Metadata-Version: 2.1
Name: django-model-api
Version: 1.2
Summary: Create model based api without any extra code.
Home-page: https://github.com/kumar-sanchay/django-easy-api
Author: Sanchay Kumar
Author-email: dsanchaykumar@gmail.com
License: MIT
Platform: UNKNOWN
Requires-Dist: django
Requires-Dist: django-filter
Requires-Dist: djangorestframework
Requires-Dist: markdown

# 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 easy_api.models import EasyAPI
	from .models import MyModel

	urlpatterns = [
	path("myurl/", EasyAPI.as_view(), {"child_model": MyModel})
	]

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



