Integrity Error: using primary_key=True - django

Question

This is my Models:
from django.db import models
from django.contrib.auth.models import User
class Bookmark(models.Model):
id = models.PositiveIntegerField(primary_key=True)
name = models.CharField(max_length=50)
url = models.URLField()
isPublic = models.BooleanField(default=True)
dateTimePosted = models.DateTimeField(auto_now=True,verbose_name="posted at")
user = models.ForeignKey(User, on_delete=models.CASCADE)
This is my 0001_initials.py:
from __future__ import unicode_literals
from django.conf import settings
from django.db import migrations, models
import django.db.models.deletion
class Migration(migrations.Migration):
initial = True
dependencies = [
migrations.swappable_dependency(settings.AUTH_USER_MODEL),
]
operations = [
migrations.CreateModel(
name='Bookmark',
fields=[
('id', models.PositiveIntegerField(primary_key=True, serialize=False)),
('name', models.CharField(max_length=50)),
('url', models.URLField()),
('isPublic', models.BooleanField(default=True)),
('dateTimePosted', models.DateTimeField(auto_now=True, verbose_name='posted at')),
('user', models.ForeignKey(on_delete=django.db.models.deletion.CASCADE, to=settings.AUTH_USER_MODEL)),
],
),
]
When I am inserting values using:
python manage.py shell:
>>from django.contrib.auth.models import User
>>user = User.objects.create_user('test', 'test#abc.com', 'test#123')
>>from bookmark.models import Bookmark
>>bookmark = Bookmark.objects.create(name='First Bookmark', url='example.com', user=user)
this error occurs:
Traceback (most recent call last):
File "", line 1, in
File "C:\Python34\lib\site-packages\django\db\models\manager.py", line 85, in manager_method
return getattr(self.get_queryset(), name)(*args, **kwargs)
File "C:\Python34\lib\site-packages\django\db\models\query.py", line 394, in create
obj.save(force_insert=True, using=self.db)
File "C:\Python34\lib\site-packages\django\db\models\base.py", line 806, in save
force_update=force_update, update_fields=update_fields)
File "C:\Python34\lib\site-packages\django\db\models\base.py", line 836, in save_base
updated = self._save_table(raw, cls, force_insert, force_update, using, update_fields)
File "C:\Python34\lib\site-packages\django\db\models\base.py", line 922, in _save_table
result = self._do_insert(cls._base_manager, using, fields, update_pk, raw)
File "C:\Python34\lib\site-packages\django\db\models\base.py", line 961, in _do_insert
using=using, raw=raw)
File "C:\Python34\lib\site-packages\django\db\models\manager.py", line 85, in manager_method
return getattr(self.get_queryset(), name)(*args, **kwargs)
File "C:\Python34\lib\site-packages\django\db\models\query.py", line 1063, in _insert
return query.get_compiler(using=using).execute_sql(return_id)
File "C:\Python34\lib\site-packages\django\db\models\sql\compiler.py", line 1099, in execute_sql
cursor.execute(sql, params)
File "C:\Python34\lib\site-packages\django\db\backends\utils.py", line 80, in execute
return super(CursorDebugWrapper, self).execute(sql, params)
File "C:\Python34\lib\site-packages\django\db\backends\utils.py", line 65, in execute
return self.cursor.execute(sql, params)
File "C:\Python34\lib\site-packages\django\db\utils.py", line 94, in exit
six.reraise(dj_exc_type, dj_exc_value, traceback)
File "C:\Python34\lib\site-packages\django\utils\six.py", line 685, in reraise
raise value.with_traceback(tb)
File "C:\Python34\lib\site-packages\django\db\backends\utils.py", line 65, in execute
return self.cursor.execute(sql, params)
File "C:\Python34\lib\site-packages\django\db\backends\sqlite3\base.py", line 328, in execute
return Database.Cursor.execute(self, query, params)
django.db.utils.IntegrityError: NOT NULL constraint failed: bookmark_bookmark.id
how to tackle this error
I have deleted database (db.sqlite3)and pycache folder unapplied migrations and then created again using:
python manage.py makemigrations
but still error occurs

Answer 1

You are not passing the primary key value (id) when you are creating your BookMark object.
Change your create as follows,
bookmark = Bookmark.objects.create(name='First Bookmark', url='example.com', user=user, id=<some_int_which_hasnt_been_used>) # Use an id that you know you hasn't been used.
Now the reason you have to do this is because you have defined your own Primary Key, so Django will not automatically create it for you.
When you do not specify a primary key, Django uses an "id" field which is of type "models.AutoField", and automatically increments the value (without you needing to specify the value for id).
In case you are interested in knowing how it works internally you can read up on Sequences in databases (what Django does internally is create a sequence in your database for the table, and your database has a default value for your id column)
I'm not sure why you have overridden the id field that Django provides automatically (because your id is PositiveIntegerField, and AutoField would generate only positive values anyway), so if there is no specific reason for you defining the PK, I suggest you go with what Django creates for you.