PyQt5如何从工作线程发出信号以按GUI线程调用事件

2024-09-27 07:19:17 发布

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正如我在标题中提到的。我怎么能做这样的事?

class Main(QWidget):

        def __init__(self):

                super().__init__()

        def StartButtonEvent(self):

                self.test = ExecuteThread()
                self.test.start()

        def MyEvent(self):

                #MainThreadGUI

class ExecuteThread(QThread):

        def run(self):

                # A lot of work

                # Signal to main thread about finishing of job = mainthread will perform MyEvent

我在这里找到一些教程pyqt4 emiting signals in threads to slots in main thread

在这里Emit signal from pyQt Qthread

但在PyQt5中似乎不起作用:/


Tags: oftointestself标题initmain

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