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(_at_)hotmail(_dot_)com]
Sent: Wednesday, July 09, 2003 2:49 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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
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