flask-auto-api 1.0.8
Provide non-intrusive and one-click machine learning api release tool.
License: BSD License

Flask-autoapi
=============

This project is mainly for machine learning projects. Sometimes a
machine learning service is written, but it requires a lot of additional
code to deploy. Is there any way to register a service to restful api
for deployment without changing all the code of the machine learning
service? flask-autoapi gives the answer.

First, you need to put the entire code of your machine learning service
into a directory to become a module (actually add the \_\_init\_\_.py
file in this directory), and then specify which python file in the
directory contains Service, and then specify which function of this file
provides the service. Turn the above into a config and directly call
flask\_autoapi.launch(config) to deploy the service directly.

Dependencies
~~~~~~~~~~~~

::

    flask-1.1.2
    flask-RESTFUL-0.3.8

Example
=======

Step1: Suppose you have a local service named service2, and you want to
provide service2.service as a restful service.

Step2: Make the service2 directory as a python module first.

The example service2 may have its own dependcies, not worry, you don't
need to change any codes.

services/service2/service2：

::

    from dependencies import add, mul

    def service2(param1, param2, param3):
        param1 = int(param1)
        param2 = int(param2)
        param3 = int(param3)

        ans1 = add(param1, param2, param3)
        ans2 = mul(param1, param2, param3)

        return str(ans1), str(ans2)

It should be noted that the input parameters and output parameters of
the interface functions that provide services must be in the type of
strings, of course, there is no limit to the parameter number.

Step3: Make such a config dictionary:

.. code:: python

    service_config = {
        'service_dir'               : './services/service2', # Service root directory
        'service_python_filename'   : 'service2',            # service2.py
        'service_python_interface'  : 'service2',            # service2 is the interface function

        'service_input_params' : {                           # interface input params are defined here
            'param1': 'param1 describe here',
            'param2': 'param2 describe here',
            'param3': 'param3 describe here',
        },

        'service_output_params' : {                          # interface output params are defined here
            'add_result': 'value describe here',
            'mul_result': 'value describe here'
        },

        'deploy_mode'   : 'restful', # c++ deploy            # now only has restful deploy
        'deploy_port'   : '12345',                           # service port
        'service_route' : '/test_service'                    # service route
    }

It should be noted that the input and output parameters of the interface
must be written in the configuration dict in the same form, and the keys
of service\_input\_params or service\_output\_params will be the http
request parameter and the response keys respectively. The values of
service\_input\_params and service\_output\_params are describes to
those parameters.

Step4: Then launch this service api:

.. code:: python

    import flask_autoapi
    flask_autoapi.launch(service_config)

Step5: The service will be running:

::

     * Running on http://127.0.0.1:12345/ (Press CTRL+C to quit)
     * Serving Flask app "flask_autoapi.restful_service_enroll" (lazy loading)
     * Environment: production
       WARNING: This is a development server. Do not use it in a production deployment.
       Use a production WSGI server instead.
     * Debug mode: off

The test url:

::

    http://127.0.0.1:12345/test_service?param1=1&param2=2&param3=3

The response:

::

    {"status": 200, "add_result": "6", "mul_result": "6"}