Django 1.9 drop foreign key in migration
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
)
]
See SeparateDatabaseAndState. It allows you to specify a Django (state) part of the migration separately from the database part of the migration.
- Amend the field in your models file.
Create the migration, as normal. You will end up with something like:
class Migration(migrations.Migration): dependencies = [ ('my_app', '0001_whatever.py'), ] operations = [ migrations.AlterField( model_name='example', name='something', field=models.CharField(max_length=255, null=True)), ), ]
Now manually amend this to:
class Migration(migrations.Migration): dependencies = [ ('my_app', '0001_whatever.py'), ] state_operations = [ migrations.AlterField( model_name='example', name='something', field=models.CharField(max_length=255, null=True)), ), ] operations = [ migrations.SeparateDatabaseAndState(state_operations=state_operations) ]
Note that you are not specifying any database_operations
argument, so the Django relationships are amended, but the database data is unchanged.
Needless to say: take a backup before you try this.