[Jython-users] 2.1b2 PyServlet.loadServlet throws exception

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

I'm trying to run the Hello.py servlet example but I get the exception from
Tomcat 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 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());
        }
  

	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>