【发布时间】:2015-01-13 13:53:20
【问题描述】:
我正在尝试从 Kivy 应用程序中运行 django 开发服务器。到目前为止,这确实很有效。
现在我想允许用户在服务器运行时继续使用该程序。我的想法是为 httpd.serve_forever() 创建一个 multiprocessing.Process 以避免主程序完全锁定。做得很好。这是我的 internal_django 模块中的代码:
import multiprocessing
import os
import time
from wsgiref.simple_server import make_server
def django_wsgi_application():
PROJECT_ROOT = os.path.dirname(os.path.abspath(__file__))
settings_module = "djangosettings"#%s.djangosettings" % PROJECT_ROOT.split(os.sep)[-1]
os.environ.update({"DJANGO_SETTINGS_MODULE":settings_module})
from django.core.wsgi import get_wsgi_application
application = get_wsgi_application()
return application
class Singleton(type):
_instances = {}
def __call__(cls, *args, **kwargs):
if cls not in cls._instances:
cls._instances[cls] = super(Singleton, cls).__call__(*args, **kwargs)
return cls._instances[cls]
class DjangoServer():
__metaclass__ = Singleton
def start(self):
self.httpd = make_server('', 8000, django_wsgi_application())
self.server = multiprocessing.Process(target=self.httpd.serve_forever)
self.server.start()
print "Now serving on port 8000..."
print "Server Process PID = %s" %self.server.pid
def stop(self):
print("shutdown initiated")
print "Server Process PID = %s" %self.server.pid
while self.server.is_alive():
self.server.terminate()
print("Server should have shut down")
time.sleep(1)
print("Server is_alive: %s" %self.server.is_alive())
self.server.join()
print("server process joined")
if __name__ == "__main__":
server = DjangoServer()
server.start()
time.sleep(3)
server.stop()
当我运行这段代码时,一切都按预期工作。这是控制台中显示的内容:
Now serving on port 8000...
Server Process PID = 1406
shutdown initiated
Server Process PID = 1406
Server should have shut down
Server is_alive: False
server process joined
下一步是提供一种从 Kivy 应用程序中停止服务器的方法。为此,我只想像以前一样使用我的 DjangoServer 类:
from internal_django import DjangoServer
class StartScreen(Screen):
def start_server(self):
server = DjangoServer()
server.start()
class StopScreen(Screen):
def stop_server(self):
server = DjangoServer()
server.stop()
但是这样做时,进程一旦开始就永远不会退出。我的第一个想法是 Singleton 没有按预期工作,我试图退出错误的过程。但正如您在输出中看到的,PID 是相同的。服务器收到终止命令,但只是继续工作。这是控制台的样子:
Now serving on port 8000...
Server Process PID = 1406
shutdown initiated
Server Process PID = 1406
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
(and so on, until i manually kill the server process)
我是否以完全错误的方式使用多处理? Kivy 是否以某种方式干扰了该过程?
【问题讨论】:
-
多久启动/停止一次服务器?如果您只启动一次,并在退出 Kivy 应用程序时停止,那么只需将服务器设置为守护进程,不必担心停止它。在调用
self.server.start()之前执行此操作:self.server.daemon = True -
据我所知,这将使服务器进程保持活动状态,这将导致在重新启动应用程序(或任何其他试图在相关端口上提供服务的本地服务)时阻塞连接端口。不是我想要的。每次单击相应的按钮时,都应通过用户交互来启动/停止服务器。
标签: python django multiprocessing kivy