[Jython-dev] my posts to jython-users@lists.sourceforge.net are showing up

[Updike, C. <Cla...@jh...>]

I've submitted the post below twice to jyt...@li...
(last night and again just now) and it doesn't show up.  Also, I'm not
getting a bounced email either.  What up?

TIA,
Clark (please reply to mailto:cla...@jh...)

p.s.  Barry, I worked with your bro Craig at C1 until I left in September.

Subject:  2.1b2 PyServlet.loadServlet 'invalid syntax'   
Sent:  12/22/2001 10:28 AM 
 Importance:  Normal   
I'm trying to run the Hello.py servlet example but I get the exception from
PyServlet.loadServlet() shown below.  The strange thing is that I can
execute the statement execfile('C:\\Program Files\\Apache Tomcat
4.0\\webapps\\JythonServlet\\Hello.py') from the interactive interpreter
without getting the 'invalid syntax' problem (which I think shows that the
path is valid).  Anyone have any ideas?

TIA, 
Clark 

javax.servlet.ServletException: Could not create Jython servletTraceback
(innermost last): 
  (no code object) at line 0 
SyntaxError: ('invalid syntax', ('C:\\Program Files\\Apache Tomcat
4.0\\webapps\\JythonServlet\\Hello.py', 3, 9, '        class
Hello(javax.servlet.http.HttpServlet):'))

        at org.python.util.PyServlet.loadServlet(PyServlet.java) 
        at org.python.util.PyServlet.getServlet(PyServlet.java) 
        at org.python.util.PyServlet.service(PyServlet.java) 
<snip> 

P.S. 
The exception is from this code block in PyServlet.java, loadServlet(): 

         try { 
            interp.execfile(path); 
            PyObject cls = interp.get(name); 
            if (cls == null) 
                throw new ServletException("No callable (class or function)
"+ 
                                       "named " + name + " in " + path); 

            PyObject pyServlet = cls.__call__(); 
            Object o = pyServlet.__tojava__(HttpServlet.class); 
            if (o == Py.NoConversion) 
                throw new ServletException("The value from " + name + 
                                       "must extend HttpServlet"); 
            servlet = (HttpServlet)o; 
            servlet.init(getServletConfig()); 

        } catch (PyException e) { 
            throw new ServletException("Could not create "+ 
                                       "Jython servlet" + e.toString()); 
        }