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?
Related
I'm writing a Flask app and I instantiate a mongo database in main.py.
I've got a submodule called user.py that holds class User. main.py takes login credentials and sends it to the class User which handles the rest.
How can I cleanly pass my mongo instance to the User class? I've tried static variables in a config.py file but they don't work because the variables are always None when user.py tries to use them.
Right now I'm resorting to passing in mongo as a parameter, but this seems like a dirty way to do it considering there will be many modules. Here's my code;
# Setup app and database
app = Flask(__name__)
app.config['MONGO_URI'] = 'mongodb://'
mongo = PyMongo(app)
You can import mongo directly into your user.py module.
To avoid circular import error, you just have to move the imports to the bottom of the file. As long as you import User into main after app is defined, it should resolve the circular import:
user.py
from .main import mongo
# class User():
# ... Your Code
main.py
from flask import Flask
from flask_pymongo import PyMongo
app = Flask(__name__)
mongo = PyMongo(app)
from .user import User
Moving imports to the bottom are normally not a good idea, but in Flask is quite common. Below is an example of a similar scenario from the official flask documentation:
# app.py
from flask import Flask
app = Flask(__name__)
import app.views
# views.py
from app import app
#app.route('/')
def index():
return 'Hello World!'
http://flask-.readthedocs.io/en/0.6/patterns/packages/#simple-packages
Do the importing at the bottom of the file [...]
Every Python programmer hates them, and yet we just added some: circular imports (That’s when two modules depend on each other. In this case views.py depends on init.py). Be advised that this is a bad idea in general but here it is actually fine. The reason for this is that we are not actually using the views in init.py and just ensuring the module is imported and we are doing that at the bottom of the file.
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.
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 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))
# ...
I want to structure my Flask app something like:
./site.py
./apps/members/__init__.py
./apps/members/models.py
apps.members is a Flask Blueprint.
Now, in order to create the model classes I need to have a hold of the app, something like:
# apps.members.models
from flask import current_app
from flaskext.sqlalchemy import SQLAlchemy
db = SQLAlchemy(current_app)
class Member(db.Model):
# fields here
pass
But if I try and import that model into my Blueprint app, I get the dreaded RuntimeError: working outside of request context. How can I get a hold of my app correctly here? Relative imports might work but they're pretty ugly and have their own context issues, e.g:
from ...site import app
# ValueError: Attempted relative import beyond toplevel package
The flask_sqlalchemy module does not have to be initialized with the app right away - you can do this instead:
# apps.members.models
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
class Member(db.Model):
# fields here
pass
And then in your application setup you can call init_app:
# apps.application.py
from flask import Flask
from apps.members.models import db
app = Flask(__name__)
# later on
db.init_app(app)
This way you can avoid cyclical imports.
This pattern does not necessitate the you place all of your models in one file. Simply import the db variable into each of your model modules.
Example
# apps.shared.models
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
# apps.members.models
from apps.shared.models import db
class Member(db.Model):
# TODO: Implement this.
pass
# apps.reporting.members
from flask import render_template
from apps.members.models import Member
def report_on_members():
# TODO: Actually use arguments
members = Member.filter(1==1).all()
return render_template("report.html", members=members)
# apps.reporting.routes
from flask import Blueprint
from apps.reporting.members import report_on_members
reporting = Blueprint("reporting", __name__)
reporting.route("/member-report", methods=["GET","POST"])(report_on_members)
# apps.application
from flask import Flask
from apps.shared import db
from apps.reporting.routes import reporting
app = Flask(__name__)
db.init_app(app)
app.register_blueprint(reporting)
Note: this is a sketch of some of the power this gives you - there is obviously quite a bit more that you can do to make development even easier (using a create_app pattern, auto-registering blueprints in certain folders, etc.)
an original app.py: https://flask-sqlalchemy.palletsprojects.com/en/2.x/quickstart/
...
app = flask.Flask(__name__)
app.config['DEBUG'] = True
app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:////tmp/test.db'
db = flask.ext.sqlalchemy.SQLAlchemy(app)
class Person(db.Model):
id = db.Column(db.Integer, primary_key=True)
...
class Computer(db.Model):
id = db.Column(db.Integer, primary_key=True)
...
# Create the database tables.
db.create_all()
...
# start the flask loop
app.run()
I just splitted one app.py to app.py and model.py without using Blueprint. In that case, the above answer dosen't work. A line code is needed to work.
before:
db.init_app(app)
after:
db.app = app
db.init_app(app)
And, the following link is very useful.
http://piotr.banaszkiewicz.org/blog/2012/06/29/flask-sqlalchemy-init_app/