Re: [xsl] Get Document File name

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Subject: Re: Get Document File name
From: "Jay Bryant" <jay@xxxxxxxxxxxx>
Date: Sun, 22 Oct 2006 16:43:33 -0500

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: 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

Current Thread
Get Document File name Philip Vallone - 22 Oct 2006 20:09:24 -0000 Jay Bryant - 22 Oct 2006 21:44:04 -0000 <= Frank Marent - 23 Oct 2006 06:12:08 -0000 Abel Braaksma - 23 Oct 2006 06:36:09 -0000 Michael Kay - 23 Oct 2006 07:58:46 -0000 Philip Vallone - 23 Oct 2006 22:58:03 -0000

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