Question

我正在使用以下示例

class bar(QObject):
    def mySlot(self,p):
        print "This is my slot " + str(p)


class Foo(QObject):
    trigger = pyqtSignal()

    def my_connect_and_emit_trigger(self):
        self.trigger.emit(12)

    def handle_trigger(self):
        # Show that the slot has been called.
        print "trigger signal received"


b = bar()
a = Foo()
a.trigger.connect(int,b.mySlot) <---how to fix this
a.connect_and_emit_trigger()

我正在尝试将带有一个int参数的插槽b.mySlot附加到信号a.trigger我的问题是我做错了什么。我找不到任何有助于信号参数的材料。

Answer 1

这是正确的：

class bar(QObject):
    def mySlot(self,p):
        print "This is my slot " + str(p)


class Foo(QObject):
    trigger = pyqtSignal(int)

    def my_connect_and_emit_trigger(self):
        self.trigger.emit(12)

    def handle_trigger(self):
        # Show that the slot has been called.
        print "trigger signal received"


b = bar()
a = Foo()
a.trigger.connect(b.mySlot)
a.my_connect_and_emit_trigger()

Doc是here

在PyQt中创建带参数的信号

1 个答案: