Hello, I am still new to Django, so I apologize if this seems like a silly question. I've not been able to find what I need on Google, StackOverflow, or the Django Docs.
I am using Django 2.0.5. I understand that the correct way to run startup code is to subclass AppConfig and override ready(). I also understand that Django launches two separate instances of the application--one to check the models, and the other to launch the server--on separate processes. The official recommendation to prevent startup code from being run more than once is to implement a flag--which can't work if there are multiple instances of the application being created. The docs also say that ready() will be re-called only rarely; but this doesn't help at all if all of the instances are calling ready() only once. I have found the manage.py runserver --noreload command, but this won't be enough to prevent multiple instances on a production server. What can I do to enforce one and only one instance of my application being run at a time? If it helps, here is the reason I need to override this multi-instantiation behavior: my application launches a multiprocessing.Process at startup to monitor and run background tasks. Having more than one background Process running at once is going to wreak havoc on the application. I've looked at other options to accomplish similar purposes, but those would all be instantiated multiple times, also. Any suggestions? Thank you, Heather -- You received this message because you are subscribed to the Google Groups "Django users" group. To unsubscribe from this group and stop receiving emails from it, send an email to django-users+unsubscr...@googlegroups.com. To post to this group, send email to django-users@googlegroups.com. Visit this group at https://groups.google.com/group/django-users. To view this discussion on the web visit https://groups.google.com/d/msgid/django-users/6d8729fe-18c4-4e53-8479-c2b9dd178b89%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
