Hi, Phil,
The usual answer to this is to pass the filename as a parameter to the
transform. Here's the FAQ entry:
http://www.dpawson.co.uk/xsl/sect2/N3663.html
However, you could use some variety of string manipulation to get just the
filename from the path.
The following transform does it:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:transform version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml" version="1.0"/>
<xsl:template match="/">
<filename><xsl:value-of select="tokenize(document-uri(.),
'/')[last()]"/></filename>
</xsl:template>
</xsl:transform>
Jay Bryant
Bryant Communication Services
----- Original Message -----
From: "Philip Vallone" <philip(_dot_)vallone(_at_)verizon(_dot_)net>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Sunday, October 22, 2006 3:08 PM
Subject: [xsl] Get Document File name
Hi Everyone,
Is there a way to get the file name of the document you are processing? If
I
use Document-uri() it returns the whole file path.
e.g.
'C:\temp\filename.xml'
I would like to get:
'filename.xml'
Is this possible?
Thanks,
Phil
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