Where to put Django startup code?


I'd like to have these lines of code executed on server startup (both development and production):

from django.core import management
management.call_command('syncdb', interactive=False)

Putting it in settings.py doesn't work, as it requires the settings to be loaded already.

Putting them in a view and accessing that view externally doesn't work either, as there are some middlewares that use the database and those will fail and not let me access the view.

Putting them in a middleware would work, but that would get called each time my app is accessed. An possible solution might be to create a middleware that does all the job and then removes itself from MIDDLEWARE_CLASSES so it's not called anymore. Can I do that without too much monkey-patching?

5/6/2010 1:18:46 PM

Accepted Answer

Write middleware that does this in __init__ and afterwards raise django.core.exceptions.MiddlewareNotUsed from the __init__, django will remove it for all requests :). __init__ is called at startup by the way, not at the first request, so it won't block your first user.

There is talk about adding a startup signal, but that won't be available soon (a major problem for example is when this signal should be sent)

Related Ticket: https://code.djangoproject.com/ticket/13024

Update: Django 1.7 includes support for this. (Documentation, as linked by the ticket)

6/25/2014 3:09:46 AM

If you were using Apache/mod_wsgi for both, use the WSGI script file described in:


Add what you need after language translations are activated.


import sys

sys.path.insert(0, '/usr/local/django/mysite')

import settings

import django.core.management
utility = django.core.management.ManagementUtility()
command = utility.fetch_command('runserver')


import django.conf
import django.utils


# Your line here.
django.core.management.call_command('syncdb', interactive=False)

import django.core.handlers.wsgi

application = django.core.handlers.wsgi.WSGIHandler()

Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow