rss_converter_instagram.com.xsl: fix the channel link
authorAntonio Ospite <ao2@ao2.it>
Fri, 20 May 2016 16:00:50 +0000 (18:00 +0200)
committerAntonio Ospite <ao2@ao2.it>
Fri, 20 May 2016 16:00:50 +0000 (18:00 +0200)
rss_converter_instagram.com.xsl

index f12ecbb..97d39d2 100644 (file)
 
     <xsl:template match="/">
         <xsl:variable name="channel-title" select="concat('Instagram / ', $screen-name)"/>
 
     <xsl:template match="/">
         <xsl:variable name="channel-title" select="concat('Instagram / ', $screen-name)"/>
-        <xsl:variable name="channel-link" select="concat($BaseURL, //__path)"/>
+        <xsl:variable name="channel-link" select="concat($BaseURL, '/', $user-name)"/>
 
         <rss version="2.0">
             <xsl:attribute name="xml:base"><xsl:value-of select="$BaseURL" /></xsl:attribute>
 
         <rss version="2.0">
             <xsl:attribute name="xml:base"><xsl:value-of select="$BaseURL" /></xsl:attribute>