AttributeError when running a flask app in flask shell

Question

I have finished a flask app. When I run it by python run.py, the app can work perfectly.

But when I want to open flask shell by flask shell or even just flask, it tell me:

Traceback (most recent call last):
  File "f:programsanacondaenvsweblibsite-packagesflaskcli.py", line 556, in list_commands
    rv.update(info.load_app().cli.list_commands(ctx))
  File "f:programsanacondaenvsweblibsite-packagesflaskcli.py", line 388, in load_app
    app = locate_app(self, import_name, name)
  File "f:programsanacondaenvsweblibsite-packagesflaskcli.py", line 257, in locate_app
    return find_best_app(script_info, module)
  File "f:programsanacondaenvsweblibsite-packagesflaskcli.py", line 83, in find_best_app
    app = call_factory(script_info, app_factory)
  File "f:programsanacondaenvsweblibsite-packagesflaskcli.py", line 117, in call_factory
    return app_factory(script_info)
  File "C:UserszkhpDesktopflask-bigger-masterbackendstartup.py", line 41, in create_app
    app.config['SECRET_KEY'] = config.get('secret', '!secret!')
AttributeError: 'ScriptInfo' object has no attribute 'get'

The last sentence is here:

def create_app(config):
    app = Flask(
        __name__,
        template_folder=template_folder,
        static_folder=static_folder
    )
    app.config['SECRET_KEY'] = config.get('secret', '!secret!')

The config is a dictionary, which is given by:

def start_server(run_cfg=None, is_deploy=False):
    config = {
        'use_cdn': False,
        'debug': run_cfg.get('debug', False),
        'secret': md5('!secret!'),
        'url_prefix': None,
        'debugtoolbar': True
    }
    app = create_app(config)

I am confused with how the dictionary config is transformed to be a ScriptInfo?

And what should I do to solve the problem?

Asked By: blueice

||

Source

Answer 1

Now I solve the problem.

I have a file manage.py to deal all shell command lines. So the right operation is input:

python manage.py shell

And now it works normally. (OK, I still don’t know why……)

Answered By: blueice

Answer 2

seeing that you’ve already resolved your initial query, i wanted to suggest a better structured config write up for your future flask apps that would also make it easier to add more config variables in the case your app becomes bigger.

Consider having the configs in a module of their own, preferably in a folder name instance in the app’s root folder. Here’s a sample.

"""
This module sets the configurations for the application
"""
import os


class Config(object):
    """Parent configuration class."""
    DEBUG = False
    CSRF_ENABLED = True
    SECRET_KEY = os.getenv("SECRET_KEY")
    DATABASE_URL = os.getenv("DATABASE_URL")
    BUNDLE_ERRORS = True


class DevelopmentConfig(Config):
    """Development phase configurations"""
    DEBUG = True


class TestingConfig(Config):
    """Testing Configurations."""
    TESTING = True
    DEBUG = True
    DATABASE_URL = os.getenv("DATABASE_TEST_URL")


class ReleaseConfig(Config):
    """Release Configurations."""
    DEBUG = False
    TESTING = False


app_config = {
    'development': DevelopmentConfig,
    'testing': TestingConfig,
    'release': ReleaseConfig,
}

Answered By: mburii_ya_mwitu

AttributeError when running a flask app in flask shell

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