在kivy应用程序中将类实例作为进程：kivy应用程序被卡住

Question

我确实有一个存储设置和启动进程的界面。但是，一旦进程启动，我无法关闭所有内容，因为在调用process.run后，kivy卡住了。下面是一个小例子：

#! /usr/bin/env python 

""" 
Activate the touch keyboard. It is important that this part is on top 
because the global config should be initiated first. 
""" 
from kivy.config import Config 
Config.set('kivy', 'keyboard_mode', 'multi') 

# the main app 
from kivy.app import App 

# The Builder is used to define the main interface. 
from kivy.lang import Builder 

# The start screen is defined as BoxLayout 
from kivy.uix.boxlayout import BoxLayout 

# The pHBot class 
from phbot import pHBot 

# The pHBot is defined as process. Otherwise it would not be possible to use the close button. 
from multiprocessing import Process, Queue 



# Definition of the Main interface 
Builder.load_string(''' 
<Interface>: 
    orientation: 'vertical' 
    Button: 
     text: 'Run pH-Bot' 
     font_size: 40 
     on_release: app.run_worker() 
    Button: 
     text: 'Close pH-Bot' 
     font_size: 40 
     on_release: app.stop_phbot() 
''') 


# Interface as a subclass of BoxLayout without any further changes. This part is used by kivy. 
class Interface(BoxLayout): 
    pass 


class SettingsApp(App): 
    """ 
    The settings App is the main app of the pHBot application. It is initiated by kivy and contains the functions 
    defining the main interface. 
    """ 

    def build(self): 
     """ 
     This function initializes the app interface and has to be called "build(self)". It returns the user interface 
     defined by the Builder. 
     """ 

     # A queque for the control all processes. 
     self.qu_stop = Queue() 

     # returns the user interface defined by the Builder 
     return Interface() 

    def run_worker(self): 
     """ 
     The pHBot application is started as a second process. 
     """ 
     bot = pHBot(self.qu_stop) 
     phbot = Process(target=bot.ph_control()) 
     # start the process 
     phbot.run() 

    def stop_phbot(self): 
     self.qu_stop.put('STOP') 

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

第二类是一个叫phbot.py文件中：

import time 


class pHBot: 

    def __init__(self, qu_stop_in): 
     self.qu_stop_in = qu_stop_in 

    def ph_control(self): 

     while True: 
      if self.qu_stop_in.full(): 
       if self.qu_stop_in.get() == 'STOP': 
        break 
      print('Back and forth forever ...') 
      time.sleep(2) 

缺少什么我在这里？

Answer 1

请注意，Process以start()开头。调用run()确实会立即从同一进程启动该工作，并因此阻塞。因此run_worker相关的行应该是：

bot = pHBot(self.qu_stop) 
    phbot = Process(target=bot.ph_control) 
    # start the process 
    phbot.start() 

此外，在你的工人，不检查Queue是否已满。相反，做一个非阻塞get和处理Queue.Empty例外：

import Queue 
... 
    def ph_control(self): 

     while True: 
      try: 
       item = self.qu_stop_in.get(False) 
       if item == 'STOP': 
        break 
      except Queue.Empty: 
       print "Nothing to see" 
      print('Back and forth forever ...') 
      time.sleep(2) 

Answer 2

phbot = Process(target=bot.ph_control()) 

您调用 bot.ph_control - 这是()语法做什么。我猜你需要通过函数本身（Process(target=bot.ph_control))），但我不熟悉多处理。