使用 Kivy 切换屏幕：AttributeError: 'NoneType' object has no attribute 'transition'

Question

我在尝试使用 python 中的 kivy 库从另一个类中调用定义在一个类中的“屏幕切换”函数时遇到了 AttributionError。

我的代码结构如下：

from kivy.uix.screenmanager import ScreenManager, Screen
from kivy.app import App

def init_grid(width, height, mines):
    global grid_field
    global flagged
    # Initiate the gird field with empty grids
    grid_field = [[ClassA() for i in range(width)] for j in range(height)]
    flagged = 0
    ...


class Main(App):
   def build(self):
      sm = ScreenManager()

      current = Current(name="current")
      screenA = ScreenA(name="screenA")
      screenB = ScreenB(name="screenB")
      ...
      ...

      sm.add_widget(current)
      sm.add_widget(screenA)
      sm.add_widget(screenB)
      ...
      ...

      sm.current = "current"
      return sm

...

class ClassA():
    # a class my code is at when the error arises, 
    # which is not the starting "current" screen class 
    # but a different class that serves a certain purpose to my program
    def __init__(self, **kwargs):
        ...
        ...
        self.button = Button()
        self.button.bind(on_touch_down=self.method)


    def method(self, instance, touch):
        if (a condition):
            ScreenA().change_to_B()

    ...




...


class ScreenA(Screen):
    def __init__(self, **kwargs):
        Screen.__init__(self, **kwargs)
        ...
        ...
        init_grid(10, 10, 10)

    def change_to_B(self):
        self.manager.transition.direction = "left"
        self.manager.current = "screenB"


...


class ScreenB(Screen):
    def __init__(self, **kwargs):
        Screen.__init__(self, **kwargs)
        ...
        ...


if __name__ == '__main__':
    Main().run()

错误信息是：

in change_to_B
     self.manager.transition.direction = "left"
 AttributeError: 'NoneType' object has no attribute 'transition'

在上面的代码中，我在ScreenA 类中定义了屏幕切换函数change_to_B，并在ClassA 类中调用了这个函数。我在我创建的ScreenA 和ScreenB 两个类中都继承了Screen 父类。但奇怪的是，错误消息似乎表明对象screenA 是NoneType 对象，这不应该是这种情况，因为我在通过继承Screen 父级创建类时声明它是屏幕对象类。

谁能指出我错过了什么错误？提前谢谢！！！

编辑：

我想我需要弄清楚我的代码。所以我基本上是在尝试写扫雷游戏，ClassA是扫雷游戏板组成的网格，它有一个Button属性来实现点击功能method（如上代码所示） .在点击函数method中，我试图调用类ScreenA中定义的屏幕切换函数change_to_B()，这是在ScreenManager的build方法中声明的屏幕对象（我认为）。

代码流程为：Current(name="current")（与本题无关） --> ScreenA(name="screenA")，其中ClassA()是在全局列表变量中创建的函数 init_grid() --> ClassA()，点击 UI 调用绑定到 ClassA() 对象的 Button 属性的函数。

ClassA 对象在函数init_grid 中创建，其中创建了一个全局变量grid_field 以将ClassA 对象存储在一个二维列表中。

Answer 1

常见错误 - 每个屏幕使用多个对象，因此您将一个对象添加到 ScreenManager 并且该对象具有属性 manager。但是然后你在这里创建另一个ScreenA 类的对象：

def __init__(self, **kwargs):
        ...
        ...
        ScreenA().change_to_B()

屏幕管理器对该对象一无所知。所以你应该记住的是你将objects添加到屏幕管理器，而不是classes。 您没有提供整个代码，所以我不知道您在哪里定义 ClassA 类的对象，但您应该从 build 方法传递对象以使其工作。像这样的：

class ScreenA(Screen):
    def __init__(self, **kwargs):
        Screen.__init__(self, **kwargs)
        ...
        ...
        # here we pass the object of the ScreenA class (which is self) to init grid to be able to pass it to ClassA
        init_grid(10, 10, 10, self)

# add object to arguments
def init_grid(width, height, mines, screenA):
    global grid_field
    global flagged
    # pass the object to ClassA
    grid_field = [[ClassA(screenA) for i in range(width)] for j in range(height)]

class classA():
    # don't forget to add object to arguments
    def __init__(self, screenA, **kwargs):
        ...
        ...
        self.screenA = screenA

    def method(self, instance, touch):
        if (a condition):
            self.screenA.change_to_B()

或者使用全局变量：

current = Current(name="current")
screenA = ScreenA(name="screenA")
screenB = ScreenB(name="screenB")

class Main(App):
   def build(self):
      sm = ScreenManager()

      global current
      global screenA
      global screenB
      ...
      ...

      sm.add_widget(current)
      sm.add_widget(screenA)
      sm.add_widget(screenB)
      ...
      ...

      sm.current = "current"
      return sm

class ClassA():
    def method(self, instance, touch):
        global screenA
        if (a condition):
            screenA.change_to_B()