Peter Doggart

Yes, you just need to use the [flask application factory structure](https://flask.palletsprojects.com/en/2.2.x/patterns/appfactories/) to create your API. There is a good example of how to do this with `flask-restx` here: https://github.com/a-luna/flask-api-tutorial

I think the issue is that `api.expect()` and `namespace.expect()` expect you to pass a single expected model (or a list of the same expected model) or request parser? I can't...

This sounds like it's maybe related to https://github.com/vimalloc/flask-jwt-extended/issues/86#issuecomment-335509456 ?

Haven't had a chance to try and run your code but I notice you have a `schema_model` defined: ```python crawler_rss_arg_json_model = api.schema_model('RSSCrawlerRequest', { "properties":{ config.RSS_URL: { "type":"string" }, }, "type":...

Think this is a duplicate of https://github.com/python-restx/flask-restx/issues/422 ?

It looks like the android request is maybe setting an `accept` header that the flask-restx app doesn't support. The bit of code in question in `api.py` is: ```python mediatype =...

Is there a reason you want to hide the UI page specifically? If you are trying to protect from people from using your endpoints without permission, you would be far...

There is also an issue with flask-restx and werkzeug 2.1.x (https://github.com/python-restx/flask-restx/issues/422) so I would maybe recommend setting `werkzeug==2.0.*` for now.

@abompard To get around the lack of customization, I just render a custom template (that flask-restx fully supports) and then you can pass whatever configurations you like to Swagger UI...

@radiantone I think you just need to pass `strict=True` as part of your `api.model()` call? Source: https://flask-restx.readthedocs.io/en/latest/_modules/flask_restx/namespace.html#Namespace.model