I am following the Flask SQLalchemy Quickstart which has all of the code in a single file:
Here is my initial index.py:
from flask import Flask
from flask.ext.sqlalchemy import SQLAlchemy
app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:////tmp/test.db'
db = SQLAlchemy(app)
# [snip] - some classes related to SQLAlchemy are here
if __name__ == '__main__':
app.run(host='0.0.0.0')
I want to split the code up a bit, so I created a separate file called database.py which will contain all of the database related code, and be imported as a module.
I modified my index.py to look like this:
from flask import Flask
# Import my database module
import database
app = Flask(__name__)
if __name__ == '__main__':
app.run(host='0.0.0.0')
And the file database.py:
from flask.ext.sqlalchemy import SQLAlchemy
app.config['SQLALCHEMY_DATABASE_URI]'] = 'sqlite:////tmp/test.db'
db = SQLAlchemy(app)
# [snip] - some classes related to SQLAlchemy are here
Obviously when I try to run this code I get the following error:
File "database.py", line 5, in <module>
app.config['SQLALCHEMY_DATABASE_URI]'] = 'sqlite:////tmp/test.db'
NameError: name 'app' is not defined
I can see that this is because the app object only exists within the parent module.
I could put the following lines back into index.py:
app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:////tmp/test.db'
db = SQLAlchemy(app)
But this creates a similar problem, whereby db is not available within the database.py file.
What is the correct way to code this?
