Access Apache SetEnv variable from Django wsgi.py file

Question

I\'m trying to separate out Django\'s secret key and DB pass into environmental variables, as widely suggested, so I can use identical code bases between local/production se

Answer 1

Here's an alternative solution that's as future-proof as get_wsgi_application. It even lets you set environment variables to use in your Django initialization.

# in wsgi.py

KEYS_TO_LOAD = [
    # A list of the keys you'd like to load from the WSGI environ
    # into os.environ
]

def loading_app(wsgi_environ, start_response):
    global real_app
    import os
    for key in KEYS_TO_LOAD:
        try:
            os.environ[key] = wsgi_environ[key]
        except KeyError:
            # The WSGI environment doesn't have the key
            pass
    from django.core.wsgi import get_wsgi_application
    real_app = get_wsgi_application()
    return real_app(wsgi_environ, start_response)

real_app = loading_app

application = lambda env, start: real_app(env, start)

I'm not 100% clear how mod_wsgi manages its processes, but I assume it doesn't re-load the WSGI app very often. If so, the performance penalty from initializing Django will only happen once, inside the first request.

Alternatively, if you don't need to set the environment variables before initializing Django, you can use the following :

# in wsgi.py

KEYS_TO_LOAD = [
    # A list of the keys you'd like to load from the WSGI environ
    # into os.environ
]

from django.core.wsgi import get_wsgi_application
django_app = get_wsgi_application()

def loading_app(wsgi_environ, start_response):
    global real_app
    import os
    for key in KEYS_TO_LOAD:
        try:
            os.environ[key] = wsgi_environ[key]
        except KeyError:
            # The WSGI environment doesn't have the key
            pass
    real_app = django_app
    return real_app(wsgi_environ, start_response)

real_app = loading_app

application = lambda env, start: real_app(env, start)