django-model-api 1.4

Create model based api without any extra code.

# django-easy-api

Now there is no need to create apis for common CRUD operation for any model. By using django-easy-api its really easy. Kindly follow the below steps for including this library into your project.

1. Create Your django project.

2. Add easy_api in settings.py

   INSTALLED_APPS = [ . . . ‘easy_api’ ]

3. Create your django app

4. Now in models.py follow create your model by inheriting EasyAPI model from easy_api app.

   from django.db import models from easy_api.models import EasyAPI

   class MyModel(EasyAPI):

   # Your Requried field here.

5. Migrate your models.

6. This is the last step. You need to add this model in your app’s urls.py

   from django.urls import path from .models import MyModel

   urlpatterns = [

   path(‘myurl/’, MyModel.as_view()),

   ]

7. That’s it. Now you will have common GET/POST/PUT/DELETE methods on your model. Also you can attach query_params only for id in your url. For example:

http://localhost:8000/myurl/?id=1

8. You can also override get, post, put and delete method according to your need. For example:

from django.db import models from easy_api.models import EasyAPI

class MyModel(EasyAPI):

# Your Requried field here.

def get(self, request):

#Your logic