models.py getting huge, what is the best way to break it up?

Question

Directions from my supervisor: "I want to avoid putting any logic in the models.py. From here on out, let's use that as only classes for accessing the database, and keep all logic in external classes that use the models classes, or wrap them."

I feel like this is the wrong way to go. I feel that keeping logic out of the models just to keep the file small is a bad idea. If the logic is best in the model, that's where it really should go regardless of file size.

So is there a simple way to just use includes? In PHP-speak, I'd like to propose to the supervisor that we just have models.py include() the model classes from other places. Conceptually, this would allow the models to have all the logic we want, yet keep file size down via increasing the number of files (which leads to less revision control problems like conflicts, etc.).

So, is there a simple way to remove model classes from the models.py file, but still have the models work with all of the Django tools? Or, is there a completely different yet elegant solution to the general problem of a "large" models.py file? Any input would be appreciated.

Answer 1

It's natural for model classes to contain methods to operate on the model. If I have a Book model, with a method book.get_noun_count(), that's where it belongs--I don't want to have to write "get_noun_count(book)", unless the method actually intrinsically belongs with some other package. (It might--for example, if I have a package for accessing Amazon's API with "get_amazon_product_id(book)".)

I cringed when Django's documentation suggested putting models in a single file, and I took a few minutes from the very beginning to figure out how to split it into a proper subpackage.

site/models/__init__.py site/models/book.py 

__init__.py looks like:

from .book import Book 

so I can still write "from site.models import Book".

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The following is only required for versions prior to Django 1.7, see https://code.djangoproject.com/ticket/3591

The only trick is that you need to explicitly set each model's application, due to a bug in Django: it assumes that the application name is the third-to-last entry in the model path. "site.models.Book" results in "site", which is correct; "site.models.book.Book" makes it think the application name is "models". This is a pretty nasty hack on Django's part; it should probably search the list of installed applications for a prefix match.

class Book(models.Model):     class Meta: app_label = "site" 

You could probably use a base class or metaclass to generalize this, but I haven't bothered with that yet.