I have a Python Flask app that accesses the Github API. For this I need to store an access token. What is the common practice to store that data and how do I access that inside my app?
from flask import Flask, request
app = Flask(__name__)
app.config['DEBUG'] = True
#app.route('/',methods=['POST'])
def foo():
...
Flask has a custom context to store app variables:
http://flask.pocoo.org/docs/1.0/appcontext/
You can use g object to store your variables:
from flask import g
g.github_token = 'secret'
And after initialization:
from flask import g
token = g.github_token
The simpliest way is to place it into configuration module (regular python .py file) and then import and use it in your code as suggested by this snippet on Flask site.
Related
In flask (example - https://flask.palletsprojects.com/en/1.1.x/patterns/appfactories/#basic-factories), following is the code provided:
def create_app(config_filename):
app = Flask(__name__)
app.config.from_pyfile(config_filename)
from yourapplication.model import db
db.init_app(app)
from yourapplication.views.admin import admin
from yourapplication.views.frontend import frontend
app.register_blueprint(admin)
app.register_blueprint(frontend)
return app
Is there any advantage of using the parameter for the factory method over for example we could even - not use the parameter and replace app.config.from_pyfile(config_filename) with app.config.from_pyfile(os.getenv('env_var', 'dev'))? Is there any limitation to using the env var approach?
I am getting confused with configurations and imports once I started using the Flask factory application pattern.
I am creating an application with the function create_app in #app/init.py
I have a config file for setting the development/testing/production variables, and an instance folder with another config file.
def create_app(config_name):
app=Flask(__name__, instance_relative_config=True)
app.config.from_object(app_config[config_name])
app.config.from_pyfile('config.py')
etc...
return app
I am using blueprints and have an authentication view in #app/auth/views.py
I am trying to set up email confirmation tokens using URLSafeTimedSerializer...
from itsdangerous import URLSafeTimedSerializer
#auth.route('/register', methods=['GET','POST'])
def register():
ts = URLSafeTimedSerializer(app.config['SECRET_KEY'])
token = ts.dumps(self.email, salt='email-confirm-key')
etc...
Now my problem is, my variable 'ts' needs the app.config['SECRET_KEY'] set. But I am unable to define the app variable (as is shown in all online tutorials). I get an error when I try to import...(in #app/auth/views.py)
from .. import app
and when I try to import like...
from .. import create_app
Can someone shine light on how to initialize modules using 'app' and app.config outside the flask app factory create_app?
Hope you understand my question.
In this scenario you should use Flask.current_app
from flask import current_app
...
ts = URLSafeTimedSerializer(current_app.config['SECRET_KEY'])
From the documentation:
flask.current_app
Points to the application handling the request. This
is useful for extensions that want to support multiple applications
running side by side. This is powered by the application context and
not by the request context, so you can change the value of this proxy
by using the app_context() method.
This link aso explains further details about the Flask application factory methodology, in particular using current_app to access the app configuration.
I have two application factory functions - one creates the "customer" app, and the other creates the "admin" backend app. Both of the factory functions essentially do what is described here - create a flask app and register some extensions to it and then add some blueprints(with a url_prefix). I glue the two apps together via the create_combined_app() from below. It is the return value of that function which I register with my Flask-Script's Manager.
def create_combined_app(config_name):
customer_app = create_customer_app(config_name)
admin_app = create_admin_app(config_name)
from werkzeug.wsgi import DispatcherMiddleware
customer_app.wsgi_app = DispatcherMiddleware(customer_app.wsgi_app, {
'/admin': admin_app
})
return customer_app
And then this is how I run it.
def make_me_an_app():
return create_combined_app(config)
manager = Manager(make_me_an_app)
...
if __name__ == '__main__':
manager.run()
I want to do some testing which involves getting all GET routes of my app and making sure they load. I followed the example from here, but I only see the urls of the customer app, and none of the urls from the admin backend.
#main.route("/site-map")
def site_map():
from flask import current_app, jsonify
links = []
app = current_app
for rule in app.url_map.iter_rules():
if "GET" in rule.methods and has_no_empty_params(rule):
url = url_for(rule.endpoint, **(rule.defaults or {}))
links.append((url, rule.endpoint))
return jsonify(links)
The admin backend works when I try to access it from the browser - it all works nicely, except that I don't see the admin's urls when I call /site-map.
