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Easiest way to rename a model using Django/South?

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?

See also stackoverflow.com/questions/3235995/… for renaming a model field rather than a model.
Optimized solution for Django >= 1.8 stackoverflow.com/questions/25091130/…

a
awidgery

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.


I was able to avoid the error message that Leopd by creating the schema migration using --auto instead of --empty. I then edited the migration file, changing the deletion/creation of the tables into a db.rename_table() call. This seems to have worked very well.
I used this technique on 9/2/2011 without getting any errors. Maybe a newer version of South solved the problem with the errors.
Thanks for keeping this updated! Jian's answer below says it is important to keep the "send_create_signal" calls, do you have any knowledge about that? If you agree, it would be great to update your example migration.
Beware that this will not rename indexes on that table. If you create a new table in the future with the same name as the old table, you can get errors from index names colliding. We were using this technique, but from now on we are going to explicitly create the new table, migrate the data, then delete the old table.
Column names in automatically generated tables, such as M2M tables to the original model, are also not migrated by this method.
J
Jian

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)

I get the error, KeyError: "The model 'contenttype' from the app 'contenttypes' is not available in this migration." Also, I have a django_content_type table, but not a contenttypes table. (Django 1.6)
@Seth I worked around this by doing the updating of the ContentType models in a separate data migration and by adding the contenttypes.ContentType model to the frozen models by using the --frozen flag to ./manage.py datamigration. For example: ./manage.py datamigration --frozen contenttypes myapp update_contenttypes. Then edit myapp_migrations/NNNN_update_contenttypes.py with the content type updating code as specified above.
@GeoffreyHing I think the parameter is freeze no frozen. south.readthedocs.io/en/latest/ormfreezing.html But thank you so much for your help, it was really helpful.
D
Dominic Rodger

South can't do it itself - how does it know that Bar represents what Foo used to? This is the sort of thing I'd write a custom migration for. You can change your ForeignKey in code as you've done above, and then it's just a case of renaming the appropriate fields and tables, which you can do any way you want.

Finally, do you really need to do this? I've yet to need to rename models - model names are just an implementation detail - particularly given the availability of the verbose_name Meta option.


Alternatively, rename the model in code but use the db_table Meta option to keep the database table name the same.
@Daniel - do you know if db_table is used to derive foreign key names?
i belive it is. If you change the model name and set db_table everything should still work as expected.
@DanielRoseman this is the best solution in the whole thread!
g
gowthaman

I followed Leopd's solution above. But, that did not change the model names. I changed it manually in the code (also in related models where this is referred as FK). And done another south migration, but with --fake option. This makes model names and table names to be same.

Just realized, one could first start with changing model names, then edit the migrations file before applying them. Much cleaner.