Easiest way to rename a model using Django/South?

Question

I\'ve been hunting for an answer to this on South\'s site, Google, and SO, but couldn\'t find a simple way to do this.

I want to rename a Django model using South.

Answer 1

Make the changes in models.py and then run

./manage.py schemamigration --auto myapp

When you inspect the migration file, you'll see that it deletes a table and creates a new one

class Migration(SchemaMigration):

    def forwards(self, orm):
        # Deleting model 'Foo'                                                                                                                      
        db.delete_table('myapp_foo')

        # Adding model 'Bar'                                                                                                                        
        db.create_table('myapp_bar', (
        ...
        ))
        db.send_create_signal('myapp', ['Bar'])

    def backwards(self, orm):
        ...

This is not quite what you want. Instead, edit the migration so that it looks like:

class Migration(SchemaMigration):

    def forwards(self, orm):
        # Renaming model from 'Foo' to 'Bar'                                                                                                                      
        db.rename_table('myapp_foo', 'myapp_bar')                                                                                                                        
        if not db.dry_run:
            orm['contenttypes.contenttype'].objects.filter(
                app_label='myapp', model='foo').update(model='bar')

    def backwards(self, orm):
        # Renaming model from 'Bar' to 'Foo'                                                                                                                      
        db.rename_table('myapp_bar', 'myapp_foo')                                                                                                                        
        if not db.dry_run:
            orm['contenttypes.contenttype'].objects.filter(app_label='myapp', model='bar').update(model='foo')

In the absence of the update statement, the db.send_create_signal call will create a new ContentType with the new model name. But it's better to just update the ContentType you already have in case there are database objects pointing to it (e.g., via a GenericForeignKey).

Also, if you've renamed some columns which are foreign keys to the renamed model, don't forget to

db.rename_column(myapp_model, foo_id, bar_id)