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

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Hi Josh,

Try this:

<xsl:stylesheet version="1.0"
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

<xsl:template match="/">
  <xsl:copy-of select="/">
</xsl:template>

</xsl:stylesheet>

Does it give you what you want?

If not, please explain how its result varies from your requirement to 
"[take] the input XML and [turn] exactly what was input into output XML". 
Depending on exactly what you mean by this, it may or may not be possible 
to do in XSLT. Examining the output of the above tranformation will tell.

There are other stylesheets that will simply copy input to output, but this 
is the simplest.

Cheers,
Wendell

At 03:49 PM 7/9/2003, you wrote:
> 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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