Thanks! :)
I think DispatcherMiddleware create separate apps. Which mean you created customer_app and admin_app. Those 2 live as standalone. They don't know each others, therefor current_app is just the show customer_app.
Here is the describe from Flask http://flask.pocoo.org/docs/0.12/patterns/appdispatch/
I'm currently using the Flask Application Factory pattern with Blueprints. The issue that I'm having is how do I access the app.config object outside of the application factory?
I don't need all the configuration options from the Flask app. I just need 6 keys. So the current way I do this is when the create_app(application factory) is called, I basically create a global_config dictionary object and I just set the global_config dictionary to have the 6 keys that I need.
Then, the other modules that need those configuration options, they just import global_config dictionary.
I'm thinking, there has to be a better way to do this right?
So, on to the code
My current init.py file:
def set_global_config(app_config):
global_config['CUPS_SAFETY'] = app_config['CUPS_SAFETY']
global_config['CUPS_SERVERS'] = app_config['CUPS_SERVERS']
global_config['API_SAFE_MODE'] = app_config['API_SAFE_MODE']
global_config['XSS_SAFETY'] = app_config['XSS_SAFETY']
global_config['ALLOWED_HOSTS'] = app_config['ALLOWED_HOSTS']
global_config['SQLALCHEMY_DATABASE_URI'] = app_config['SQLALCHEMY_DATABASE_URI']
def create_app(config_file):
app = Flask(__name__, instance_relative_config=True)
try:
app.config.from_pyfile(config_file)
except IOError:
app.config.from_pyfile('default.py')
cel.conf.update(app.config)
set_global_config(app.config)
else:
cel.conf.update(app.config)
set_global_config(app.config)
CORS(app, resources=r'/*')
Compress(app)
# Initialize app with SQLAlchemy
db.init_app(app)
with app.app_context():
db.Model.metadata.reflect(db.engine)
db.create_all()
from authenication.auth import auth
from club.view import club
from tms.view import tms
from reports.view import reports
from conveyor.view import conveyor
# Register blueprints
app.register_blueprint(auth)
app.register_blueprint(club)
app.register_blueprint(tms)
app.register_blueprint(reports)
app.register_blueprint(conveyor)
return app
An example of a module that needs access to those global_config options:
from package import global_config as config
club = Blueprint('club', __name__)
#club.route('/get_printers', methods=['GET', 'POST'])
def getListOfPrinters():
dict = {}
for eachPrinter in config['CUPS_SERVERS']:
dict[eachPrinter] = {
'code': eachPrinter,
'name': eachPrinter
}
outDict = {'printers': dict, 'success': True}
return jsonify(outDict)
There has to be a better way then passing a global dictionary around the application correct?
There is no need to use global names here, that defeats the purpose of using an app factory in the first place.
Within views, such as in your example, current_app is bound to the app handling the current app/request context.
from flask import current_app
#bp.route('/')
def example():
servers = current_app.config['CUPS_SERVERS']
...
If you need access to the app while setting up a blueprint, the record decorator marks functions that are called with the state the blueprint is being registered with.
#bp.record
def setup(state):
servers = state.app.config['CUPS_SERVERS']
...
I'm having some trouble understanding how to incorporate Flask-Pymongo. My app is initiated from my rrapp.py Inside of this file, I have
rrapp.py
#
# Imports up here
#
app = Flask(__name__)
mongo = PyMongo(app)
# Code down here
Now, to use this, I simply do mongo.db.users.find(). This works fine.
Now, say I have another file called userservice.py that I call methods from one of my endpoints within rrapp.py. How do I incorporate PyMongo(app) in my userservice.py file if I don't have access to the app object? Or am I missing something obvious here?
you should first define mongo oustside create_app to have access to it from inside other files.
then init_app with that like the following:
from flask import Flask, current_app
from flask_pymongo import PyMongo
mongo = PyMongo()
def create_app(config_name):
app = Flask(__name__, instance_relative_config=False)
app.config.from_object(app_config[config_name])
# INIT EXTENSIONS ----------------------
mongo.init_app(app)
return app
then in any file you can import mongo from above file. for example:
from ../factory import mongo