How to get Django AutoFields to start at a higher number

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

from django.db.models.signals import post_syncdb
from project.app 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
                            """)
    transaction.commit_unless_managed()

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.

Answer 4

Here is what I did..

def update_auto_increment(value=5000, app_label="remrate_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)
        cursor.execute(alter_str)
        transaction.commit_unless_managed()

Answer 5

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 views.py, 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.

Answer 6

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:

1.) models.py

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: views.py

from app.models import Product

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

Products.objects.create(**product)

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

models.py

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...