Metadata-Version: 2.1
Name: django-model-api
Version: 1.4
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 .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





