目标是在我的主屏幕上有一个按钮,当按下延迟屏幕时将打开。然后需要5秒的时间延迟,之后屏幕将返回到家。
我的问题是与按下按钮相关的任何时间延迟,停止延迟屏幕打开
from kivy.app import App
from kivy.lang import Builder
from kivy.uix.screenmanager import ScreenManager, Screen, FadeTransition
from kivy.uix.floatlayout import FloatLayout
import time
Builder.load_string("""
<MainScreen>:
name: "main"
FloatLayout:
canvas:
Rectangle:
pos: self.pos
size: self.size
Button:
background_color: 1, .8, .3, 1
text:'Home'
size_hint: .8,.2
pos_hint: {'center_x':.5,'y':.1}
on_press:root.manager.current = "delay"
<TimeDelay>:
name: "delay"
FloatLayout:
Button:
color: 0,1,0,1
font_size: 25
size_hint: 0.3,0.2
text: "delay"
pos_hint: {'center_x':0.5,'top':0.5}
on_press:root.manager.current = "main"
""")
class MainScreen(Screen):
pass
class TimeDelay(Screen):
time.sleep(5)
sm.switch_to('main')
sm = ScreenManager()
sm.add_widget(MainScreen(name='main'))
sm.add_widget(TimeDelay(name='delay'))
class timedelayapp(App):
def build(self):
return sm
if __name__ == "__main__":
timedelayapp().run()
答案 0 :(得分:1)
如果您不想阻止用户界面,则不想使用time.sleep()。请改用kivy.clock。
我不清楚是否要在按下按钮后5秒钟返回屏幕,或者在进入屏幕后5秒钟返回屏幕。按下按钮后,此示例将返回:
from kivy.app import App
from kivy.lang import Builder
from kivy.clock import Clock, mainthread
from kivy.uix.screenmanager import ScreenManager, Screen, FadeTransition
from kivy.uix.floatlayout import FloatLayout
from kivy.uix.button import Button
kv_string = """
ScreenManager:
MainScreen:
TimeDelay:
<MainScreen>:
name: "main"
FloatLayout:
canvas:
Rectangle:
pos: self.pos
size: self.size
Button:
background_color: 1, .8, .3, 1
text:'Home'
size_hint: .8,.2
pos_hint: {'center_x':.5,'y':.1}
on_press: root.manager.current = "delay"
<TimeDelay>:
name: "delay"
FloatLayout:
ButtonDelay:
color: 0,1,0,1
font_size: 25
size_hint: 0.3,0.2
text: "delay"
pos_hint: {'center_x':0.5,'top':0.5}
##on_press: root.manager.current = "main"
on_press: self.clocked_switch()
"""
class MainScreen(Screen):
pass
class TimeDelay(Screen):
pass
class ButtonDelay(Button):
def clocked_switch(self):
Clock.schedule_once(self.switch_to_main, 5)
def switch_to_main(self, *args):
app = App.get_running_app()
app.root.current = "main"
class TimeDelayApp(App):
def build(self):
root_widget = Builder.load_string(kv_string)
return root_widget
if __name__ == "__main__":
TimeDelayApp().run()
如果您想在输入屏幕后返回5秒,请使用类似的时钟功能调用,但是来自Screen.on_enter()。