Execute code when Django starts ONCE only?

Question

Update: Django 1.7 now has a hook for this

file: myapp/apps.py

from django.apps import AppConfig
class MyAppConfig(AppConfig):
    name = 'myapp'
    verbose_name = "My Application"
    def ready(self):
        pass # startup code here

file: myapp/__init__.py

default_app_config = 'myapp.apps.MyAppConfig'

For Django < 1.7

The number one answer does not seem to work anymore, urls.py is loaded upon first request.

What has worked lately is to put the startup code in any one of your INSTALLED_APPS init.py e.g. myapp/__init__.py

def startup():
    pass # load a big thing

startup()

When using ./manage.py runserver ... this gets executed twice, but that is because runserver has some tricks to validate the models first etc ... normal deployments or even when runserver auto reloads, this is only executed once.

Update from Pykler's answer below: Django 1.7 now has a hook for this

---

Don't do it this way.

You don't want "middleware" for a one-time startup thing.

You want to execute code in the top-level urls.py. That module is imported and executed once.

urls.py

from django.confs.urls.defaults import *
from my_app import one_time_startup

urlpatterns = ...

one_time_startup()

This question is well-answered in the blog post Entry point hook for Django projects, which will work for Django >= 1.4.

Basically, you can use <project>/wsgi.py to do that, and it will be run only once, when the server starts, but not when you run commands or import a particular module.

import os
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "{{ project_name }}.settings")

# Run startup code!
....

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

As suggested by @Pykler, in Django 1.7+ you should use the hook explained in his answer, but if you want that your function is called only when run server is called (and not when making migrations, migrate, shell, etc. are called), and you want to avoid AppRegistryNotReady exceptions you have to do as follows:

file: myapp/apps.py

import sys
from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name = 'my_app'

    def ready(self):
        if 'runserver' not in sys.argv:
            return True
        # you must import your modules here 
        # to avoid AppRegistryNotReady exception 
        from .models import MyModel 
        # startup code here

If it helps someone, in addition to pykler's answer, "--noreload" option prevents runserver from executing command on startup twice:

python manage.py runserver --noreload

But that command won't reload runserver after other code's changes as well.

Note that you cannot reliability connect to the database or interact with models inside the AppConfig.ready function (see the warning in the docs).

If you need to interact with the database in your start-up code, one possibility is to use the connection_created signal to execute initialization code upon connection to the database.

from django.dispatch import receiver
from django.db.backends.signals import connection_created

@receiver(connection_created)
def my_receiver(connection, **kwargs):
    with connection.cursor() as cursor:
        # do something to the database

Obviously, this solution is for running code once per database connection, not once per project start. So you'll want a sensible value for the CONN_MAX_AGE setting so you aren't re-running the initialization code on every request. Also note that the development server ignores CONN_MAX_AGE, so you WILL run the code once per request in development.

99% of the time this is a bad idea - database initialization code should go in migrations - but there are some use cases where you can't avoid late initialization and the caveats above are acceptable.