我想部署一个我用Python和Flask构建的简单Web应用程序。
我的应用程序结构如下:
/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
为了部署目的,我正在按照这个链接进行操作:http://vladikk.com/2013/09/12/serving-flask-with-nginx-on-ubuntu/ 在这个教程中,Flask应用程序只包含一个hello.py文件。他配置uwsgi文件的方式如下(/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
我尝试将相同的逻辑应用到我的uwsgi.ini
文件上,但是我做错了什么。这是我的文件样子:
[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
当我在本地开发应用程序时,我运行以下命令启动服务器:./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:鉴于我的Flask应用程序不止一个文件,我的uwsgi.ini文件应该是什么样子的?