Hi,

> Is there an  xsl function for obtaining the path of an xsl 
> template and then
> parsing that path into its various directory names.

No.
 
> Does anyone please have an example of how to do this.

Pass the information using xsl:param or with an external source tree accessed using document().

Cheers,

Jarno - Wumpscut: Mother (Maternal Instinct)

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list