RE: Get path of xsl template

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)

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