Site matching query does not exist

Question

Python noob, as in this is my first project, so excuse my unfamiliarity.

The site was working very well until I clicked \"log out\" on my app. After that, the websit

Answer 1

it seems you fotgot to add SITE_ID=1 in settings.py

Answer 2

I fixed it without using python manage.py shell

At first I tried using the commands above using manage.py shell:

from django.contrib.sites.models import Site
new_site = Site.objects.create(domain='foo.com', name='foo.com')
print(new_site.id)

But it gave out an INTEGRITY ERROR

The short answer is add SITE_ID = 1 to your settings.py If you want to know what your site id is, then go into the actual database, I downloaded sqliteman to see what my table had. So whatever site id you have in the table is what gets assigned to SITE_ID

This is because there is a function get_current that goes looking for SITE_ID and it doesn't find it in your settings.py

tables
-django_site
--Columns
---id

it should have id as 1, name as example.com, domain as example.com

Answer 3

Table django_site must contain a row with the same value than id (by default equals to 1), as SITE_ID is set to (inside your settings.py).

Answer 4

I see answers to create a new site and reference id for that. But if you already have a site or somehow you deleted and created the site again from UI then the id keeps on incrementing. To solve this you can get the id as follows:

python manage.py shell
from django.contrib.sites.models import Site
print(Site.objects.get(name='example.com').id)

Get the id you get here into setting.py . Eg.

SITE_ID = 8

Basically ID in table corresponding yo tour site and in the settings.py should match.

Answer 5

Query the ID in your database django_site tables and set the right one in your Django settings.py, for example: SITE_ID = 3

Answer 6

Add SITE_ID = 1 to settings.py in your django project