CODEWITHSUNDEEP
pythondjango
Bob_94
Bob_94

Reputation: 2855

Execute code when Django starts ONCE only?

I'm writing a Django Middleware class that I want to execute only once at startup, to initialise some other arbritary code. I've followed the very nice solution posted by sdolan here, but the "Hello" message is output to the terminal twice. E.g.

from django.core.exceptions import MiddlewareNotUsed
from django.conf import settings

class StartupMiddleware(object):
    def __init__(self):
        print "Hello world"
        raise MiddlewareNotUsed('Startup complete')

and in my Django settings file, I've got the class included in the MIDDLEWARE_CLASSES list.

But when I run Django using runserver and request a page, I get in the terminal

Django version 1.3, using settings 'config.server'
Development server is running at http://127.0.0.1:8000/
Quit the server with CONTROL-C.
Hello world
[22/Jul/2011 15:54:36] "GET / HTTP/1.1" 200 698
Hello world
[22/Jul/2011 15:54:36] "GET /static/css/base.css HTTP/1.1" 200 0

Any ideas why "Hello world" is printed twice? Thanks.

Upvotes: 266

Views: 153149

Answers (11)

matthew strange
matthew strange

Reputation: 11

For those looking to only run their code once on startup in a production environment using gunicorn, you can make use of the the --preload command given by gunicorn.

By default Django uses multiple workers for any startup code, but by passing the --preload command Django will only run the startup command in the parent worker.

This is a good post explaining how to add the --preload command. Then just add your code to run on startup under the ready function like belo

from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name = 'app_name'

    def ready(self):
       code_to_run()

Upvotes: 1

J_Zar
J_Zar

Reputation: 2456

Standard solution

With Django 3.1+ you can write this code to execute only once a method at startup. The difference from the other questions is that the main starting process is checked (runserver by default starts 2 processes, one as an observer for quick code reload):

import os 
from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name = 'app_name'

    def ready(self):
        if os.environ.get('RUN_MAIN'):
            print("STARTUP AND EXECUTE HERE ONCE.")
            # call here your code

Another solution is avoiding the environ check but call --noreload to force only one process.

Alternative method

The first question to answer is why we need to execute code once: usually we need to initialize some services, data in the database or something one-shot. The 90% of the time it is some database initialization or job queue.

The approach to use the AppConfig.ready() method is not reliable, not always reproducible in production and it cannot guarantee to be executed exactly once (but at least once that is not the same). To have something quite predictable and executed exactly one time the best approach is developing a Django BaseCommand and call it from a startup script.

So for example, we can code in your "myapp", in the file "app/management/commands/init_tasks.py":

from django.core.management.base import BaseCommand
from project.apps.myapp.tasks import scheduler
from project import logger, initialize_database_data

class Command(BaseCommand):
    help = "Init scheduler or do some staff in the database."

    def handle(self, *args, **options):
        scheduler.reload_jobs()
        initialize_database_data()
        logger.info("Inited")

And finally we can have a start script "Start.bat" (in the example a windows batch) to setup the full application start:

start /b python manage.py qcluster
start /b python manage.py runserver 0.0.0.0:8000
start /b python manage.py init_tasks

Upvotes: 20

Fahad Alduraibi
Fahad Alduraibi

Reputation: 382

I used the accepted solution from here which checks if it was run as a server, and not when executing other managy.py commands such as migrate

apps.py:

from .tasks import tasks

class myAppConfig(AppConfig):
    ...

    def ready(self, *args, **kwargs):
        is_manage_py = any(arg.casefold().endswith("manage.py") for arg in sys.argv)
        is_runserver = any(arg.casefold() == "runserver" for arg in sys.argv)

        if (is_manage_py and is_runserver) or (not is_manage_py):
            tasks.is_running_as_server = True

And since that will still get executed twice when in development mode, without using the parameter --noreload, I added a flag to be triggered when it is running as a server and put my start up code in urls.py which is only called once.

tasks.py:

class tasks():
    is_running_as_server = False

    def runtask(msg):
        print(msg)

urls.py:

from . import tasks

task1 = tasks.tasks()

if task1.is_running_as_server:
    task1.runtask('This should print once and only when running as a server')

So to summarize, I am utilizing the read() function in AppConfig to read the arguments and know how the code was executed. But since in the development mode the ready() function is run twice, one for the server and one for the reloading of the server when the code changes, while urls.py is only executed once for the server. So in my solution i combined the two to run my task once and only when the code is executed as a server.

Upvotes: 0

nathan
nathan

Reputation: 1

In my case, I use Django to host a site, and using Heroku. I use 1 dyno (just like 1 container) at Heroku, and this dyno creates two workers. I want to run a discord bot on it. I tried all methods on this page and all of them are invalid.

Because it is a deployment, so it should not use manage.py. Instead, it uses gunicorn, which I don't know how to add --noreload parameter. Each worker runs wsgi.py once, so every code will be run twice. And the local env of two workers are the same.

But I notice one thing, every time Heroku deploys, it uses the same pid worker. So I just

if not sys.argv[1] in ["makemigrations", "migrate"]: # Prevent execute in some manage command
    if os.getpid() == 1: # You should check which pid Heroku will use and choose one.
        code_I_want_excute_once_only()

I'm not sure if the pid will change in the future, hope it will be the same forever. If you have a better method to check which worker is it, please tell me.

Upvotes: 0

Pykler
Pykler

Reputation: 14865

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.

Upvotes: 357

Alberto Pianon
Alberto Pianon

Reputation: 438

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

Upvotes: 33

Oscar
Oscar

Reputation: 41

if you want print "hello world" once time when you run server, put print ("hello world") out of class StartupMiddleware

from django.core.exceptions import MiddlewareNotUsed
from django.conf import settings

class StartupMiddleware(object):
    def __init__(self):
        #print "Hello world"
        raise MiddlewareNotUsed('Startup complete')

print "Hello world"

Upvotes: 2

RichardW
RichardW

Reputation: 969

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.

Upvotes: 15

S.Lott
S.Lott

Reputation: 391992

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()

Upvotes: 148

AnaPana
AnaPana

Reputation: 2058

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.

Upvotes: 23

augustomen
augustomen

Reputation: 9759

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()

Upvotes: 47

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