Kivy:在python中添加的小部件中获取父级

Kivy: get parent inside widget which is added in python(Kivy:在python中添加的小部件中获取父级)

问题描述

如何在不是由 kvlang 添加但在 python 中添加的小部件中获取对父级的引用.通常你只需调用 self.parent 但是如果在 python 中将小部件添加到父级，则返回 Null.

How do I get the reference to a parent inside a widget that is not added by kvlang but in python. Normally you would just call self.parent however that returns Null if the widget is added in python to the parent.

一个例子:

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()

这将返回:

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

推荐答案

使用 add_widget 添加的小部件实际上具有对其父级的有效引用:

Widgets added with add_widget actually has a valid reference to their parent:

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()

它只是在他们的 __init__ 方法中不可用.

It's just not available in their __init__ method.

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