How to get Django AutoFields to start at a higher number

你。 提交于 2019-11-27 18:50:02

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.

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.

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()

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.

William Keller

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" — djangoproject.com

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.

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