SimpleXMLRPCServer interruptable?

[Bret]

On Dec 6, 7:04 am, Bret <bret.wort... at gmail.com> wrote:
> On Dec 5, 10:00 pm, Edward Kozlowski <ekozlows... at gmail.com> wrote:
>
>
>
> > On Dec 5, 10:19 pm, Edward Kozlowski <ekozlows... at gmail.com> wrote:
>
> > > 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
>
> > Sorry about the first post, I shot from the hip and had to make a few
> > more modifications to make it 'working' code.  The example below works
> > fine for me.
>
> > import SimpleXMLRPCServer
>
> > class myServer:
> >     def __init__(self, host, port):
> >         self.done = False
> >         self.server = SimpleXMLRPCServer.SimpleXMLRPCServer((host,
> > port))
>
> >     def shutdown(self):
> >         self.done = True
> >         return 0
>
> >     def getName(self):
> >         return "Hey, I'm Ed"
>
> >     def serverWrapper(self):
> >         self.server.register_function(self.getName)
> >         self.server.register_function(self.shutdown)
> >         try:
> >             while not self.done:
> >                 self.server.handle_request()
> >         except:
> >             pass
>
> > if __name__ == "__main__":
> >     myServer('localhost', 6058).serverWrapper()
>
> > ---->>> s.getName()
> > "Hey, I'm Ed"
> > >>> s.shutdown()
>
> > 0
>
> Thanks to all!  I'm now keeping a file of my own snippets in hardcopy
> so I won't lose them next time I change jobs.  :-)
>
> Bret

Oops, this actually misses part of the problem -- I need to construct
this server programmatically, so imbedding the call to start the
server in the "if __name__" construct doesn't work.  If I start it
within the __init__, then the object that's creating it gets stuck
waiting for it to finish, which it never will.  I need this to be
autonomous, which is why I was trying to do the start of the server in
a separate thread, but that doesn't seem to work either (that's what's
causing my process to need the second call; the thread completes the
loop and enters the next handle_request() call before the event or
boolean gets set).

Basically, I need a way for one server to start other autonomous
servers without waiting on them, and still retain the ability to shut
them down when their work is done.

Back to the keyboard.  ;-)


Bret