RE: [xsl] presenting xml elements as they appear in the source file

[Rod Kane <rkane@xxxxxxxxxxxxxxxxxx>]

Many thanks for everyone for the fast responses.

Rod

-----Original Message-----
From: Michael Kay [mailto:mike@xxxxxxxxxxxx]
Sent: Tuesday, April 27, 2010 2:37 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE: [xsl] presenting xml elements as they appear in the source file


If you're using Saxon there's saxon:serialize() which returns a string
representation of the lexical XML, which you can insert into your HTML
output.

Another trick that works in most environments, despite being deprecated, is
to use the old <xmp> tag in HTML:

<xmp>
 <xsl:copy-of select="$element"/>
</xmp>

Regards,

Michael Kay
http://www.saxonica.com/
http://twitter.com/michaelhkay

> -----Original Message-----
> From: Rod Kane [mailto:rkane@xxxxxxxxxxxxxxxxxx]
> Sent: 27 April 2010 18:55
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] presenting xml elements as they appear in the
> source file
>
> Hello,
>
> In general can anyone suggest how to display xml nodes as
> they appear in the source file? If there is an easy way of
> doing this it will assist with this problem:
>  
> I'd like to use xsl to prepare an html document but at
> certain points the xml becomes an arbitrary mix of nodes,
> think of a number of a number of legal/valid elements that
> can be arranged in arbitrary ways, such as a nested query.
>  
> Since the composition of the query is unpredictable it is
> necessary to just present the nodes as they are in the source
> xml, complete with the < > brackets, to make things easy.
>  
> I see that there are examples of presenting xml in trees and
> that may be a bit time consuming for a beginner to implement
> so simply listing the elements verbatim is fine for now.
>  
> Thank you
>  
> Rod