Django 1.9 drop foreign key in migration

Question

I have a Django model that has a foreign key to another model:

class Example(models.Model)
   something = models.ForeignKey(SomeModel, db_index=True)

I want to keep the underlying DB column as a field, but to get rid of the foreign key constraint in the database.

So the model will change to:

class Example(models.Model):
   something_id = models.IntegerField() 

And, to be clear, something_id is the column that Django had created for the foreign key field.

I do not want to drop the column and re-create it (this is what Django does when I auto-generate migrations after changing the model as above).

I want to keep the field but I want to remove the foreign key constraint in the database with a migration. It's not clear to me how to do this with a Django migration - is there some built in support for it or do I have to run some raw SQL and, if so, how do I programatically get the name of the constraint?

Answer 1

This is how I managed to do it, it's based on nimasmi's answer above:

class Migration(migrations.Migration):
    dependencies = [
        ('my_app', '0001_initial'),
    ]

    # These *WILL* impact the database!
    database_operations = [
        migrations.AlterField(
            model_name='Example',
            name='something',
            field=models.ForeignKey('Something', db_constraint=False, db_index=True, null=False)
        ),
    ]

    # These *WON'T* impact the database, they update Django state *ONLY*!
    state_operations = [
        migrations.AlterField(
            model_name='Example',
            name='something',
            field=models.IntegerField(db_index=True, null=False)
        ),
        migrations.RenameField(
            model_name='Example',
            old_name='something',
            new_name='something_id'
        ),
    ]

    operations = [
        migrations.SeparateDatabaseAndState(
            database_operations=database_operations,
            state_operations=state_operations
        )
    ]