Python – Easiest way to rename a model using Django/South

djangodjango-modelsdjango-southpythonrename

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.
Say you have the following:

class Foo(models.Model):
    name = models.CharField()

class FooTwo(models.Model):
    name = models.CharField()
    foo = models.ForeignKey(Foo)

and you want to convert Foo to Bar, namely

class Bar(models.Model):
    name = models.CharField()

class FooTwo(models.Model):
    name = models.CharField()
    foo = models.ForeignKey(Bar)

To keep it simple, I'm just trying to change the name from Foo to Bar, but ignore the foo member in FooTwo for now.

What's the easiest way to do this using South?

I could probably do a data migration, but that seems pretty involved.
Write a custom migration, e.g. db.rename_table('city_citystate', 'geo_citystate'), but I'm not sure how to fix the foreign key in this case.
An easier way that you know?

Best Answer

To answer your first question, the simple model/table rename is pretty straightforward. Run the command:

./manage.py schemamigration yourapp rename_foo_to_bar --empty

(Update 2: try --auto instead of --empty to avoid the warning below. Thanks to @KFB for the tip.)

If you're using an older version of south, you'll need startmigration instead of schemamigration.

Then manually edit the migration file to look like this:

class Migration(SchemaMigration):

    def forwards(self, orm):
        db.rename_table('yourapp_foo', 'yourapp_bar')


    def backwards(self, orm):
        db.rename_table('yourapp_bar','yourapp_foo')

You can accomplish this more simply using the db_table Meta option in your model class. But every time you do that, you increase the legacy weight of your codebase -- having class names differ from table names makes your code harder to understand and maintain. I fully support doing simple refactorings like this for the sake of clarity.

(update) I just tried this in production, and got a strange warning when I went to apply the migration. It said:

The following content types are stale and need to be deleted:

    yourapp | foo

Any objects related to these content types by a foreign key will also
be deleted. Are you sure you want to delete these content types?
If you're unsure, answer 'no'.

I answered "no" and everything seemed to be fine.

Command History

django-admin.py startproject southtest
manage.py startapp movies
manage.py startapp tv
manage.py syncdb
manage.py startmigration movies --initial
manage.py startmigration tv --initial
manage.py migrate
manage.py shell          # added some fake data...
manage.py startapp media
manage.py startmigration media --initial
manage.py migrate
# edited code, wrote new models, but left old ones intact
manage.py startmigration movies unified-videofile --auto
# create a new (blank) migration to hand-write data migration
manage.py startmigration movies videofile-to-movievideofile-data 
manage.py migrate
# edited code, wrote new models, but left old ones intact
manage.py startmigration tv unified-videofile --auto
# create a new (blank) migration to hand-write data migration
manage.py startmigration tv videofile-to-movievideofile-data
manage.py migrate
# removed old VideoFile model from apps
manage.py startmigration movies removed-videofile --auto
manage.py startmigration tv removed-videofile --auto
manage.py migrate

For space sake, and since the models invariably look the same in the end, I'm only going to demonstrate with 'movies' app.

movies/models.py

from django.db import models
from media.models import VideoFile as BaseVideoFile

# This model remains until the last migration, which deletes 
# it from the schema.  Note the name conflict with media.models
class VideoFile(models.Model):
    movie = models.ForeignKey(Movie, blank=True, null=True)
    name = models.CharField(max_length=1024, blank=True)
    size = models.IntegerField(blank=True, null=True)
    ctime = models.DateTimeField(blank=True, null=True)

class MovieVideoFile(BaseVideoFile):
    movie = models.ForeignKey(Movie, blank=True, null=True, related_name='shows')

movies/migrations/0002_unified-videofile.py (schema migration)

from south.db import db
from django.db import models
from movies.models import *

class Migration:

    def forwards(self, orm):

        # Adding model 'MovieVideoFile'
        db.create_table('movies_movievideofile', (
            ('videofile_ptr', orm['movies.movievideofile:videofile_ptr']),
            ('movie', orm['movies.movievideofile:movie']),
        ))
        db.send_create_signal('movies', ['MovieVideoFile'])

    def backwards(self, orm):

        # Deleting model 'MovieVideoFile'
        db.delete_table('movies_movievideofile')

movies/migration/0003_videofile-to-movievideofile-data.py (data migration)

from south.db import db
from django.db import models
from movies.models import *

class Migration:

    def forwards(self, orm):
        for movie in orm['movies.videofile'].objects.all():
            new_movie = orm.MovieVideoFile.objects.create(movie = movie.movie,)
            new_movie.videofile_ptr = orm['media.VideoFile'].objects.create()

            # videofile_ptr must be created first before values can be assigned
            new_movie.videofile_ptr.name = movie.name
            new_movie.videofile_ptr.size = movie.size
            new_movie.videofile_ptr.ctime = movie.ctime
            new_movie.videofile_ptr.save()

    def backwards(self, orm):
        print 'No Backwards'

South is awesome!

Ok standard disclaimer: You're dealing with live data. I've given you working code here, but please use the --db-dry-run to test your schema. Always make a backup before trying anything, and generally be careful.

COMPATIBILITY NOTICE

I'm going to keep my original message intact, but South has since changed the command manage.py startmigration into manage.py schemamigration.

Python – Django – How to rename a model field using South

You can use the db.rename_column function.

class Migration:

    def forwards(self, orm):
        # Rename 'name' field to 'full_name'
        db.rename_column('app_foo', 'name', 'full_name')




    def backwards(self, orm):
        # Rename 'full_name' field to 'name'
        db.rename_column('app_foo', 'full_name', 'name')

The first argument of db.rename_column is the table name, so it's important to remember how Django creates table names:

Django automatically derives the name of the database table from the name of your model class and the app that contains it. A model's database table name is constructed by joining the model's "app label" -- the name you used in manage.py startapp -- to the model's class name, with an underscore between them.

In the case where you have a multi-worded, camel-cased model name, such as ProjectItem, the table name will be app_projectitem (i.e., an underscore will not be inserted between project and item even though they are camel-cased).