Using pytest, how can I test the configuration of my Flask app with and without a config file?

Question

I'm currently writing a simple flask application and want to test that my configuration options are working.

Here's my config.py:

import logging
import logging.handlers
import os


class BaseConfig(object):
    DEBUG = False
    TESTING = False
    ...


class DevelopmentConfig(BaseConfig):
    DEBUG = True
    TESTING = True
    ...


class TestingConfig(BaseConfig):
    DEBUG = False
    TESTING = True
    ...

config = {
    'development': 'app.config.DevelopmentConfig',
    'testing': 'app.config.TestingConfig',
    'default': 'app.config.BaseConfig'
}


def configure_app(app):
    config_name = os.environ.get('FLASK_CONFIGURATION', 'default')
    app.config.from_object(config[config_name])
    app.config.from_pyfile('config.py', silent=True)
    apikey = os.environ.get('TOKEN', None)
    if apikey:
        app.config['APIKEY'] = apikey

I call configure_app in my __init__.py right after I create the Flack object.

app = Flask(__name__)
configure_app(app)

config.py:

APIKEY = 'filesecret'
HOOKS = {'testpush': 'scripts/test.sh'}

What I want to do is be able to unittest using py.test the various configuration options. My current attempt is to try and use mock_open, but the configuration keep reading what's in the file instead of skipping it.

class TestConfig:

    def setup(self):
        self.app = app.app
        self.app.config['TESTING'] = True

    def test_default(self, mocker):
        m = mocker.mock_open(read_data='')
        configure_app(self.app)
        assert self.app.config['APIKEY'] == "secret"

results of test:

    def test_default(self, mocker):
        m = mocker.mock_open(read_data='')
        configure_app(self.app)
>       assert self.app.config['APIKEY'] == "secret"
E       assert 'filesecret' == 'secret'

I'm not sure how to do this when I don't know the internals of flask. Is there a way to fake/mock the file? I want to be able to write the various test cases of the configuration file being present, not present, and the environment variable being set to make sure priority is maintained.

score 0 · Answer 1 · edited May 23 '17 at 12:33

So I was going about this all wrong. I looked at Flask's source code and saw how they were testing their configuration functions, and they simply instantiated a new Flask object with their own environment variables.

Flask project test_config.py

Using the context manager found on this stackoverflow question (currently the bottom answer with contextlib) to set my environment variables I got this all working great.

Final test:

@pytest.mark.parametrize(
    'envtoken, flask_conf, test_path, apikey, debug, testing', [
        ('', 'default', 'missing', 'secret', False, False),
        ('', 'development', 'missing', 'secret', True, True),
        ('', 'testing', 'missing', 'secret', False, True),
        ('', 'default', '../tests/data', 'filesecret', False, False),
        ('envsecret', 'default', '../tests/data/', 'envsecret', False, False)])
def test_config_app(envtoken, flask_conf, test_path, apikey, debug, testing):
    with set_env(DOCKERHOOK_TOKEN=envtoken, FLASK_CONFIGURATION=flask_conf):
        # app = Flask(__name__)
        base_path = get_app_base_path()
        instance_path = os.path.join(base_path, test_path)

        app = Flask(__name__,
                    instance_path=instance_path,
                    instance_relative_config=True)

        configure_app(app)

        assert app.config['APIKEY'] == apikey
        assert app.config['DEBUG'] == debug
        assert app.config['TESTING'] == testing

Using pytest, how can I test the configuration of my Flask app with and without a config file?

1 Answers1