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Home/ Questions/Q 8959935
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T15:33:30+00:00 2026-06-15T15:33:30+00:00

I am doing an app whit wxPython and XMLRPC i need that the window

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I am doing an app whit wxPython and XMLRPC i need that the window does an action every time the XMLRPC server has a request

How could i do it without blocking the main Window?

I tried with threads but it doesnt work also I tried calling the run method of the thread in the Frame’s constructor neither it worked

Sorry for the language
I hope to be clear
Thanks

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  1. Editorial Team
    Editorial Team
    2026-06-15T15:33:31+00:00Added an answer on June 15, 2026 at 3:33 pm

    Here’s an example of a threaded XMLRPC server using SimpleXMLRPCServer. Note the wx.CallAfter to call into the wx main thread and the “return 0” (though you can configure the server so that return values of None are OK.)

    from SimpleXMLRPCServer import SimpleXMLRPCServer
    import threading
    
    class XMLRPCServerThread(threading.Thread):
        def __init__(self, remoteObject, host='localhost', port=8000):
            self.remoteObject = remoteObject
            self.host = host
            self.port = port
            threading.Thread.__init__(self)
    
        def stop(self):
            self.server.shutdown()    
    
        def run(self):
            self.server = SimpleXMLRPCServer( (self.host, self.port), logRequests=False )
            self.server.register_instance( self.remoteObject )
            self.server.serve_forever()
    
    class MyRemoteCalls(object):
    
        def __init__(self, obj):
            self.obj = obj
    
        def exampleCall(self, arg):
            wx.CallAfter(self.obj.method, arg)
            return 0  
    
    def getRPCThread(obj, host='localhost', port=8000):
        remoteObj = MyRemoteCalls(obj)
        rpcThread = XMLRPCServerThread(remoteObj, host, port)
        rpcThread.start()
        return rpcThread
    
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