How do I split Flask models out of app.py without passing db object all over?

Question

I\'d like to use Flask-Migrate and am looking at their example:

from flask import Flask
from flask.ext.sqlalchemy import SQLAlchemy
from flask.ext.script imp         


        

Answer 1

Partitioning such a small application into modules is tricky, because you find a lot of cases where the two modules that you create need to mutually import each other, creating circular dependencies.

I recommend that you look at how you can structure a larger application properly, using an app factory function and delayed initialization of all extensions. An example application that does this is the Flasky app featured in my book.

All that said, it is possible to separate the application into two parts, you just need to be careful with where you place the import statements. In the example below, I decided to move the creation of the db instance and the User model into a models.py file.

Here is the main application's module:

from flask import Flask
from flask.ext.script import Manager
from flask.ext.migrate import Migrate, MigrateCommand

app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///app.db'

from models import db  # <-- this needs to be placed after app is created
migrate = Migrate(app, db)

manager = Manager(app)
manager.add_command('db', MigrateCommand)

if __name__ == '__main__':
    manager.run()

And here is models.py:

from __main__ import app
from flask.ext.sqlalchemy import SQLAlchemy

db = SQLAlchemy(app)

class User(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(128))

Here the main module will create app, and only then will import models.py. When models.py tries to import app from the main module, it has already been created. If you move from models import db to the top of the file with the other imports this code breaks.