Question

我有一个连接到CG服务器（CasparCG）的Python QT应用程序。 QT应用程序触发一个QThread，它使用模块 pynput 侦听热键-并向CasparCG发送命令以针对每个按下的键播放不同的视频文件。

在主GUI中，我可以将视频文件分配给热键列表，然后从子菜单项触发热键侦听线程。

self.actionStart_Hotkeys = QtWidgets.QAction(MainWindow)
self.menuCasparCG.addAction(self.actionStart_Hotkeys)    
self.actionStart_Hotkeys.triggered.connect(self.StartHotkeys)

Main应用程序和Ui_Window代码非常长，没有问题-它的功能应该像它应该的那样。视频也可以按预期的方式播放，方法是通过按按键激活热键，但是在播放了一些视频文件后，应用程序的主窗口冻结了-我不确定为什么主GUI在热键之后没有响应输入线程已启动。

到目前为止的代码看起来像这样。

from pynput import keyboard

class HotKeys(QThread):
    def __init__(self, parent):
        QThread.__init__(self, parent)
        self.COMBINATIONS = [
            {keyboard.KeyCode(char='0')},
            {keyboard.KeyCode(char='1')},
            {keyboard.KeyCode(char='2')},
            {keyboard.KeyCode(char='3')},  
            ]

        self.caspar = None
        self.current = set()
        self.Connect()
        self.Listen()

    def exit(self, i):
        if not self.caspar == None:
            self.caspar.close
        sys.exit(i)

    def Connect(self):
        try:
            self.caspar = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
            self.caspar.connect(("127.0.0.1", 5250))
            print("Connected to caspar")

        except socket.error:
            print("CasparCG not running, or incorrect settings.xml")
            self.exit(0)

    def execute(self, k=None): # k is videofile 
        movie = bytes("PLAY 1-20 {} \r\n".format(k), 'utf8')
        self.caspar.send(movie)

    def on_press(self, key):
        if any([key in COMBO for COMBO in self.COMBINATIONS]):
             self.current.add(key)
            if any(all(k in self.current for k in COMBO) for COMBO in self.COMBINATIONS):
                self.execute(key)

    def on_release(self, key):
        if any([key in COMBO for COMBO in self.COMBINATIONS]):
            self.current.remove(key)

    def Listen(self):
        with keyboard.Listener(on_press=self.on_press, on_release=self.on_release) as listener:
            listener.join()

我像这样在应用程序的Main类中触发此热键QThread ...

class Main(QMainWindow, Ui_MainWindow):
    def __init__(self):
        QMainWindow.__init__(self)
        self.setupUi(self) # from Ui_MainWindow class       

    def StartHotkeys(self):
        hotkey_thread = HotKeys(self)
        hotkey_thread.start()

和类似的应用程序...

if __name__ == "__main__":
    import sys
    app = QtWidgets.QApplication(sys.argv) 
    gui = Main()
    gui.show()
    sys.exit(app.exec_())

那Main为什么会冻结？

Answer 1

QThread不是线程，它是线程处理程序，如果要在另一个线程中执行任务，则必须使用run()方法来执行，该方法是唯一的部分。在另一个线程中执行。在您的情况下，Listen()任务正在阻塞，您可以在构造函数中调用它，而QThread构造函数在GUI线程中运行，这就是GUI冻结的原因。解决方案是将“连接并监听”方法移至run()：

class HotKeys(QThread):
    def __init__(self, parent=None):
        QThread.__init__(self, parent)
        self.COMBINATIONS = [
            {keyboard.KeyCode(char='0')},
            {keyboard.KeyCode(char='1')},
            {keyboard.KeyCode(char='2')},
            {keyboard.KeyCode(char='3')},  
            ]

        self.caspar = None
        self.current = set()

    def run(self):
        self.Connect()
        self.Listen()

    def exit(self, i):
        if self.caspar:
            self.caspar.close()
        sys.exit(i)

    def Connect(self):
        try:
            self.caspar = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
            self.caspar.connect(("127.0.0.1", 10000))
            print("Connected to caspar")

        except socket.error:
            print("CasparCG not running, or incorrect settings.xml")
            self.exit(0)

    def execute(self, k=None): # k is videofile 
        if self.caspar:
            movie = bytes("PLAY 1-20 {} \r\n".format(k), 'utf8')
            self.caspar.send(movie)

    def on_press(self, key):
        if any([key in COMBO for COMBO in self.COMBINATIONS]):
            self.current.add(key)
            if any(all(k in self.current for k in COMBO) for COMBO in self.COMBINATIONS):
                self.execute(key)

    def on_release(self, key):
        if any([key in COMBO for COMBO in self.COMBINATIONS]):
            self.current.remove(key)

    def Listen(self):
        with keyboard.Listener(on_press=self.on_press, on_release=self.on_release) as listener:
            listener.join()

PyQt-主要挂在QThread

1 个答案: