Kivy：如何在 __init__ 上更改 ScreenManager 的“当前”属性

Question

My App is a ScreenManager. Depending on the circumstances, I want the active screen to either be "Screen 1" or "Screen 2" when the app opens. How would I do this the most elegant way? I thought that this is as trivial as changing the current property in the initialization of the app. Sadly, this does not work. Here's what should work imo:

main.py:

MyApp(App):
     def build(self):
         return Builder.load_file("MyApp.kv")

     def __init__(self, **kwargs):
         super(MyApp, self).__init__(**kwargs)

         if foo: # Here's the problem:
             self.root.current = "Screen 1"
         else:
             self.root.current = "Screen 2"

MyApp.kv:

ScreenManager:
    Screen1:
        name: "Screen 1"
    Screen2:
        name: "Screen 2"

<Screen1@Screen>
etc...

But it doesn't. Throws the following error:

    self.root.current = "Screen 1"
AttributeError: 'NoneType' object has no attribute 'current'

My guess is that I set the current attribute to early, before root is set. An idea of mine is to 1) create a property-var for MyApp, 2) set current to be that property, 3) change that property in the init method. That's a lot of effort and code-cluttering just to change a screen on initialization.

How would I do it? Thanks a lot in advance!

Answer 1

That's simply because you don't have self.root object specified. Why would you need to change Screens during __init__ ? You should use build function for that.

My example:

import random

from kivy.app import App
from kivy.lang import Builder
from kivy.uix.screenmanager import ScreenManager

Builder.load_string('''
<Root>:
    Screen:
        name: "Screen 1"
        Label:
            text: "Screen 1!"

    Screen:
        name:"Screen 2"
        Label:
            text: "Screen 2!"
''')

class Root(ScreenManager):
    pass

class MyApp(App):
    def build(self):
        self.root = Root()

        foo = random.randint(0,1)
        if foo:
            self.root.current = "Screen 1"
        else:
            self.root.current = "Screen 2"

        return self.root



MyApp().run()

self.root.cureent_screen property will be changed before self.root object will be visible

Answer 2

Only this is working. I said above statement because I tryed many codes(but not works)

class NextScreen(ScreenManager):

    def __init__(self,**kwargs):
        super(NextScreen,self).__init__(**kwargs)
        #code goes here and add:
        Window.bind(on_keyboard=self.Android_back_click)

    def Android_back_click(self,window,key,*largs):
        # print(key)
        if key == 27:

            if self.current_screen.name == "Screen1" or self.current_screen.name == "Screen_main":
                return False

            elif self.current_screen.name == "Screen2":
                try:
                    self.current = "Screen1"
                except:
                    self.current = "Screen_main"
          
            return True

Use 'esc' button to get key(27) which means back in android