Re: [xsl] Get Document File name

["Jay Bryant" <jay@xxxxxxxxxxxx>]

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.vallone@xxxxxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
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