我想为python应用程序生成openAPI规范。我可以生成swagger.json

import logging
import json
from apispec import APISpec
from init import get_slack_url
from flask import Flask
from apispec.ext.marshmallow import MarshmallowPlugin
from apispec_webframeworks.flask import FlaskPlugin
logging.basicConfig(level=logging.DEBUG)
hook_url = get_slack_url()
spec = APISpec(
    title="Notification Service",
    version="1.0.0",
    openapi_version="3.0.2",
    plugins=[FlaskPlugin(), MarshmallowPlugin()]
)

app = Flask(__name__)
@app.route('/slack_notification', methods=['POST'])
def send_slack_notification():
    return 'wathever'

with app.test_request_context():
    spec.path(view=send_slack_notification)
with open('swagger.json', 'w') as f:
    json.dump(spec.to_dict(), f)

if __name__ == '__main__':
    app.run(host='0.0.0.0', port=5001, debug=False)

swagger.json被生成。

{"paths": {"/slack_notification": {}}, "info": {"title": "Notification Service", "version": "1.0.0"}, "openapi": "3.0.2"}

但我无法加载swagger或apidocs端点。我收到错误。

INFO:werkzeug: * Running on http://0.0.0.0:5001/ (Press CTRL+C to quit)
INFO:werkzeug:127.0.0.1 - - [27/Oct/2021 21:03:50] "GET /swagger/ HTTP/1.1" 404 -
INFO:werkzeug:127.0.0.1 - - [27/Oct/2021 21:03:53] "GET /openapi HTTP/1.1" 404 -
INFO:werkzeug:127.0.0.1 - - [27/Oct/2021 21:03:57] "GET / HTTP/1.1" 404 -

关于如何在UI上看到这一点，有什么建议吗？
我尝试的另一种选择是Flasgger。我可以看到UI，但规格有swagger: 2.0，我特别想要OpenApi 3.0。
参考：https://donofden.com/blog/2020/06/14/Python-Flask-automatically-generated-Swagger-3-0-openapi-Document