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

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Rod,

If all you need to do is output an element's name in "tags", that's as easy as:

<xsl:template match="*">
  <span class="tag">
    <xsl:text>&lt;</xsl:text>
    <xsl:value-of select="name()"/>
    <xsl:text>&gt;</xsl:text>
  </span>
  <xsl:apply-templates/>
  <span class="tag">
    <xsl:text>&lt;/</xsl:text>
    <xsl:value-of select="name()"/>
    <xsl:text>&gt;</xsl:text>
  </span>
</xsl:template>

On the other hand, your requirements are probably more complex. Maybe 
you want line feeds and indenting, maybe you need to show attributes 
or other node types, etc. etc.

Eventually you may want an entire XML serializer implemented in XSLT. 
This has been done too, but some problems on the edge such as 
escaping < and & characters, or serializing namespace desclarations, 
can make it fairly tricky. At that point you may want to check out 
someone else's library, like Evan Lenz's at 
http://lenzconsulting.com/xml-to-string/.

Cheers,
Wendell

At 01:54 PM 4/27/2010, you wrote:
> 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





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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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