在on_start方法中访问kivy屏幕ID

Question

我正在尝试使用on_start方法将小部件添加到屏幕，但是我遇到了尝试访问的屏幕ID出现键错误的情况。因此，我打印出self.root.ids，它是一个空字典。

这与on_start方法中未填充的屏幕有关吗？

main.py

class SelectChat(Screen):
    pass

class MyApp(MDApp):
    def build(self):
        kv = Builder.load_file("kivy.kv")
        self.sm = WindowManager()

        screens = [SelectChat(name="select_chat")]
        for screen in screens:
            self.sm.add_widget(screen)

        self.sm.current = "select_chat"
        return self.sm


    def on_start(self):
        self.docs = self.my_firestore.db.collection(u'users').where(u'value', u'==', True).stream()
        print(self.root.ids) #This is producing the empty dict
        #select_chat = self.root.ids['select_chat'].ids['select_chat_grid'] # This is where I get the key error

main.kv

<SelectChat>:
    name: "select_chat"
    id: select_chat

    GridLayout:
        cols: 1
        id: select_chat_grid

Answer 1

我不得不改变

select_chat = self.root.ids['select_chat'].ids['select_chat_grid']

至：

select_chat = self.root.get_screen("select_chat").ids['select_chat_grid']

所以我必须引用屏幕名称，而不是屏幕ID