Allow only one instance of a model in Django

Question

I would like to control some configuration settings for my project using a database model. For example:

class JuicerBaseSettings(models.Model):
    max_rpm = model.IntegerField(default=10)
    min_rpm = model.IntegerField(default=0)

There should only be one instance of this model:

juicer_base = JuicerBaseSettings()
juicer_base.save()

Of course, if someone accidentally creates a new instances, it's not the end of the world. I could just do JuicerBaseSettings.objects.all().first(). However, is there a way to lock it down such that it's impossible to create more than 1 instance?

I found two related questions on SO. This answer suggests using 3rd party apps like django-singletons, which doesn't seem to be actively maintained (last update to the git repo is 5 years ago). Another answer suggests using a combination of either permissions or OneToOneField. Both answers are from 2010-2011.

Given that Django has changed a lot since then, are there any standard ways to solve this problem? Or should I just use .first() and accept that there may be duplicates?

Answer 1

You can override save method to control number of instances:

class JuicerBaseSettings(models.Model):

    def save(self, *args, **kwargs):
        if not self.pk and JuicerBaseSettings.objects.exists():
        # if you'll not check for self.pk 
        # then error will also raised in update of exists model
            raise ValidationError('There is can be only one JuicerBaseSettings instance')
        return super(JuicerBaseSettings, self).save(*args, **kwargs)

Answer 2

Either you can override save and create a class function JuicerBaseSettings.object()

class JuicerBaseSettings(models.Model):

    @classmethod
    def object(cls):
        return cls._default_manager.all().first() # Since only one item

    def save(self, *args, **kwargs):
        self.pk = self.id = 1
        return super().save(*args, **kwargs)

============= OR =============

Simply, Use django_solo.

https://github.com/lazybird/django-solo

Snippet Courtsy: django-solo-documentation.

# models.py


from django.db import models
from solo.models import SingletonModel

class SiteConfiguration(SingletonModel):
    site_name = models.CharField(max_length=255, default='Site Name')
    maintenance_mode = models.BooleanField(default=False)

    def __unicode__(self):
        return u"Site Configuration"

    class Meta:
        verbose_name = "Site Configuration"

# admin.py

from django.contrib import admin
from solo.admin import SingletonModelAdmin
from config.models import SiteConfiguration

admin.site.register(SiteConfiguration, SingletonModelAdmin)

# There is only one item in the table, you can get it this way:
from .models import SiteConfiguration
config = SiteConfiguration.objects.get()

# get_solo will create the item if it does not already exist
config = SiteConfiguration.get_solo()

Answer 3

If your model is used in django-admin only, you additionally can set dynamic add permission for your model:

# some imports here
from django.contrib import admin
from myapp import models

@admin.register(models.ExampleModel)
class ExampleModelAdmin(admin.ModelAdmin):
    
    # some code...

    def has_add_permission(self, request):
        # check if generally has add permission
        retVal = super().has_add_permission(request)
        # set add permission to False, if object already exists
        if retVal and models.ExampleModel.objects.exists():
            retVal = False
        return retVal