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']
...
Related
I am trying to define a mongodb object inside main flask app. And I want to send that object to one of the blueprints that I created. I may have to create more database objects in main app and import them in different blueprints. I tried to do it this way.
from flask import Flask, render_template
import pymongo
from admin_component.bp1 import bp_1
def init_db1():
try:
mongo = pymongo.MongoClient(
host='mongodb+srv://<username>:<passwrd>#cluster0.bslkwxdx.mongodb.net/?retryWrites=true&w=majority',
serverSelectionTimeoutMS = 1000
)
db1 = mongo.test_db1.test_collection1
mongo.server_info() #this is the line that triggers exception.
return db1
except:
print('Cannot connect to db!!')
app = Flask(__name__)
app.register_blueprint(bp_1, url_prefix='/admin') #only if we see /admin in url we gonna extend things in bp_1
with app.app_context():
db1 = init_db1()
#app.route('/')
def test():
return '<h1>This is a Test</h1>'
if __name__ == '__main__':
app.run(port=10001, debug=True)
And this is the blueprint and I tried to import the init_db1 using current_app.
from flask import Blueprint, render_template, Response, request, current_app
import pymongo
from bson.objectid import ObjectId
import json
bp_1 = Blueprint('bp1', __name__, static_folder='static', template_folder='templates')
print(current_app.config)
db = current_app.config['db1']
But it gives this error without specifying more details into deep.
raise RuntimeError(unbound_message) from None
RuntimeError: Working outside of application context.
This typically means that you attempted to use functionality that needed
the current application. To solve this, set up an application context
with app.app_context(). See the documentation for more information.
Can someone point out what am I doing wrong here??
The idea you are attempting is correct; however it just needs to be done a little differently.
First, start by declaring your mongo object in your application factory:
In your app/__init__.py:
import pymongo
from flask import Flask
mongo = pymongo.MongoClient(
host='mongodb+srv://<username>:<passwrd>#cluster0.bslkwxdx.mongodb.net/?retryWrites=true&w=majority',
serverSelectionTimeoutMS = 1000
)
# Mongo is declared outside of function
def create_app(app):
app = Flask(__name__)
return app
And then in your other blueprint, you would call:
from app import mongo # This right here will get you the mongo object
from flask import Blueprint
bp_1 = Blueprint('bp1', __name__, static_folder='static', template_folder='templates')
db = mongo
I have created a small Flask application which stores its data in an sqlite database that I access via flask-sqlalchemy.
However, when I run it, I get the following error:
RuntimeError: No application found. Either work inside a view function or push an application context. See http://flask-sqlalchemy.pocoo.org/contexts/.
I have debugged my application and now know that this error stems from these two functions:
def user_exists(email):
if User.query.filter_by(email = email).count() == 0:
return False
else:
return True
def get_user(email):
user = User.query.filter_by(email = email).first()
return user
Now I am wondering: Is it impossible to access the database via flask-sqlalchemy outside of view functions?
For further context, I added the files in which I configure my flask app:
presentio.py
from app import create_app
app = create_app(os.getenv("FLASK_CONFIG", "default"))
app/init.py
from flask_mail import Mail
from flask_sqlalchemy import SQLAlchemy
from config import config
mail = Mail()
db = SQLAlchemy()
def create_app(config_name):
app = Flask(__name__)
app.config.from_object(config[config_name])
config[config_name].init_app(app)
mail.init_app(app)
db.init_app(app)
from .main import main as main_blueprint
app.register_blueprint(main_blueprint)
from .auth import auth as auth_blueprint
app.register_blueprint(auth_blueprint, url_prefix = "/auth")
from .text import text as text_blueprint
app.register_blueprint(text_blueprint, url_prefix = "/text")
return app
You need to give the flask app a context after you create it.
