Re: [xsl] most efficient way to get XML source's parent dir path

[Robert Koberg <rob@xxxxxxxxxx>]

On Feb 11, 2009, at 4:51 AM, Owen Rees wrote:

> --On Tuesday, February 10, 2009 11:53:24 -0500 Robert Koberg wrote:
>
> > I want to get the path to the parent directory of the XML used as the
> > source in a transformation.
>
> Even if an absolute URI is known for the document supplied as the  
> source, the concept of 'parent directory' is not necessarily well  
> defined with respect to that URI.
>
> What are you expecting to be able to do with the string you are  
> generating?

I want to resolve references to other documents the are relative to  
the source XML. E.g.

<div>
  <p>blah</p>
  <include ref="blah.xml"/>
</div>
...
match="include"
<xsl:apply-templates select="doc(concat($parent-dir, '/', @ref))/*"/>

What do you see as the problem?

-Rob