Django - manage.py sql APPNAME not generating model SQL

Question

I have a relatively large flat application that I'm working on. To maintain separation of concerns, I've split the model and view files into auth_models, dashboard_models, taxonomy_models and more. These have been placed in the folder structure as

APPNAME/
  app/
    models/ 
      __init__.py
      auth_models.py
      dashboard_models.py
      taxonomy_models.py
      ...
    views/
      __init__.py
      dashboard_views.py
      taxonomy_views.py
      ...

My app/models/__init__.py has the following:

from auth_models import *
from dashboard_models import *
from taxonomy_models import *

However, when I run ./manage.py sql app, I get no output whatsoever. No warnings, nothing at all.

The reason for this question is I'm implementing a database modification and migration in which South cannot handle several aspects. So I'm essentially starting over with a fresh schema and will later implement a data conversion script to migrate the existing dataset. To do this, I need the schema to create the model tables.

See this answer stackoverflow.com/a/1160735/302243. Django makes assumptions about your app name from the path in which the models live, so you're forced in this situation to add a class Meta: app_label = 'app' to every model in the *_models files. — Austin Phillips
– Austin Phillips, Commented Aug 20, 2013 at 0:06
@AustinPhillips, that was the solution. If you make an answer, I'll accept it. Thanks! — Jason
– Jason, Commented Aug 20, 2013 at 0:52

Community · Accepted Answer · 2017-05-23 12:28:55Z

3

See this answer.

Django makes assumptions about your app name from the path in which the models live, so you're forced in this situation to add an app label to every imported model like this:

class MyModel(Model):
    # Model fields
    class Meta:
        app_label = 'app'

Background:

As of this writing, Django has the following code to detect an app label for a model:

    if getattr(meta, 'app_label', None) is None:
        # Figure out the app_label by looking one level up.
        # For 'django.contrib.sites.models', this would be 'sites'.
        model_module = sys.modules[new_class.__module__]
        kwargs = {"app_label": model_module.__name__.split('.')[-2]}

From this, we see that it infers the app_label from the modules name, which may exist deep in the app hierarchy.

edited May 23, 2017 at 12:28

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answered Aug 20, 2013 at 1:02

Austin Phillips

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Comments

Joseph Victor Zammit · Accepted Answer · 2013-08-19 23:38:50Z

2

If your auth_models.py contains a model class MyUserModel, you need to put the following in the __init__.py under app/models:

try:
    from .auth_models import MyUserModel
except ImportError as e:
    sys.stderr.write("Error: failed to import models module ({})".format(e))

P.S. Another suggestion would be to improve the style you name your files with. Since auth_models.py, dashboard_models.py and taxonomy_models.py are all under app/models you can simply truncate the _models.py ending. I.e. rename your files under /models:

auth_models.py -> auth.py
dashboard_models.py -> dashboard.py
taxonomy_models.py -> taxonomy.py

To import the Model classes as instructed above:

from .auth import MyUserModel

edited Aug 19, 2013 at 23:38

answered Aug 19, 2013 at 23:23

Joseph Victor Zammit

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6 Comments

Jason Over a year ago

Yes, I do import the models, and may implement your renaming suggestion. I've edited my question to include the contents of the __init__.py file.

Joseph Victor Zammit Over a year ago

Did you add the try .. except part to except for ImportError?

Jason Over a year ago

Yes I did. There is no change in output. Can this be a result of using import * rather than specific model names?

Joseph Victor Zammit Over a year ago

Give it a try, because I'm out of ideas. In my experience import * is legal but a bad practice (doesn't aid readability and you're importing stuff you might not need at all).

Joseph Victor Zammit Over a year ago

By the way, if you manage to make it work, you should do except ImportError, e and handling the exception accordingly. We shouldn't except blindly as I did in my example; in fact I will update it.

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