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

["Dimitre Novatchev" <dnovatchev@xxxxxxxxx>]

This is an instance of another kind of identity template -- "one node at a
time". Sometimes (e.g. in positional grouping) it may be useful:



  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@*"/>
      <xsl:apply-templates select="node()[1]"/>
    </xsl:copy>
    <xsl:apply-templates select="following-sibling::node()[1]"/>
  </xsl:template>




=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL




"Josh Hone" <icsad@xxxxxxxxxxx> wrote in message
news:Law8-F93o5xvZDRb78O0004594e@xxxxxxxxxxxxxx
> 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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