Python WSGIREF:在某种情况下杀死服务器

问题描述 投票:0回答:1

我有下面的代码:

from wsgiref import simple_server
import falcon

app = falcon.API(middleware=Auth())
msg = Something()
app.add_route('/hello', msg)     

httpd = simple_server.make_server(127.0.0.1, 8987, app)

m_process = Thread(target=httpd.serve_forever(poll_interval=0.5), name="m_process")
m_process.start()

s = Thread(target=httpd.shutdown(), name="s_process")
s.start()

Something()类的逻辑中,我正在基于某种逻辑递增或递减存储在数据库中的计数器。我想根据数据库上的条件调用s线程。因此,实际上,我正在尝试关闭服务器。我用http://code.nabla.net/doc/wsgiref/api/wsgiref/simple_server/wsgiref.simple_server.WSGIServer.html中的解释来做到这一点。但这似乎并没有杀死服务器。我在做什么错?

python rest falconframework wsgiref
1个回答
0
投票

好吧,我得到了我问题的答案。我在设置m_process的target时犯了一个愚蠢的错误。下面正在工作

from wsgiref import simple_server
import falcon

app = falcon.API(middleware=Auth())
msg = Something()
app.add_route('/hello', msg)     

httpd = simple_server.make_server(127.0.0.1, 8987, app)

m_process = Thread(target=httpd.serve_forever, name="m_process")
m_process.start()

s = Thread(target=httpd.shutdown(), name="s_process")
s.start()
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