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

[Robert Koberg <rob@xxxxxxxxxx>]

Hi,

I want to get the path to the parent directory of the XML used as the  
source in a transformation.

Is this the best way?

<xsl:variable
  name="path-tokens"
  select="tokenize(document-uri(/), '/')" as="xs:string*"/>
<xsl:variable
  name="source-dir-path"
  select="string-join(remove($path-tokens, count($path-tokens)),  
'/')" as="xs:string"/>

I also unsuccessfully tried to use only one variable:

<xsl:variable
  name="source-dir-path-test"
  select="string-join(tokenize(document-uri(/), '/')[not(last())],  
'/')" as="xs:string"/>

which results in an empty string.

Is there a better way?

thanks,
-Rob