Python新样式信号以及线程和gui应用之间的插槽

问题描述

我是OOP和python的新手.我正在尝试使用新样式的信号和插槽将信号从Qthread发射到Qt GUI主窗口.

I am newbie to OOP and python. I am trying to emit signal from Qthread to Qt GUI main window using new style signals and slots.

这是线程.在内部，单击GUI中的运行"按钮后3秒钟，我将在GUI中发出用于更新消息对话框的信号.我不确定继承是否定义正确，或者信号是否定义正确.

This is the thread. Inside I will emit signals for updating message dialog in GUI after clicking RUN button in GUI and 3 seconds after that. I am not sure if the inheritance is defined OK or is the signal defined in the right manner.

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, time.time() - start, 'You waited 3 seconds...')

主类和应用程序部分是这样的(我省略了其他可能无关的代码).

The main class and the app part is like this (I omitted the other probably unrelevant code).

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

如果一切都是这样写的，那就行得通.但是，如果我想在Main中使用新样式进行连接:

If evertyhing is written like this, It works.Yet if I want to use new style for connecting in Main:

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

它说:

感谢您的建议(与主题和风格有关)，并为没有制作MWE表示歉意.

I appreciate your advices (related to the subject and related to the style) and apologize for not making an MWE.

推荐答案

示例的结构或多或少是正确的:但是您正在将旧样式的信号槽语法与新样式混合在一起.

The structure of your example is more or less right: but you are mixing up the old-style signal-slot syntax with the new-style.

信号定义应如下所示:

class OptimThread(QtCore.QThread):
    signalUpdateMessageDialog = QtCore.pyqtSignal(int, str)

信号应该这样发射:

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

这是信号的连接方式:

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

使用新样式的语法，永远不需要使用SIGNAL()或SLOT()，并且永远不需要指定C ++签名.

With the new-style syntax, there is never any need to use SIGNAL() or SLOT(), and it is never necessary to specify the C++ signature.

有关更多详细信息，请参阅以下网页中的新型信号和插槽支持. PyQt4参考.

For further details, see New-style Signal and Slot Support in the PyQt4 reference.

这篇关于Python新样式信号以及线程和gui应用之间的插槽的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！