python - 如何处理来自 Gtk.Switch 的 "notify::active"信号？ (MVC架构)

Question

我不知道如何处理 Gtk.Switch 的 notify::active 信号。我正在使用 MVC 架构(模式)suggested here .

我得到的错误是这样的:

TypeError:_on_switch_serial_toggled() 缺少 1 个必需的位置参数:'state'

这是我的最小工作示例(没有模型):

import sys
import gi
gi.require_version('Gtk', '3.0')
from gi.repository import Gtk, Gio, GObject


class Application(Gtk.Application):
    def __init__(self):
        app_id = "org.iea.etc"
        flags = Gio.ApplicationFlags.FLAGS_NONE

        super(Application, self).__init__(application_id=app_id, flags=flags)

    def do_activate(self):
        # c.Controller(m.Model(), v.View(application=self))
        Controller(None, View(application=self))

    def do_startup(self):
        Gtk.Application.do_startup(self)


class Controller(object):
    def __init__(self, model, view):
        self._model = model
        self._view = view

        self._view.connect('switch_serial_toggled',
                           self._on_switch_serial_toggled)

        self._view.show_all()

    def _on_switch_serial_toggled(self, switch, state):
            if switch.get_active():
                print('Switch ON')
            else:
                print('Switch OFF')


class View(Gtk.ApplicationWindow):
    __gsignals__ = {
        'switch_serial_toggled': (GObject.SIGNAL_RUN_FIRST, None, ())
    }

    def __init__(self, **kw):
        super(View, self).__init__(**kw)

        self._switch_serial = Gtk.Switch()
        self._switch_serial.connect("notify::active",
                                    self.on_switch_serial_toggled)

        self.add(self._switch_serial)

    def on_switch_serial_toggled(self, switch, state):
        self.emit('switch_serial_toggled')


if __name__ == '__main__':
    app = Application()
    exit_status = app.run(sys.argv)
    sys.exit(exit_status)

Answer 1

首先，您正在为 Gtk.Switch almost 的 active 属性处理 notify 信号适本地。问题在于处理您已添加到 View 中的自定义信号。

了解您在做什么很重要:您已经创建了一个具有属性的 View 类，该属性是 Gtk.Switch。您还创建了一个与此类关联的信号，名为 switch_serial_toggled。在类内部，您希望在内部 Gtk.Switch 更改状态时，View 触发其自己的“切换”信号。

话虽如此，让我们修复您的代码:

1 - 让我们将切换状态作为 bool 值添加到 View switch_serial_toggled

class View(Gtk.ApplicationWindow):
    __gsignals__ = {
        'switch_serial_toggled': (GObject.SIGNAL_RUN_FIRST, None, (bool,))
    }

2 - 让我们为 View 内部信号处理程序的参数指定适当的名称，并将状态添加到发出的信号:

def on_switch_serial_toggled(self, obj, pspec):
    self.emit('switch_serial_toggled', self._switch_serial.get_active ())

GObject.Object 通知信号处理程序是 described here

3 - 现在，让我们转到 Controller 并正确处理 View 信号:

def _on_switch_serial_toggled(self, widget, active):
        if active is True:
            print('Switch ON')
        else:
            print('Switch OFF')

widget参数是 Controller 实例中的 View 实例，state 是随信号发射传递的 bool 值。

包含上述修复的完整代码应如下所示:

import sys
import gi
gi.require_version('Gtk', '3.0')
from gi.repository import Gtk, Gio, GObject


class Application(Gtk.Application):
    def __init__(self):
        app_id = "org.iea.etc"
        flags = Gio.ApplicationFlags.FLAGS_NONE

        super(Application, self).__init__(application_id=app_id, flags=flags)

    def do_activate(self):
        # c.Controller(m.Model(), v.View(application=self))
        Controller(None, View(application=self))

    def do_startup(self):
        Gtk.Application.do_startup(self)


class Controller(object):
    def __init__(self, model, view):
        self._model = model
        self._view = view

        self._view.connect('switch_serial_toggled',
                           self._on_switch_serial_toggled)

        self._view.show_all()

    def _on_switch_serial_toggled(self, widget, active):
            if active is True:
                print('Switch ON')
            else:
                print('Switch OFF')


class View(Gtk.ApplicationWindow):
    __gsignals__ = {
        'switch_serial_toggled': (GObject.SIGNAL_RUN_FIRST, None, (bool,))
    }

    def __init__(self, **kw):
        super(View, self).__init__(**kw)

        self._switch_serial = Gtk.Switch()
        self._switch_serial.connect("notify::active", self.on_switch_serial_toggled)

        self.add(self._switch_serial)

    def on_switch_serial_toggled(self, obj, pspec):
        self.emit('switch_serial_toggled', self._switch_serial.get_active ())


if __name__ == '__main__':
    app = Application()
    exit_status = app.run(sys.argv)
    sys.exit(exit_status)

结果是这样的:

PS:请注意，在第 2 步中，obj 是您连接信号处理程序的对象，类似于 self._switch_serial。这意味着您可以改用 obj.get_active ():

def on_switch_serial_toggled(self, obj, pspec):
    self.emit('switch_serial_toggled', obj.get_active ())