Re: most efficient way to get XML source's parent dir

--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?

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