django-easy-services 0.2

A simple Django app to handle services request.

Services is a simple Django app to consume simple json services in a simple way. You can make requests directly, async or recursive using failover services automatically while the services return errors. Example:

from services.models import Service

service_example = Service(name='simple_service', method='get', url='http://example.com/?query_param=<data>')
url_data = {'<data>': 'example_data'}
service_response = service_example.request(url_data=get_data)

if service_response['success']:
    respose_content = service_response['content'] # content is a dict
    print(response_content)

Post example:

from services.models import Service

service_example = Service(name='simple_service', method='post', url='http://example.com/', headers='{"Content-Type": "<content_type>"}', parameters= '{"query_param": "<data>"}')
get_data = {'<data>': 'example_data'}
header_data = {'<content_type>': 'application/json'}
service_response = service_example.request(get_data=get_data, header_data=header_data)

if service_response['success']:
    respose_content = service_response['content'] # content is a dict
    print(response_content)

Or simpler:

from services.models import Service

# This headers will be use for every request to this service
service_example = Service(name='simple_service', method='post', url='http://example.com/', headers='{"Content-Type": "application/json"}')
# parameters will be put directly to the body of the post request to the service
parameters = {'query_param': 'example_data'}
service_response = service_example.request(parameters=parameters)

if service_response['success']:
    respose_content = service_response['content'] # content is a dict
    print(response_content)

You can also configure failover services, so if the primary service return some error code (for now >= 400) the failover/s will we call one by one untill some of them return a non error code:

from services.models import Service, ServiceFailover

service_01 = Service(name='registration_precheck', method='get', url='http://example_01.com/?query_param=<data>').save()
service_02 = Service(name='registration_precheck', method='get', url='http://example_02.com/?query_param=<data>').save()
service_03 = Service(name='registration_precheck', method='get', url='http://example_03.com/?query_param=<data>').save()

ServiceFailover(service=service_01, failover=service_02, order=1).save()
ServiceFailover(service=service_01, failover=service_03, order=2).save()

url_data = {'<data>': 'example_data'}

# Now if service_01 fails service_02 will be execute with the same context data,
# the same for service_03 if service_02 fails in this call.
service_response = service_example.request_recursive(url_data=get_data)

if service_response['success']:
    respose_content = service_response['content'] # content is a dict
    print(response_content)

If you want to publish information to some service but you don’t want to wait you can make async request:

from services.models import Service

service_example = Service(name='simple_service', method='post', url='http://example.com/', headers='{"Content-Type": "application/json"}')
parameters = {'query_param': 'example_data'}
# max_retry is the amount of retries do you want to execute the request while the answer is a error (0 if you want to retry "forever").
# retry_interval is the amount of seconds you want to wait between retries.
service_example.request_async(max_retry=10, retry_interval=1, parameters=parameters)

Quick start

1. Run pip install django-easy-services

2. Add “services” to your INSTALLED_APPS setting like this:

   INSTALLED_APPS = [
           ...
           'services',
   ]

3. Run python manage.py migrate to create the services models.

4. Start the development server and visit http://127.0.0.1:8000/admin/services to create a services (you’ll need the Admin app enabled).