SimpleXMLRPCServer interruptable?

[Edward Kozlowski]

On Dec 5, 6:22 pm, "Gabriel Genellina" <gagsl-... at yahoo.com.ar> wrote:
> En Wed, 05 Dec 2007 18:20:35 -0300, Bret <bret.wort... at gmail.com> escribió:
>
> > I just tried changing this so that I now have a threading.Event()
> > called self.done, which is set within the body of the shutdown()
> > method.  The serverWrapper loop now looks like this:
>
> > def serverWrapper():
> >     while True:
> >         server.handle_request()
> >         if self.done.isSet():
> >             break
>
> > This works, but only if I follow the shutdown() rpc call a few seconds
> > later with ... something else to cause handle_request() to complete
> > again.  Obviously, not quite the right approach....
>
> You could try setting a reasonable timeout on the listening socket; I
> would override the server_bind method, calling self.socket.settimeout(xxx)
> before calling the inherited method. I've never actually done it with a
> SimpleXMLRPCServer, but *should* work. Don't use a very small timeout,
> because it affects *all* subsequent operations.
>
> --
> Gabriel Genellina

Try this:

def __init__(self, host, port):
    self.done = False
    server = SimpleXMLRPCServer((host, port))
    :
    : Bunch of server.register_function calls
    :
    def serverWrapper():
        try:
            while not self.done:
                server.handle_request()
        except:
            pass

Your shutdown method becomes:

def shutdown(self):
    self.done = True

HTH

-Edward Kozlowski