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RE: RE: How to get output XML same as input XML?

2003-07-09 13:24:42
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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