Middleware advice/documentation
danpalmer opened this issue · comments
I'm trying to use pytest-flask to test a flask server that has some WSGI middleware wrapping it, but it seems this isn't supported as an exception is raised from within pytest-flask:
@pytest.fixture(autouse=True)
def _monkeypatch_response_class(request, monkeypatch):
"""Set custom response class before test suite and restore the original
after. Custom response has `json` property to easily test JSON responses::
@app.route('/ping')
def ping():
return jsonify(ping='pong')
def test_json(client):
res = client.get(url_for('ping'))
assert res.json == {'ping': 'pong'}
"""
if 'app' not in request.fixturenames:
return
app = request.getfuncargvalue('app')
monkeypatch.setattr(app, 'response_class',
> _make_test_response_class(app.response_class))
E AttributeError: 'function' object has no attribute 'response_class'
This makes sense, as the flask app is no longer a flask app with all the properties that pytest-flask is expecting, but I'm unsure how to go about testing this as a result.
Is there a way to get around this with pytest-flask? Is there a different API I should be using for my middleware (or a different way of registering them) than WSGI middleware? Am I missing something obvious here?
Thanks for the help!
I just had the exact same error message after mistakenly returning the app
function itself inside my app
fixture, not the local app_
instance create from the create_app
call.
@pjwerneck would you be able to provide a code snippet? I'm running into the same error.
Here's my conftest.py
import pytest
from server import server as flask_app
@pytest.fixture
def app():
yield flask_app
@pytest.fixture
def client(app):
return app.test_client()
@oneilk It was something like this
import pytest
from myapp import create_app
@pytest.fixture
def app():
app_ = create_app()
return app # should be `return app_`
Hey @danpalmer, thanks for taking the time to open the issue.
This problem can probably be avoided by applying the wsgi middleware as suggested by the official flask docs as follows:
wsgi_app
is the actual WSGI application. This is not implemented in __call__()
so that Middlewares can be applied without losing a reference to the app object.
Instead of doing this:
app = MyMiddleware(app)
It’s a better idea to do this instead:
app.wsgi_app = MyMiddleware(app.wsgi_app)
I'll be closing this for now, feel free to reopen if the problem still persists (try including a minimal, reproducible example in this case).