How to get Django AutoFields to start at a higher number

For our Django App, we'd like to get an AutoField to start at a number other than 1. There doesn't seem to be an obvious way to do this. Any ideas?

Asked by: Roman575 | Posted: 28-01-2022

Answer 1

Like the others have said, this would be much easier to do on the database side than the Django side.

For Postgres, it'd be like so: ALTER SEQUENCE sequence_name RESTART WITH 12345; Look at your own DB engine's docs for how you'd do it there.

Answered by: Chester684 | Posted: 01-03-2022

Answer 2

For MySQL i created a signal that does this after syncdb:

from django.db.models.signals import post_syncdb
from import models as app_models

def auto_increment_start(sender, **kwargs):
    from django.db import connection, transaction
    cursor = connection.cursor()
    cursor = cursor.execute("""
                                ALTER table app_table AUTO_INCREMENT=2000

post_syncdb.connect(auto_increment_start, sender=app_models)

After a syncdb the alter table statement is executed. This will exempt you from having to login into mysql and issuing it manually.

EDIT: I know this is an old thread, but I thought it might help someone.

Answered by: Sawyer593 | Posted: 01-03-2022

Answer 3

A quick peek at the source shows that there doesn't seem to be any option for this, probably because it doesn't always increment by one; it picks the next available key: "An IntegerField that automatically increments according to available IDs" —

Answered by: Carina896 | Posted: 01-03-2022

Answer 4

Here is what I did..

def update_auto_increment(value=5000, app_label="xxx_data"):
    """Update our increments"""
    from django.db import connection, transaction, router
    models = [m for m in get_models() if m._meta.app_label == app_label]
    cursor = connection.cursor()
    for model in models:
        _router = settings.DATABASES[router.db_for_write(model)]['NAME']
        alter_str = "ALTER table {}.{} AUTO_INCREMENT={}".format(
            _router, model._meta.db_table, value)

Answered by: Carina605 | Posted: 01-03-2022

Answer 5

I found a really easy solution to this! AutoField uses the previous value used to determine what the next value assigned will be. So I found that if I inserted a dummy value with the start AutoField value that I want, then following insertions will increment from that value.

A simple example in a few steps:


class Product(models.Model):
    id = model.AutoField(primaryKey=True) # this is a dummy PK for now
    productID = models.IntegerField(default=0)
    productName = models.TextField()
    price = models.DecimalField(max_digits=6, decimal_places=2)
  • makemigrations
  • migrate

Once that is done, you will need to insert the initial row where "productID" holds a value of your desired AutoField start value. You can write a method or do it from django shell.

From view the insertion could look like this:

from app.models import Product

dummy = {
   'productID': 100000,
   'productName': 'Item name',
   'price': 5.98,


Once inserted you can make the following change to your model:

class Product(models.Model):
    productID = models.AutoField(primary_key=True)
    productName = models.TextField()
    price = models.DecimalField(max_digits=6, decimal_places=2)

All following insertions will get a "productID" incrementing starting at 100000...100001...100002...

Answered by: Vanessa643 | Posted: 01-03-2022

Answer 6

The auto fields depend, to an extent, on the database driver being used.

You'll have to look at the objects actually created for the specific database to see what's happening.

Answered by: Patrick293 | Posted: 01-03-2022

Answer 7

I needed to do something similar. I avoided the complex stuff and simply created two fields:

id_no = models.AutoField(unique=True)
my_highvalue_id = models.IntegerField(null=True)

In, I then simply added a fixed number to the id_no:

my_highvalue_id = id_no + 1200

I'm not sure if it helps resolve your issue, but I think you may find it an easy go-around.

Answered by: Anna257 | Posted: 01-03-2022

Answer 8

In the model you can add this:

def save(self, *args, **kwargs):
       if not User.objects.count():
 = 100
 = User.objects.last().id + 1
       super(User, self).save(*args, **kwargs)

This works only if the DataBase is currently empty (no objects), so the first item will be assigned id 100 (if no previous objects exist) and next inserts will follow the last id + 1

Answered by: Kellan187 | Posted: 01-03-2022

Similar questions

Still can't find your answer? Check out these communities...

PySlackers | Full Stack Python | NHS Python | Pythonist Cafe | Hacker Earth | Discord Python