我如何直接到下一页，并在Kivy中单击按钮上调用功能

Question

我有一个应用程序，上面有两个按钮。我希望两个按钮都可以路由到下一页，并同时调用它们各自的功能。任何人都可以帮助我链接或代码。预先感谢。

Answer 1

以下示例使用Kivy ScreenManager，Button窗口小部件和按钮的on_release和on_press事件。

main.py

from kivy.app import App
from kivy.lang import Builder
from kivy.uix.screenmanager import ScreenManager, Screen
# Create both screens. Please note the root.manager.current: this is how
# you can control the ScreenManager from kv. Each screen has by default a
# property manager that gives you the instance of the ScreenManager used.
Builder.load_string("""
<ScreenManagement>:
    MenuScreen:
        id: name
        name: 'menu'
        
    SettingsScreen:
        id: settings
        name: 'settings'
        
<MenuScreen>:
    BoxLayout:
        Button:
            text: 'Goto settings & invoke function abc'
            on_release: 
                root.manager.current = 'settings'
                root.manager.ids.settings.func_abc(self)    # optional: passing Button instance
                
        Button:
            text: 'Goto settings & invoke function xyz'
            on_release: 
                root.manager.current = 'settings'
                root.manager.ids.settings.func_xyz(self)    # optional: passing Button instance
<SettingsScreen>:
    BoxLayout:
        Button:
            text: 'My settings button'
        Button:
            text: 'Back to menu'
            on_press: root.manager.current = 'menu'
""")

# Declare both screens
class MenuScreen(Screen):
    pass

class SettingsScreen(Screen):
    def func_abc(self, instance):
        print(f"func_abc: Called from Button with text={instance.text}")
    def func_xyz(self, instance):
        print(f"func_xyz: Called from Button with text={instance.text}")

# Create the screen manager
class ScreenManagement(ScreenManager):
    pass

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

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