This is done automatically in view functions, but outside those, you need to do this after you create the app:
app.app_context().push()
See the docs: https://flask-sqlalchemy.palletsprojects.com/en/2.x/contexts/
I am using the app factory pattern to set up my Flask application. My app uses the Flask-Babel extension, and that is set up in the factory as well. However, I want to access the extension in a blueprint in order to use it,
The factory is in __init__.py.
def create_app(object_name):
app = Flask(__name__)
app.config.from_object(object_name)
babel = Babel(app)
app.register_blueprint(main_blueprint)
app.register_blueprint(category_blueprint)
app.register_blueprint(item_blueprint)
db.init_app(app)
return app
I want to add the following to main.py:
#babel.localeselector
def get_locale():
if 'locale' in session:
return session['locale']
return request.accept_languages.best_match(LANGUAGES.keys())
#application.route('/locale/<locale>/', methods=['GET'])
def set_locale(locale):
session['locale'] = locale
redirect_to = request.args.get('redirect_to', '/')
return redirect(redirect_to) # Change this to previous url
Unfortunately, main.py doesn't have access to the babel variable from the application factory. How should I go about solving this?
Flask extensions are designed to be instantiated without an app instance for exactly this case. Outside the factory, define your extensions. Inside the factory, call init_app to associate the app with the extension.
babel = Babel()
def create_app():
...
babel.init_app(app)
...
Now the babel name is importable at any time, not just after the app has been created.
You already appear to be doing this correctly with the db (Flask-SQLAlchemy) extension.
In the case of your specific babel.localeselector example, it might make more sense to put that next to babel since it's being defined there.
In order to simplify the __init__.py main module, I want to push helper functionality to a different file/class. This requires passing many flask extensions instances when initializing the class, which seems inelegant. My current structure is as follows:
__init__.py:
from flask import Flask, render_template,request
from flask.ext.sqlalchemy import SQLAlchemy
from flask_mail import Mail
from FEUtils import FEUtils
# .. and more imports of various extensions ..
db = SQLAlchemy()
app = Flask(__name__)
db.init_app(app)
mail = Mail(app)
fe_utils = FEUtils(db,mail,app.config)
# Flask code..
if __name__ == '__main__':
app.run()
and FEUtils.py:
from models import User
class FEUtils(object):
def __init__(self,db,mail,config):
self.session = db.session # to access database
self.mail = mail # to send emails
self.config = config # to access app config dictionary
def count_users(self): # example helper method
return self.session.query(User).count()
This all works fine, but seems cumbersome. I'd like the helper class to inherit the various extension instances from the main module, and be able to access the flask config parameters from within the helper class, without passing each when the helper class is instantiated.
Asked differently, is there a way to have the helper class behave as if each of its methods was defined in the main module in an elegant way?
I am using an app factory to initialize my app. In it, I import all the blueprints and register them one by one. Is there a way I can move the import and register statements to a different file or inform the factory about them without referencing them all individually?
def create_app(config_filename):
app = Flask(__name__)
app.config.from_object(config_filename)
from app.users.models import db
db.init_app(app)
from app.users.views import users
from app.posts.views import posts
app.register_blueprint(posts, url_prefix='/posts')
app.register_blueprint(users, url_prefix='/users')
return app
In my project I'm actually generating the blueprints with another script, so I'd like to be able to generate the registration too by appending to a file or something, rather than trying to modify code in the factory.
Yes, you can import and register the blueprints in some other module. But there's no practical point to this approach, it just moves the imports and register calls somewhere else.
myapp/blueprints.py:
from app.users.views import users
from app.posts.views import posts
def init_app(app):
app.register_blueprint(users, prefix='/users')
app.register_blueprint(posts, prefix='/posts')
myapp/__init__.py:
def create_app():
app = Flask(__name__)
# ...
from myapp import blueprints
blueprints.init_app(app)
# ...
Something more useful might be to tell the app what packages to import from and have the app expect to find a blueprint in some standard location for each package. Assuming the blueprint variable will always have the same name as the package, is defined in views, and has the same prefix as the name:
from werkzeug.utils import import_string
def create_app():
app = Flask(__name__)
# ...
for name in ('users', 'posts'):
bp = import_string('myapp.{0}.views:{1}'.format(name, name))
app.register_blueprint(bp, prefix='/{0}'.format(name))
# ...