RE: [xsl] How to get output XML same as input XML?

["Tengshe, Ashish" <Tengshe.Ashish@xxxxxxxxxxxxx>]

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

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