How do I split Flask models out of app.py without

问题:

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

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

db = SQLAlchemy(app)
migrate = Migrate(app, db)

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

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

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

This works great as a simple play example, but I have more than a single model and I don't want to define the models in both this script and the one that defines my application code. Thus, I want to pull them into a model file that I can share between the two.

I attempting to do this, by placing the User class into a models.py and then importing User from there. Unfortunately, that throws a NameError: name 'db' is not defined.

My question(s) are:

• Do I need to use db = SQLAlchemy(app) in my models.py, and if so will this be available in both my migration script and the flask application itself?
• If I can't (or shouldn't) put it in models.py, how do I utilize my models in their own file without passing db all over?

回答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.

回答2:

Here is an idea. You can create a package model in which you define classes as individual .py files. For example, user.py contains the User class.

In the __init__.py file you can define the db variable. You may also include from user import User statement so that it would be possible to access User object directly outside the package. For example you can import the model package somewhere else in the program using import model and then use model.User to directly use the User class.

To use db variable directly in the user.py use from ..model import db which imports db variable defined in the __init__.py file.