from kivy.app import App
from kivy.uix.label import Label
from kivy.clock import Clock
import socket
class MyApp(App):
def build(self):
return Label(text="test app")
def timer_callback(dt):
try:
data, addr = sock.recvfrom( 1024 )
if data == "exit\n":
sock.close()
Clock.unschedule(timer_callback)
quit() # how do i quit GUI (MyApp) ???
except socket.error, e:
if e.errno != 10035:
raise e
if __name__ == '__main__':
IP = "127.0.0.1"
PORT = 1234
sock = socket.socket(socket.AF_INET, # Internet
socket.SOCK_DGRAM) # UDP
sock.setblocking(0)
sock.bind((IP, PORT))
Clock.schedule_interval(timer_callback, 1)
MyApp().run()
Hello, in timer_callback() i wanna quit the app if 'exit' is received.
But my GUI still is open after quit() (Kivy on Windows)
How do i exit an Kivy-App with code ??
greets Janette