[Django]-Django 'Sites' Model – what is and why is 'SITE_ID = 1'?

Things would be much easier to understand if Django’s default SiteAdmin included the id field in the list_display fields.

To do this, you can redefine SiteAdmin (anywhere in your app, but I’d recommend your admin.py or maybe your urls.py) like this:

from django.contrib import admin
from django.contrib.sites.models import Site

admin.site.unregister(Site)
class SiteAdmin(admin.ModelAdmin):
    fields = ('id', 'name', 'domain')
    readonly_fields = ('id',)
    list_display = ('id', 'name', 'domain')
    list_display_links = ('name',)
    search_fields = ('name', 'domain')
admin.site.register(Site, SiteAdmin)

After including this code snippet, the ID for each “Site” will be shown in the first column of the admin list and inside the form as a read only field. These ‘id’ fields are what you need to use as SITE_ID:

The concept is that each different site runs in a different application server instance, launched using a different yourdomain_settings.py that then includes a base_settings.py with the rest of the common configuration.

Each of these yourdomain_settings.py will define its own SITE_ID and all other different settings.py parameters that they need to look and be different from each other (static resources, templates, etc.) then you’ll define a DJANGO_SETTINGS_MODULE environment variable pointing to that specific yourdomain_settings.py file when launching the application server instance for that domain.

A further note: get_current_site(request) does need request to be available for it to work. If your code doesn’t have one, you can use Site.objects.get_current() that however will need a SITE_ID properly defined in the running application server’s settings.

This is a late answer but for anyone else having SITE_ID issues and Site problems.
Inside the database, django has a django_site table with(id, domain, name). This is where django stores the SITE_IDs. Mine was actually 5 in the database but i had it set to SITE_ID=1 in the settings.

Knowing that, i can now go back to the database and clear it to get back to zero or use the actual id in the database.

This is covered in the documentation for the Sites framework:

In order to serve different sites in production, you’d create a
separate settings file with each SITE_ID (perhaps importing from a
common settings file to avoid duplicating shared settings) and then
specify the appropriate DJANGO_SETTINGS_MODULE for each site.

But if you didn’t want to do it that way, you can not set SITE_ID at all and just look up the current site based on the domain name in your views using get_current_site:

from django.contrib.sites.shortcuts import get_current_site

def my_view(request):
    current_site = get_current_site(request)
    if current_site.domain == 'foo.com':
        # Do something
        pass
    else:
        # Do something else.
        pass

This link explains it:

You’ll want to create separate settings files for each domain you’re adding; each one will need its own MEDIA_URL and other settings. You’ll also want to do two things to make sure everything works out properly for administering the different sites:

Create a new Site object in your admin for each domain, and put the id of that Site into its settings file as SITE_ID so Django knows which site in the database corresponds to this settings file.

In the settings file for your original site (the one with id 1), add the other sites’ settings files to the ADMIN_FOR setting, to let Django know that this one instance of the admin application will handle all of the sites.

Also, if you wanna figure out how to modify models and set the views You may take a look at this link:
https://django.cowhite.com/blog/managing-multiple-websites-with-a-common-database-in-django-the-sites-framework/

"Sites" was initially intended for multiple servers sharing a DB, but was extended with feature #15089 around version 1.4 to support basic multitenancy.

Nowadays you can simply not define SITE_ID if you always use methods that take a request.