【发布时间】:2015-02-24 06:18:42
【问题描述】:
我正在尝试制作一个 android 应用程序...我正在尝试根据用户输入从菜单切换到服务器或客户端...但我无法从一个类切换到另一个类...主要的事情让我生气的是,ServerScreen 中的打印语句正在工作,但其他部分却没有工作……不知道为什么
代码:---
import kivy
from kivy.app import App
from kivy.uix.widget import Widget
from kivy.uix.label import Label
from kivy.uix.button import Button
from kivy.uix.screenmanager import ScreenManager, Screen
class MenuScreen(Screen):
def __init__(self, **kwargs):
super(MenuScreen, self).__init__(**kwargs)
def Server(instance):
self.clear_widgets()
self.add_widget(Label ( text = 'Inside server function'))
server = ServerScreen()
#return server
#server.function()
self.add_widget(Label ( text = 'What Type Of Service You Want...???'))
button1 = Button(text = 'Server',size_hint = (None,None),pos = (0,0))
self.add_widget(button1)
button1.bind(on_press = Server)
button2 = Button(text = 'Client',size_hint = (None,None),pos = (100,0))
self.add_widget(button2)
#button2.bind(on_press = Client)
class ServerScreen(Screen):
def __init__(self, **kwargs):
super(ServerScreen, self).__init__(**kwargs)
print('Inside server screen')
self.clear_widgets()
self.add_widget(Label (text = 'Working As A Server'))
print("Hellooooooooooo")
sm = ScreenManager()
sm.add_widget(MenuScreen(name='Menu'))
sm.add_widget(ServerScreen(name='Server'))
#sm.add_widget(ClientScreen(name='Client'))
class FileApp(App):
def build(self):
#return Menu()
return sm
if __name__ == '__main__':
FileApp().run()
我在哪个部分犯了错误...请指出该部分并提供我该如何解决...
输出--->
内部服务器屏幕
喂喂喂喂
【问题讨论】:
标签: python-2.7 kivy