Ytukyg

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?

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)

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

http://blog.dscpl.com.au/2010/03/improved-wsgi-script-for-use-with.html

Add what you need after language translations are activated.

Thus:

import sys



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



import settings



import django.core.management

django.core.management.setup_environ(settings)

utility = django.core.management.ManagementUtility()

command = utility.fetch_command('runserver')



command.validate()



import django.conf

import django.utils



django.utils.translation.activate(django.conf.settings.LANGUAGE_CODE)



# Your line here.

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



import django.core.handlers.wsgi



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

You can create a custom command and write your code in the handle function. details here https://docs.djangoproject.com/en/dev/howto/custom-management-commands/

Then you can create a startup script that runs the django server then executes your new custom command.

If you are using mod_wsgi you can put it in the wsgi start app

Here is how I work around the missing startup signal for Django:
https://github.com/lsaffre/djangosite/blob/master/djangosite/models.py
The code that is being called there is specific to my djangosite project, but the trick to get it called by writing a special app (based on an idea by Ross McFarland) should work for other environments.
Luc

In Django 1.7+ if you want to run a startup code and,

1. Avoid running it in migrate, makemigrations, shell sessions, ...

2. Avoid running it twice or more

A solution would be:

file: myapp/apps.py

from django.apps import AppConfig



def startup():

    # startup code goes here



class MyAppConfig(AppConfig):

    name = 'myapp'

    verbose_name = "My Application"

    def ready(self):

        import os

        if os.environ.get('RUN_MAIN'):

            startup()

file: myapp/__init__.py

default_app_config = 'myapp.apps.MyAppConfig'

This post is using suggestions from @Pykler and @bdoering