如何在不阻止我的代码的情况下启动事件驱动的WSGI服务器

Question

从python-socketio的此片段开始

import socketio
import eventlet
from flask import Flask, render_template

sio = socketio.Server()
app = Flask(__name__)

@app.route('/')
def index():
    """Serve the client-side application."""
    return render_template('index.html')

@sio.on('connect')
def connect(sid, environ):
    print('connect ', sid)

@sio.on('my message')
def message(sid, data):
    print('message ', data)

@sio.on('disconnect')
def disconnect(sid):
    print('disconnect ', sid)

if __name__ == '__main__':
    # wrap Flask application with socketio's middleware
    app = socketio.Middleware(sio, app)

    # deploy as an eventlet WSGI server
    eventlet.wsgi.server(eventlet.listen(('', 8000)), app)     <====

代码的最后一行在那儿阻止了我的代码，但我想将其作为守护程序运行并继续执行更多逻辑。那有可能吗？

我已经阅读了这篇文章，并尝试threading按建议使用，但我担心这样做不正确。

wst = threading.Thread(target=eventlet.wsgi.server(eventlet.listen(('', 7000)), app))
wst.daemon = True
wst.start()

有指针吗？

更新

我试过了

    # wrap Flask application with socketio's middleware
    app = socketio.Middleware(sio, app)

    # deploy as an eventlet WSGI server
    # eventlet.wsgi.server(eventlet.listen(('', 7000)), app)

    wst = threading.Thread(target=self.serve_app, args=(app))
    wst.daemon = True
    wst.start()

def serve_app(app):
    eventlet.wsgi.server(eventlet.listen(('', 7000)), app)

但

@4000000056a6bbe61979723c Exception in thread Thread-1:
@4000000056a6bbe619797df4 Traceback (most recent call last):
@4000000056a6bbe6197981dc   File "/usr/lib/python2.7/threading.py", line 810, in __bootstrap_inner
@4000000056a6bbe6197985c4     self.run()
@4000000056a6bbe6197989ac   File "/usr/lib/python2.7/threading.py", line 763, in run
@4000000056a6bbe619798d94     self.__target(*self.__args, **self.__kwargs)
@4000000056a6bbe61979917c TypeError: serve_app() argument after * must be a sequence, not Middleware
@4000000056a6bbe6197a164c 

什么是正确的方法？

更新2

更改了一点代码

    wst = threading.Thread(target=self.serve_app, args=(sio))
    wst.daemon = True
    wst.start()

def serve_app（sio）：app = socketio.Middleware（sio，app）eventlet.wsgi.server（eventlet.listen（（''，7000）），app）

如建议通过一个元组，(app,)但现在出现此错误

@4000000056a7543717fb2324   File "/usr/lib/python2.7/threading.py", line 810, in __bootstrap_inner
@4000000056a7543717fb270c     self.run()
@4000000056a7543717fb270c   File "/usr/lib/python2.7/threading.py", line 763, in run
@4000000056a7543717fb2af4     self.__target(*self.__args, **self.__kwargs)
@4000000056a7543717fb2edc TypeError: serve_app() takes exactly 1 argument (2 given)

这看起来像我缺少Python语法

更新3-已解决

我忘记了 (self,

       wst = threading.Thread(target=self.serve_app, args=(sio,app))
        wst.daemon = True
        wst.start()

def serve_app(self, _sio, _app):
    app = socketio.Middleware(_sio, _app)
    eventlet.wsgi.server(eventlet.listen(('', 7000)), app)