Recommended way to serve .yaml/.json

I’m trying to find the best way to serve locally my swagger.json and set it up to get_swaggerui_blueprint. I’m not Flask fluent and the project I’m working on configured a resource that read the json file, load it into a dict and then returns it. Surely there is a better way.

I believe would be better to have an example serving the swagger.json locally on the README.md, since it’s probably the most common use.

[CaveSven]

For a static yaml file you can do

from flask_swagger_ui import get_swaggerui_blueprint
from yaml import Loader, load
from settings import SWAGGER_URL, swagger_path

swagger_yml = load(open(swagger_path, 'r'), Loader=Loader)

blueprint = get_swaggerui_blueprint(SWAGGER_URL, swagger_path, config={'spec': swagger_yml})

[sveint]

Sorry, yeah, I meant API documentation.

In principle they are decoupled. For instance, in the application where I use this package myself, the swagger json is dynamically created from the REST API endpoints’ documentation. I don’t have an actual file on my system. I would imagine this being common for many (most?) people wanting to use this package.

Also, there can other tools using the swagger.json, which are not related to this package at all.

Creating a resource to serve the file can be as simple as creating a dumb flask resource:

# Import flask and create app, then:

@app.route('/api/swagger.json')
def swagger_json():
    # Read before use: http://flask.pocoo.org/docs/0.12/api/#flask.send_file
    return send_file('path/to/swagger.json')  

swaggerui_blueprint = get_swaggerui_blueprint(
    SWAGGER_URL,
    '/api/swagger.json',
    ...
)
app.register_blueprint(swaggerui_blueprint, url_prefix=SWAGGER_URL)

Have you tried this? Is there anything preventing this from working?