Duhohh! I typed "identity template" first, then, in a fit of nincompoopery, I
changed it. Well it seems that another list member has spilled the beans on the
details of the identity template, but erred in suggesting xsl:copy, which does
not copy attributes as xsl:copy-of does.
--
Charles Knell
cknell(_at_)onebox(_dot_)com - email
-----Original Message-----
From: cknell(_at_)onebox(_dot_)com
Sent: Wed, 09 Jul 2003 16:05:12 -0400
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] How to get output XML same as input XML?
Why not just use the input XML?
If you have a good answer for that, then search for "XSLT default template" and
the xsl:copy-of element.
--
Charles Knell
cknell(_at_)onebox(_dot_)com - email
-----Original Message-----
From: "Josh Hone" <icsad(_at_)hotmail(_dot_)com>
Sent: Wed, 09 Jul 2003 15:49:17 -0400
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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