

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@xxxxxxxxxxxx
work: Michael.Kay@xxxxxxxxxxxxxx

> -----Original Message-----
> From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Garrick
> Besterwitch
> Sent: 19 February 2002 06:09
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> 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
>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list



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