def resource_name(res_pkg_name, name, my_pkg=None):
 if (res_pkg_name != my_pkg):
  return ((res_pkg_name + PRN_SEPARATOR) + name)
 return name

def resource_name ( res_pkg_name , name , my_pkg = None ) : if ( res_pkg_name != my_pkg ) : return ( ( res_pkg_name + PRN_SEPARATOR ) + name ) return name
convert package name + resource into a fully qualified resource name .