How to split flask app.py into modules in Zappa?

Question

Amazon Web Services

Hello!
There is app/__init__.py

from flask import Flask

app = Flask(__name__)


class HelloWorldService:

    def greeting(self):
        print('greeting app')
        return 'Hello World'


@app.route('/')
def index():
    hello = HelloWorldService()
    return hello.greeting(), 200


# include this for local dev
if __name__ == '__main__':
    app.run()

Lambda works successfully, everything is OK, but if I want to take out the routing in app/routes.py

from app import app

@app.route('/')
def index():
    hello = HelloWorldService()
    return hello.greeting(), 200

Or put HelloWorldService in app/services.py and in __init__ import from app.services import HelloWorldService
Then I get an error while executing 'zappa update':
Error: Warning! Status check on the deployed lambda failed. A GET request to '/' yielded a 502 response code.
Please tell me what is the problem? As I understand it, zappa loses its routing when importing.
Running 'flask run' locally does not suffer in all cases and everything is OK.

Answer 1

The repository for the example is https://github.com/tonyin/flask-zappa-base.
According to him, I was able to correctly set up the project structure so that flask with blueprints would work successfully with zappa.

Answer 2

Throw out Flask, throw out Zappa, take the Serverless Framework and refuse routing inside the lambda. Get used to AWS API Gateway