Return-Path: Mailing-List: contact cocoon-users-help@xml.apache.org; run by ezmlm Delivered-To: mailing list cocoon-users@xml.apache.org Received: (qmail 29118 invoked from network); 9 Jan 2001 15:07:32 -0000 Received: from sylt.pixelpark.com (HELO mail.pixelpark.com) (62.52.66.77) by h31.sny.collab.net with SMTP; 9 Jan 2001 15:07:32 -0000 Received: from pixelpark.com (dhcp-62-52-63-152.pixelpark.com [62.52.63.152]) by mail.pixelpark.com (8.11.1/8.11.1) with ESMTP id f09F7Tr13434 for ; Tue, 9 Jan 2001 16:07:29 +0100 (MET) Message-ID: <3A5B2930.614BEA0E@pixelpark.com> Date: Tue, 09 Jan 2001 16:07:28 +0100 From: Johannes Koch Organization: Pixelpark AG X-Mailer: Mozilla 4.7 [de] (WinNT; I) X-Accept-Language: de MIME-Version: 1.0 To: cocoon-users@xml.apache.org Subject: Re: How to get the URI of an XML document in an XSL stylesheet? References: <3A5B25CF.D7189F01@congenio.de> Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: 8bit X-Spam-Rating: h31.sny.collab.net 1.6.2 0/1000/N "Dr. Uwe Meyer-Gruhl" schrieb: > > Hi, > [...] > So I guess I have to find a way of generating the source document file > name from the stylesheet prepended to the "?aspekt=" part but have been > unlucky at finding out how. Is there a function or pre-defined variable > in Cocoon or Xalan or a method for that in standard XSLT? You can get the XML file URI using XSP (request.getRequestURI()) -- Johannes Koch . IT Developer Pixelpark AG . http://www.pixelpark.com Rotherstra�e 8 . 10553 Berlin . Germany phone: +49 30 5058 - 1288 . fax: - 1355