Kivy屏幕更改

Question

玉家伙我有一个问题我想点击弹出菜单的按钮，然后后，我点击它必须transistion到另一个屏幕弹出....这里是我的代码 只是想获得一个新的新鲜屏幕后我点击弹出按钮

from kivy.uix.popup import Popup 
from kivy.app import App 
from kivy.uix.gridlayout import GridLayout 
from kivy.uix.label import Label 
from kivy.uix.textinput import TextInput 
from kivy.uix.button import Button 
# from kivy.uix.boxlayout import BoxLayout 
# from kivy.uix.stacklayout import StackLayout 
from kivy.uix.screenmanager import ScreenManager, Screen 


class LoginScreen(GridLayout, Screen): 
    def __init__(self, sm, **kwargs): 
     super(LoginScreen, self).__init__(**kwargs) 
     self.sm = sm 
     self.cols = 2 
     self.row = 2 
     self.add_widget(Label(text='User Name', font_size='20sp')) 
     self.username = TextInput(multiline=False) 
     self.add_widget(self.username) 
     self.add_widget(Label(text='password')) 
     self.password = TextInput(password=True, multiline=False) 
     self.add_widget(self.password) 
     self.hello = Button(text="hello", on_press=lambda a: self.save(), size=(100, 100), 
          size_hint=(0.3, 0.3)) 
     self.add_widget(self.hello) 

    def save(self): 
     print("s") 
     id_name = self.username._get_text() 
     id_num = self.password._get_text() 
     if id_name == "Hendricko" and id_num == "stokkies123": 
      content = Button(text="Press Here", size=(100, 100), size_hint=(0.3, 0.3)) 
      popup = Popup(title="You May Proceed ", 
         content=content, 
         size=(50, 50), 
         size_hint=(0.3, 0.3), 
         auto_dismiss=False) 
      content.bind(on_press=lambda b: self.check_menu_press) 
      popup.open() 

    def check_menu_press(self, button, *args): 
     if button.state == 'normal': 
      self.sm.current = "SecondScreen" 


class SecondScreen(GridLayout,Screen): 
    def __init__(self, sm, **kwargs): 
     super(SecondScreen, self).__init__(**kwargs) 
     self.row = 2 
     self.cols = 2 
     self.add_widget(Label(text="hello", font_size="20sp")) 

     self.sm = sm 

    def on_touch_down(self, touch): 
     self.sm.current = "LoginScreen" 


class MyApp(App): 
    def build(self): 
     sm = ScreenManager() 
     sm.add_widget(LoginScreen(sm, name="SecondScreen")) 
     sm.add_widget(SecondScreen(sm, name="LoginScreen")) 
     return sm 




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

Answer 1

content.bind（on_press =拉姆达b：self.check_menu_press）

此lambda函数什么都不做。你大概的意思content.bind(on_press=lambda b: self.check_menu_press(b))。我认为这将是整洁使用functools.partial不过，如果你想这样做事。