RSS Transformation-- Namespace Issue?

[Jacob Hookom]

Hello,   I’m trying to transform the Slashdot RSS feed via an XSL File posted at IBM’s DevZone:   http://slashdot.org/slashdot.rss   http://www-106.ibm.com/developerworks/xml/library/x-tiphdln.html?dwzone=xml   I’m able to pull in the feed from Slashdot easily and I’m able to grab the first node and output its namespace-uri as a test, but, trying to select any of the channels or items returns nothing:   <?xml version="1.0"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output method="html"/>   <xsl:template match="/">   <h1><xsl:value-of select="namespace-uri(.)"/></h1>   <xsl:apply-templates select="//channel"/>   <ul>     <xsl:apply-templates select="//item"/>   </ul> </xsl:template>   <xsl:template match="channel"> <xsl:apply-templates select="../image"/>  <h2><xsl:value-of select="title"/></h2>  <h3><xsl:value-of select="description"/></h3> </xsl:template>   <xsl:template match="item">   <li>     <xsl:element name="a">       <xsl:attribute name="href"><xsl:value-of select="link"/></xsl:attribute>       <xsl:value-of select="title" />     </xsl:element>   </li> </xsl:template>   <xsl:template match="image">   <xsl:element name="img">     <xsl:attribute name="src"><xsl:value-of select="url"/></xsl:attribute>     <xsl:attribute name="style">float:left; padding: 10px;</xsl:attribute>   </xsl:element> </xsl:template>   <xsl:template match="language"> </xsl:template>   </xsl:stylesheet>   Returns:   <h1>http://www.w3.org/1999/02/22-rdf-syntax-ns#</h1> <ul></ul>   I’m wondering if I need to put a namespace item in my XSL sheet or something.   Any help would be greatly appreciated! Jacob Hookom