我如何才能转到下一页，还可以在按钮中单击按钮来调用函数？

Question

我在第一个屏幕上有一个带有两个按钮的应用程序。我希望两个按钮都路由到下一页，同时调用它们各自的功能。 任何人都可以通过链接或代码帮助我。预先感谢。

Answer 1

以下示例说明了如何使用Kivy ScreenManager，Button小部件以及按钮的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()