Subject: RE: [xsl] How to get output XML same as input XML? From: "Tengshe, Ashish" <Tengshe.Ashish@xxxxxxxxxxxxx> Date: Wed, 9 Jul 2003 15:00:51 -0500 |
I think all you are looking for is an identity template: <?xml version="1.0" encoding="iso-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" indent="yes"/> <xsl:template match="node()|@*"> <xsl:copy> <xsl:apply-templates select="@*|node()" /> </xsl:copy> </xsl:template> </xsl:stylesheet> Thanks, Ashish -----Original Message----- From: Josh Hone [mailto:icsad@xxxxxxxxxxx] Sent: Wednesday, July 09, 2003 2:49 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: [xsl] How to get output XML same as input XML? Hi all - This is my first post on this list, but I have been using XSL for over a year. I searched the archive and FAQ but found no answer to my question. I am afraid there is not a favorable answer to my question, but here it is. I need to see if I can write a stylesheet which simply takes the input XML and turns exactly what was input into output XML. I know that you can easily do this with Xalan/DOM code, but my system requires that this situation be solved by stylesheets so that no extra code is written. Is this possible? Thank you very much. Josh Hone _________________________________________________________________ Tired of spam? Get advanced junk mail protection with MSN 8. http://join.msn.com/?page=features/junkmail 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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