transform RDF to HTML via XSL and PHP

Hello :) I wish to add the FreeBSD news in my site but I can't transform the RDF document: http://www.freebsd.org/news/news.rdf I have a little knowledge in XML XSL but it seems that isn't enough :) here is my XSL script: ======================================================================= <?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"> <xsl:output method='html' version='1.0' encoding='UTF-8' indent='yes'/> <xsl:template match="/"> <html> <body> <table border="1"> <xsl:for-each select="//rdf:RDF/item"> <tr> <td><xsl:value-of select="title"/></td> <td><xsl:value-of select="link"/></td> </tr> </xsl:for-each> </table> </body> </html> </xsl:template> </xsl:stylesheet> /====================================================================== why doesn't this work ? it outputs this: ======================================================================= <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "http://www.w3.org/TR/REC-html40/loose.dtd"> <html><body><table border="1"></table></body></html> /====================================================================== I believe that my mistake is in the '<xsl:for-each select="//rdf:RDF/item">' tag I've tried also '//rdf:item', '/rdf:item', 'rdf:item', '/rdf:RDF/item', 'rdf:RDF/item' but with the same result how can I catch this ? can anyone help me please ? thank you :) ----------------------------------------------------------------- http://www.sportni.bg/worldcup/ - Германия 2006 - Световното първенство по футбол наближава!