Flask: How to manage different environment databases?

Question

I am working on an app which looks similar to

facebook/
         __init__.py
         feed/
             __init__.py
             business.py
             vi         


        

Answer 1

Solution I use:

#__init__.py
app = Flask(__name__)
app.config.from_object('settings')
app.config.from_envvar('MYCOOLAPP_CONFIG',silent=True)

On the same level from which application loads:

#settings.py
SERVER_NAME="dev.app.com"
DEBUG=True
SECRET_KEY='xxxxxxxxxx'


#settings_production.py
SERVER_NAME="app.com"
DEBUG=False

So. If Environment Variable MYCOOLAPP_CONFIG does not exist -> only settings.py will load, which refers to default settings (development server as for me)
This is the reason for "silent=True", second config file not required, while settings.py default for development and with default values for common config keys

If any other settings_file will be loaded in addition to first one values inside it overrides values in original one. (in my example DEBUG and SERVER_NAME will be overrided, while SECRET_KEY stays same for all servers)

The only thing you should discover for yourself depends on the way how you launch your application
Before launching ENVVAR MYCOOLAPP_CONFIG should be set
For example I run with supervisor daemon and on production server I just put this in supervisor config file:

environment=MYCOOLAPP_CONFIG="/home/tigra/mycoolapp/settings_production.py"

With this way you can easily manage all your configuration files, moreover, with this way you can exclude this files from git or any other version control utility

default Linux way is this one in console before launching:
export MYCOOLAPP_CONFIG="/home/tigra/mycoolapp/settings_production.py"