python - 在kivy中使用python将变量从一个屏幕更改为另一个屏幕

Question

我有一个双屏 kivy 应用程序，我想使用 python 将变量从一个屏幕传递到第二个屏幕。

Python文件

from kivy.app import App
from kivy.uix.boxlayout import BoxLayout
from kivy.uix.screenmanager import ScreenManager, Screen
from kivy.properties import StringProperty,ObjectProperty
from kivy.uix.label import Label

class MenuScreen(Screen):
    label1=StringProperty()
    label2=StringProperty()
    def __init__(self,**kwargs):
        super(MenuScreen, self).__init__(**kwargs)  
        self.label1="hello"
        self.label2="world" 
    def change_text(self):          
        lbl1=self.label1+ " and "
        lbl2= "A beautiful "+self.label2
        chg=SettingsScreen()
        chg.UpdateSettings(lbl1,lbl2)

        # HERE: something like



class SettingsScreen(Screen):
    label3=StringProperty()
    label4=StringProperty()
    def __init__(self,**kwargs):
        super(SettingsScreen, self).__init__(**kwargs)  
        #some default texts
        self.label3="Nothing"
        self.label4="Nothing"
    def UpdateSettings(self,lbl1,lbl2):
        print(lbl1,lbl2)
        self.label3=lbl1
        self.label4=lbl2

class TestScreenManager(ScreenManager):
    menu_screen=ObjectProperty(None)
    settings_screen=ObjectProperty(None)

class TestApp(App):
    def build(self):            
        return  TestScreenManager()

TestApp().run()

kv文件

#: import ScreenManager kivy.uix.screenmanager.ScreenManager
#: import Screen kivy.uix.screenmanager.ScreenManager
#: import SettingsScreen screen


<TestScreenManager>:
    id: screen_manager
    menu_screen: menu_screen
    settings_screen: settings_screen

    MenuScreen:
        id: menu_screen
        name: 'menu'
        manager: screen_manager
    SettingsScreen:
        id: settings_screen
        name: 'settings_screen'
        manager: screen_manager


<MenuScreen>:
    name: 'MenuScreen'
    BoxLayout:
        Label:
            text:root.label1
        Label:
            text:root.label2
        Button:
            text: 'Goto nn'
            size_hint_y:0.2
            on_press: 
                root.manager.current = 'settings_screen'
                root.change_text()


<SettingsScreen>:
    #name: 'SettingsScreen'
    label_id: label_field
    BoxLayout:
        Label:
            text:root.label3
        Label:
            text:root.label4
        Label:
            id: label_field
            text: "some text"

它运行时没有错误，但它没有更改第二个屏幕的变量。为了调试，我在 UpdateSetting 函数中添加了 print(lbl1,lbl2) ，它打印出传递的 lbl1 和 lbl2 变量，但它不更新标签。

Answer 1

当您执行 chg=SettingsScreen() 时，您将创建 SettingsScreen 类的新实例，并修改该实例的标签实例。您需要访问 ScreenManager 使用的对象，而不是创建一个新对象。

您可以使用当前屏幕的manager属性来获取另一个屏幕的实例引用:

def change_text(self):          
    lbl1=self.label1 + " and "
    lbl2= "A beautiful "+ self.label2
    chg = self.manager.settings_screen   #<<<<<<<<<<<<<<
    chg.UpdateSettings(lbl1,lbl2)