'PySide2.QtCore.Signal'对象没有属性'connect'

时间:2018-10-30 01:23:50

标签: python pyside2

我正在尝试为我的PySide2应用程序的QRunnable对象创建自定义信号。所有示例均导致我通过以下方式创建信号:

class Foo1(QtCore.QObject):

    def __init__():
        super().__init__()
        self.thread = Foo2()
        self.thread.signal.connect(foo)

    def foo():
        # do something


class Foo2(QtCore.QRunnable):

    signal = QtCore.Signal()

但是,我在self.thread.signal.connect(foo)上遇到以下错误:

'PySide.QtCore.Signal' object has no attribute 'connect'

如何为QRunnable对象实现自定义信号?

1 个答案:

答案 0 :(得分:2)

QRunnable不是QObject,因此它不能具有信号,因此可能的解决方案是创建一个提供信号的类:

class FooConnection(QtCore.QObject):
    foosignal = QtCore.Signal(foo_type)

class Foo2(QtCore.QRunnable):
    def __init__(self):
        super(Foo2, self).__init__() 
        self.obj_connection = FooConnection()

    def run(self):
        # do something
        foo_value = some_operation()
        self.obj_connection.foosignal.emit(foo_value)

class Foo1(QtCore.QObject):
    def __init__():
        super().__init__()
        self.pool = Foo2()
        self.pool.obj_connection.foosignal.connect(foo)
        QtCore.QThreadPool.globalInstance().start(self.pool)

    @QtCore.Slot(foo_type)
    def foo(self, foo_value):
        # do something