Thanks everyone for the help.
<xsl:value-of select="tokenize(document-uri(/), '/')[last()]"/>
returns the full path.
As a fix, I added a required attribute to a top element for Filename.
I am not sure if this is the best way about it.
Regards,
Phil V
-----Original Message-----
From: Michael Kay [mailto:mike(_at_)saxonica(_dot_)com]
Sent: Monday, October 23, 2006 3:58 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] Get Document File name
i'm using
tokenize(document-uri(/), '/')[last()]
or '\' if working with backslashes.
The document-uri() function returns a URI, not a filename, and URIs always
use forwards slashes as separators.
Michael Kay
http://www.saxonica.com/
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