RE: [Jython-users] compiled servlet does not run

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Robert,

This works for you?  Don't you need to add 'self' to the doGet() 
method?  It's an instance method, no?

brian

At 09:07 AM 3/13/2001 -0600, Robert W. Bill wrote:

>On Tue, 13 Mar 2001, Eduardo Jesus Fernandez Corrales wrote:
> >
> > Sure, there it goes.
> >
> > I am posting the generated file too, in case someone wants to check if 
> it runs.
>
>"*" in import, Hmmmmmm.  I tried the following with success...
>
>
>#spare imports removed and "*" replace with explicit package name
>from javax.servlet import http
>
># prefix superclass with explicit package name "http."
>class Hello(http.HttpServlet):
>     def doGet(req, res):
>         res.setContentType("text/html")
>         out = res.getWriter()
>         out.println("<html><head><title>Hello World!</title></head>")
>         out.println("<body><h1>Hello World!</h1></body></html>")
>
>
>Another suspicion is the classpath when you compile.  Last I checked, you
>can compile something like above _without_  javax.servlet in the
>classpath and with no compile errors.  However, at runtime you get a NPE,
>so you need to make sure javax.servlet.*... is in the classpath rather
>than relying on compile time errors to remind you  (I don't remember the
>original post though, so this may not apply.)
>
>
> > >>> "brian zimmer" <bz...@zi...> 13/03/01 14:10 >>>
> > >It's hard to tell without the servlet file, but make sure the file 
> Hello.py
> > >has a single class called 'Hello' that extends from HttpServlet:
>
>I think there can be more than one class.  If you have multiple classes
>in the file, the servlet class (the subclass of HttpServlet) should be the
>same name as the file.
>
>Cheers,
>Robert
>
>
>
>
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