在pyqtBoundSignal上侦听新的信号连接

Question

我有一个PyQt5应用程序，它具有可选功能（比方说一个按钮），仅在应用程序被明确告知“打开此选项”时才显示。当调用此按钮时，应用程序会进行一些计算，然后发出带有结果的信号。

我不希望显式传递构造函数选项，而是希望程序在有人连接到信号时显示功能。

工作实例：

class ExampleApp(QWidget):
    do_something = pyqtSignal(str)

    def __init__(self, something_enabled: bool = False):
        super().__init__()

        self.button = QPushButton("Do something")
        self.button.pressed.connect(self.do_work)

        self.button.setVisible(something_enabled)

        self.setLayout(QVBoxLayout())
        self.layout().addWidget(self.button)

        self.show()

    def do_work(self):
        self.do_something.emit("Something!")

if __name__ == '__main__':
    import sys
    app = QApplication(sys.argv)
    window = ExampleApp(something_enabled=True)
    window.do_something.connect(lambda result: print(result))
    app.exec()

我在文档中找不到任何可用选项，但我无法找到此功能的实现。在我自己的代码中，我尝试直接更改connect函数，但是我得到了一个只读的AttributeError。

包装功能：

self.original_connect = self.do_something.connect

def new_connect(slot, type=None, no_receiver_check=False):
    print("New connection!")
    self.original_connect(slot, type, no_receiver_check)

self.do_something.connect = new_connect

AttributeError的：

AttributeError: 'PyQt5.QtCore.pyqtBoundSignal' object attribute 'connect' is read-only

我还创建了一个工作包装类，但我对未知后果持谨慎态度。

包装类：

class SignalWrapper(QObject):

    new_connection = pyqtSignal(object)

    def __init__(self, signal_to_wrap):
        super().__init__()
        self.wrapped_signal = signal_to_wrap
        self.signal = self.wrapped_signal.signal

    def connect(self, slot, type=None, no_receiver_check=False):
        self.new_connection.emit(slot)
        return self.wrapped_signal.connect(slot)

    def disconnect(self, slot=None): # real signature unknown; restored from __doc__
        return self.wrapped_signal.disconnect(slot)

    def emit(self, *args): # real signature unknown; restored from __doc__
        return self.wrapped_signal.emit(*args)

    def __call__(self, *args, **kwargs): # real signature unknown
        return self.wrapped_signal.__call__(*args, **kwargs)

    def __getitem__(self, *args, **kwargs): # real signature unknown
        return self.wrapped_signal.__getitem__(*args, **kwargs)

    def __repr__(self, *args, **kwargs): # real signature unknown
        return self.wrapped_signal.__repr__(*args, **kwargs)     

ExampleApp的修改部分：

class ExampleApp(QWidget):
    do_something = pyqtSignal(str)

    def __init__(self):
        super().__init__()
        self.do_something = SignalWrapper(self.do_something)

我想知道是否有一种优雅的方法来拦截PyQt中的信号连接。

Answer 1

QObject类具有虚拟connectNotify和disconnectNotify方法，可以覆盖它们以提供不同的行为。默认实现什么都不做。还有一个receivers方法，它给出了信号的当前连接数。

下面的演示使用这些方法创建一个通用连接观察器类，可以附加到任何QObject子类。它监视一组信号，并在检测到初始连接或最终断开连接时发出connected信号：

from PyQt5.QtCore import *
from PyQt5.QtWidgets import *

class ConnectionWatcher(QObject):
    connected = pyqtSignal(str, bool)

    def __init__(self, target, *signals):
        if not target.findChild(ConnectionWatcher):
            super().__init__(target)
            self._signals = set(signals)
            target.connectNotify = lambda s: self._notified(s, True)
            target.disconnectNotify = lambda s: self._notified(s, False)
        else:
            raise RuntimeError('target already has a connection watcher')

    def _notified(self, signal, connecting):
        name = str(signal.name(), 'utf-8')
        if name in self._signals:
            count = self.parent().receivers(getattr(self.parent(), name))
            if connecting and count == 1:
                self.connected.emit(name, True)
            elif not connecting and count == 0:
                self.connected.emit(name, False)

class ExampleApp(QWidget):
    do_something = pyqtSignal(str)

    def __init__(self, something_enabled: bool = False):
        super().__init__()

        self.button = QPushButton("Do something")

        watcher = ConnectionWatcher(self.button, 'pressed')
        watcher.connected.connect(self.handleConnection)

        c1 = self.button.pressed.connect(self.do_work)
        c2 = self.button.pressed.connect(self.do_work)

        self.button.pressed.disconnect(c1)
        self.button.pressed.disconnect(c2)

        self.setLayout(QVBoxLayout())
        self.layout().addWidget(self.button)

        self.show()

    def handleConnection(self, name, connecting):
        print('connecting:', name, connecting)

    def do_work(self):
        self.do_something.emit("Something!")


if __name__ == '__main__':

    import sys
    app = QApplication(sys.argv)
    window = ExampleApp(something_enabled=True)
    window.do_something.connect(lambda result: print(result))
    app.exec()

Answer 2

您不应该更改内置函数的默认行为。由于隐藏的逻辑（为什么信号会使按钮可见？），它将使您的代码难以理解和维护。

但是，您可以在一个方法中包装连接：

class ExampleApp(QWidget):
    def turnOnOption(self, callback):
        """
        Make button visible. Callback will be called when the work is done.
        """
        self.do_something.connect(callback)
        self.button.setVisible(True)

window = ExampleApp()
#window.do_something.connect(lambda result: print(result))
window.turnOnOption(lambda result: print(result))

这对其他开发者来说很明确，你的代码看起来不像是黑魔法......