I am trying to deploy a simple web app that I have built using Python and Flask.
My app has the following structure:
/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 For this purpose of deploying, I am following this link: http://vladikk.com/2013/09/12/serving-flask-with-nginx-on-ubuntu/
In this tutorial, the Flask app consists of only a hello.py file. The way he configures his uwsgi file is like this (/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 I have tried to apply the same logic to my uwsgi.ini file, but I am doing something wrong. This is how my file looks like:
[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 When I am developing my app in my local machine, I run this command to start de the server: ./run.py.
This is my run.py file:
#!flask/bin/python from app import app app.run(debug = False) Now, my question is: how should my uwsgi.ini file look like given that my Flask app consists of more than a single file?
