CODEWITHSUNDEEP
pythondjangodjango-modelsimport
diegueus9
diegueus9

Reputation: 31532

Split models.py into several files

I'm trying to split the models.py of my app into several files:

My first guess was do this:

myproject/
    settings.py
    manage.py
    urls.py
    __init__.py
    app1/
        views.py
        __init__.py
        models/
            __init__.py
            model1.py
            model2.py
    app2/
        views.py
        __init__.py
        models/
            __init__.py
            model3.py
            model4.py

This doesn't work, then i found this, but in this solution i still have a problem, when i run python manage.py sqlall app1 I got something like:

BEGIN;
CREATE TABLE "product_product" (
    "id" serial NOT NULL PRIMARY KEY,
    "store_id" integer NOT NULL
)
;
-- The following references should be added but depend on non-existent tables:
-- ALTER TABLE "product_product" ADD CONSTRAINT "store_id_refs_id_3e117eef" FOREIGN KEY     ("store_id") REFERENCES "store_store" ("id") DEFERRABLE INITIALLY DEFERRED;
CREATE INDEX "product_product_store_id" ON "product_product" ("store_id");
COMMIT;

I'm not pretty sure about this, but i'm worried aboout the part The following references should be added but depend on non-existent tables:

This is my model1.py file:

from django.db import models

class Store(models.Model):
    class Meta:
        app_label = "store"

This is my model3.py file:

from django.db import models

from store.models import Store

class Product(models.Model):
    store = models.ForeignKey(Store)
    class Meta:
        app_label = "product"

And apparently works but i got the comment in alter table and if I try this, same thing happens:

class Product(models.Model):
    store = models.ForeignKey('store.Store')
    class Meta:
        app_label = "product"

So, should I run the alter for references manually? this may bring me problems with south?

Upvotes: 136

Views: 82801

Answers (6)

Adam Luchjenbroers
Adam Luchjenbroers

Reputation: 5019

I've actually come across a tutorial for exactly what you're asking about, you can view it here:

https://web.archive.org/web/20190331105757/http://paltman.com/breaking-apart-models-in-django/

One key point that's probably relevant - you may want to use the db_table field on the Meta class to point the relocated classes back at their own table.

I can confirm this approach is working in Django 1.3

Upvotes: 13

Kevin K
Kevin K

Reputation: 2227

The relevant link for Django 3 is:

https://docs.djangoproject.com/en/3.2/topics/db/models/#organizing-models-in-a-package

Links to previous versions of documentation are broken. The example there is very succinct:

To do so, create a models package. Remove models.py and create a myapp/models/ directory with an init.py file and the files to store your models. You must import the models in the init.py file.

For example, if you had organic.py and synthetic.py in the models directory:

from .organic import Person
from .synthetic import Robot

Upvotes: 11

user1659565
user1659565

Reputation:

I wrote a script that might be useful.

github.com/victorqribeiro/splitDjangoModels

it split the models in individual files with proper naming and importing; it also create an init file so you can import all your models at once.

let me know if this helps

Upvotes: 0

Ted
Ted

Reputation: 12318

I'd do the following:

myproject/
    ...
    app1/
        views.py
        __init__.py
        models.py
        submodels/
            __init__.py
            model1.py
            model2.py
    app2/
        views.py
        __init__.py
        models.py
        submodels/
            __init__.py
            model3.py
            model4.py

Then

#myproject/app1/models.py:
    from submodels/model1.py import *
    from submodels/model2.py import *

#myproject/app2/models.py:
    from submodels/model3.py import *
    from submodels/model4.py import *

But, if you don't have a good reason, put model1 and model2 directly in app1/models.py and model3 and model4 in app2/models.py

---second part---

This is app1/submodels/model1.py file:

from django.db import models
class Store(models.Model):
    class Meta:
        app_label = "store"

Thus correct your model3.py file:

from django.db import models
from app1.models import Store

class Product(models.Model):
    store = models.ForeignKey(Store)
    class Meta:
        app_label = "product"

Edited, in case this comes up again for someone: Check out django-schedule for an example of a project that does just this. https://github.com/thauber/django-schedule/tree/master/schedule/models https://github.com/thauber/django-schedule/

Upvotes: 44

Parth Jani
Parth Jani

Reputation: 33

Easiest Steps :

  1. Create model folder in your app (Folder name should be model)
  2. Delete model.py file from app directory (Backup the file while you delete it)
  3. And after create init.py file in model folder
  4. And after init.py file in write simple one line
  5. And after create model file in your model folder and model file name should be same like as class name,if class name is 'Employee' than model file name should be like 'employee.py'
  6. And after in model file define your database table same as write like in model.py file
  7. Save it

My Code : from django_adminlte.models.employee import Employee

For your : from app_name.models.model_file_name_only import Class_Name_which_define_in_model_file

__init__.py

from django_adminlte.models.employee import Employee

model/employee.py (employee is separate model file)

from django.db import models

class Employee(models.Model):
eid = models.CharField(max_length=20)
ename = models.CharField(max_length=20)
eemail = models.EmailField()
econtact = models.CharField(max_length=15)

class Meta:
    db_table = "employee"
    # app_label = 'django_adminlte'
    
def __str__(self):
    return self.ename

Upvotes: 2

Vic
Vic

Reputation: 2706

For anyone on Django 1.9, it is now supported by the framework without defining the class meta data.

https://docs.djangoproject.com/en/1.9/topics/db/models/#organizing-models-in-a-package

NOTE: For Django 2, it's still the same

The manage.py startapp command creates an application structure that includes a models.py file. If you have many models, organizing them in separate files may be useful.

To do so, create a models package. Remove models.py and create a myapp/models/ directory with an __init__.py file and the files to store your models. You must import the models in the __init__.py file.

So, in your case, for a structure like

app1/
    views.py
    __init__.py
    models/
        __init__.py
        model1.py
        model2.py
app2/
    views.py
    __init__.py
    models/
        __init__.py
        model3.py
        model4.py

You only need to do

#myproject/app1/models/__init__.py:
from .model1 import Model1
from .model2 import Model2

#myproject/app2/models/__init__.py:
from .model3 import Model3
from .model4 import Model4

A note against importing all the classes:

Explicitly importing each model rather than using from .models import * has the advantages of not cluttering the namespace, making code more readable, and keeping code analysis tools useful.

Upvotes: 199

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