使用uwsgi和nginx部署Flask应用程序

我正在尝试部署一个使用Python和Flask构建的简单Web应用程序.

我的应用程序具有以下结构：

/var/www/watchgallery/
    + app
        + __init__.py
        + views.py
        + templates
    + flask #virtual environment for Flask
    + run.py #script I used in my machine to start the development Flask server
    + watchgallery_nginx.conf
    + watchgallery_uwsgi.ini
    + watchgallery_uwsgi.sock

为此,我遵循以下链接：http://vladikk.com/2013/09/12/serving-flask-with-nginx-on-ubuntu/

在本教程中,Flask应用程序仅包含hello.py文件.他配置uwsgi文件的方式如下所示(/var/www/demoapp/demoapp_uwsgi.ini)：

[uwsgi]
#application's base folder
base = /var/www/demoapp

#python module to import
app = hello
module = %(app)

home = %(base)/venv
pythonpath = %(base)

#socket file's location
socket = /var/www/demoapp/%n.sock

#permissions for the socket file
chmod-socket    = 666

#the variable that holds a flask application inside the module imported at line #6
callable = app

#location of log files
logto = /var/log/uwsgi/%n.log

我试图将相同的逻辑应用于uwsgi.ini文件,但是我做错了.这是我的文件的样子：

[uwsgi]
#application's base folder
base = /var/www/watchgallery

#python module to import
app = run
module = %(app)

home = %(base)/flask
pythonpath = %(base)

#socket file's location
socket = /var/www/watchgallery/%n.sock

#permissions for the socket file
chmod-socket    = 666

#the variable that holds a flask application inside the module imported at line #6
callable = app

当我在本地计算机上开发应用程序时,我运行以下命令来启动服务器：./run.py.

这是我的run.py文件：

#!flask/bin/python
from app import app
app.run(debug = False)

现在,我的问题是：考虑到我的Flask应用程序包含多个文件,我的uwsgi.ini文件应如何显示？