Pyqt和websocket客户端.在后台监听websocket

Question

我有一个PyQt Gui应用程序。这个应用程序有一个主窗口，应该在启动后打开。

这个应用程序应该监听websocket。

我试着解决它:

...
if __name__ == '__main__':
    app = QtGui.QApplication(sys.argv)
    window = Window()
    window.show()
    websocket.enableTrace(True)
    ws = websocket.WebSocketApp("ws://localhost:8080/chatsocket",
                                on_message = on_message,
                                on_error = on_error,
                                on_close = on_close)
#    ws.on_open = on_open
    ws.run_forever()
    sys.exit(app.exec_())

但是，在启动应用程序后，主窗口没有打开。

如果没有"ws.run_forever()"行，主窗口打开，但应用程序不监听websocket。

我需要在后台监听websocket吗?你能帮我吗?

PS:(抱歉我的英文)

Answer 1

谢谢enginefree。

我让它

class Window(QtGui.QDialog):
    def __init__(self, parent=None):
        super(Window, self).__init__()
        self.thread = ListenWebsocket()
        self.thread.start()
...

class ListenWebsocket(QtCore.QThread):
    def __init__(self, parent=None):
        super(ListenWebsocket, self).__init__(parent)
        websocket.enableTrace(True)
        self.WS = websocket.WebSocketApp("ws://localhost:8080/chatsocket",
                                on_message = self.on_message,
                                on_error = self.on_error,
                                on_close = self.on_close) 
    def run(self):
        #ws.on_open = on_open
        self.WS.run_forever()

    def on_message(self, ws, message):
        print message
    def on_error(self, ws, error):
        print error
    def on_close(self, ws):
        print "### closed ###"
if __name__ == '__main__':
    app = QtGui.QApplication(sys.argv)
    QtGui.QApplication.setQuitOnLastWindowClosed(False)
    window = Window()
    window.show()
    sys.exit(app.exec_())