Python新风格信号和线程与gui app之间的插槽

这是线程.在GUI中点击RUN按钮后3秒内,我将发出信号,用于更新GUI中的消息对话框.我不确定继承是否定义好或者是否以正确的方式定义了信号.

class OptimThread (QtCore.QThread):

    signalUpdateMessageDialog = QtCore.SIGNAL("updateMessageDialog(PyQt_PyObject,QString)")

    def __init__(self):
        QtCore.QThread.__init__(self)

    def run(self):

        start = time.time()

        self.emit(self.signalUpdateMessageDialog,time.time() - start,'Initialising...')     

        time.sleep(3)

        self.emit(self.signalUpdateMessageDialog,'You waited 3 seconds...')

主类和应用程序部分是这样的(我省略了其他可能不相关的代码).

class Main(QtGui.QMainWindow,Ui_MainWindow):    

    def __init__(self,parent=None):
        super(Main,self).__init__(parent)

    def updateMessageDialog(self,times,dialog):  

        hours = str(datetime.timedelta(seconds=int(times)))

        self.MessageDialog.insertHtml('<tt>' + hours + ':</tt> ' + dialog + '<br>')

        return
    def clickRun(self):


        self.optimThread = OptimThread()

        self.connect(self.optimThread,QtCore.SIGNAL("updateMessageDialog(PyQt_PyObject,QString)"),self.updateMessageDialog)

        #self.optimThread.signalUpdateMessageDialog.connect(self.updateMessageDialog)

        self.optimThread.start()

if __name__ == '__main__':
    app=QtGui.QApplication(sys.argv)
    window=Main(None)
    app.setActiveWindow(window)
    window.show()
    sys.exit(app.exec_()) # Exit from Python

如果evertyhing是这样写的,那就有效.
然而,如果我想在Main中使用新样式进行连接：

self.optimThread.signalUpdateMessageDialog.connect(self.updateMessageDialog)

它说：

self.optimThread.signalUpdateMessageDialog.connect(self.updateMessageDialog)
AttributeError: ‘str’ object has no attribute ‘connect’

我感谢您的建议(与主题相关并与风格相关),并为不制作MWE而道歉.