Kivy : 파이썬에 추가 된 위젯 내부에 부모 가져 오기

Question

kvlang이 아닌 python으로 추가 된 위젯 내부의 부모에 대한 참조를 어떻게 얻을 수 있습니까? 일반적으로 self.parent을 호출하지만 위젯이 파이썬에 부모에 추가 된 경우 Null을 반환합니다.

예 :

import kivy 
kivy.require('1.9.0') # replace with your current kivy version ! 

from kivy.app import App 
from kivy.lang import Builder 
from kivy.uix.screenmanager import ScreenManager,Screen 
from kivy.clock import Clock 

kvlang = ''' 
<ScreenManagement>: 
    ScreenOne: 

<ScreenOne>: 
    name: 'First' 


<ScreenTwo>: 
    name: 'Second' 

''' 


class ScreenManagement(ScreenManager): 
    def __init__(self,**kwargs): 
     super().__init__(**kwargs) 

     def setup(*args): 
      self.add_widget(ScreenTwo()) #add ScreenTwo later in python 

     Clock.schedule_once(setup) 

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

     def setup(*args): 
      print("Parent of ScreenOne: {}".format(self.parent))  #this is working 
     Clock.schedule_once(setup) 

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

     def setup(*args): 
      print("Parent of ScreenTwo: {}".format(self.parent))  #this is not working, self.parent will return None 
     Clock.schedule_once(setup) 



class MyApp(App): 

    def build(self): 
     Builder.load_string(kvlang) 
     return ScreenManagement() 


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

이 반환됩니다

Parent of ScreenOne: <__main__.ScreenManagement object at 0x7f98a3fddb40> 
Parent of ScreenTwo: None 

Answer 1

위젯 add_widget에 추가 실제로 부모에 대한 유효한 참조가 있습니다

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

from kivy.lang import Builder 
Builder.load_string(''' 
<ScreenTwo> 
    Label: 
     text: 'Hello, world'  
''') 

class ScreenManagement(ScreenManager): 
    def __init__(self,**kwargs): 
     super(ScreenManagement, self).__init__(**kwargs) 
     screen = ScreenTwo() 

     print(screen.parent) 
     self.add_widget(screen) 
     print(screen.parent) 


class ScreenTwo(Screen): 
    def on_touch_down(self, *args): 
     print(self.parent) 


class MyApp(App): 
    def build(self): 
     return ScreenManagement() 


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

그것은 단지 사용할 수없는을 그들의 __init__ 방법.