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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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