My folder structure is the following:
- app.py
app
- __init__.py
- database.py
in app.py I have:
from app import create_app
app = create_app()
my init.py looks something like:
from flask import Flask
from app.database import db_session, init_db
db = SQLAlchemy()
def create_app():
myapp = Flask(__name__, static_folder='static', static_url_path='/static', template_folder="templates")
myapp.config.from_object('config.Config')
db.init_app(myapp)
migrate.init_app(myapp, db)
# loading blueprints
from app.core_bp import core_bp
myapp.register_blueprint(core_bp, url_prefix='/', template_folder="templates")
return myapp
and database.py looks like this
from sqlalchemy import create_engine
from sqlalchemy.orm import scoped_session, sessionmaker
from sqlalchemy.ext.declarative import declarative_base
engine = create_engine('sqlite:///rapporteur.db')
db_session = scoped_session(sessionmaker(autocommit=False,autoflush=False,bind=engine))
That sqlite path is currently hardcoded to raporteur.db but it should be loaded from config, because I don't want it hardcoded, but unfortunately the flask app is not yet loaded. So how would I do this?
You can use flask-sqlalchemy and set the SQLALCHEMY_DATABASE_URI in the config file.
class Config:
SQLALCHEMY_DATABASE_URI = 'sqlite:///rapporteur.db'
then in your init.py
from youApp import Config
def create_app(config_class=Config)
app = Flask(__name__)
app.config.from_object(config_class)
db = SQLAlchemy()
db.init_app(app)
return app
Related
Hello everyone I have this issue I am not able to correct:
KeyError: 'A secret key is required to use CSRF.'
I am now using Flask with Blueprint.
I am not using CSRF at all but I think the LoginForm is.
I structured my project with Blueprint.
Before that, everything was find.
Here is my init.py file:
from flask import Flask
from flask_login import LoginManager
from flask_bcrypt import Bcrypt
from flask_sqlalchemy import SQLAlchemy
from flask_migrate import Migrate
from flask_admin import Admin
from flask_admin.contrib.sqla import ModelView
from flask.config import Config
from flask_wtf.csrf import CSRFProtect
db = SQLAlchemy()
migrate = Migrate(db)
bcrypt = Bcrypt()
csrf = CSRFProtect()
login_manager = LoginManager()
login_manager.login_view = "login"
login_manager.login_message_category = "info"
from Flask import models
from Flask.models import User
admin = Admin(name='Admin')
def create_app(config_class=Config):
app = Flask(__name__)
app.config.from_object(Config)
admin.init_app(app)
db.init_app(app)
csrf.init_app(app)
login_manager.init_app(app)
migrate.init_app(app)
bcrypt.init_app(app)
db.init_app(app)
from Flask.users.routes import users
app.register_blueprint(users)
return app
This is my config.py file:
import os
class Config:
SECRET_KEY = "ef2006629e09b70e55a6fb95c4e3a538"
SQLALCHEMY_DATABASE_URI = "sqlite:///site.db"
# WTF_CSRF_SECRET_KEY= "bjk567nvhbvj63vg363vghvghv3768vgfbkijvr784"
# CSRF_ENABLED = True
Thank you for your help !
You should create a SECRET_KEY=<Your secret key here> property in your configuration. It must be a difficult string.
I found the problem
I was not calling properly my config.py file
In my init.py file I change the line:
from flask.config import Config
with the line:
from Flask.config import Config
Flask is the name of my file, which is different from flask.
I should have found another name
I have the following structure:
app_dir/
| myapi/
| __init__.py
| myapi_app.py
where myapi_app.py is
from myapi import create_app, db
app = create_app()
and myapi/__init__.py is
import logging
import os
from logging.handlers import RotatingFileHandler
from flask import Flask, request, current_app
from flask_sqlalchemy import SQLAlchemy
from myapi.config import Config
db = SQLAlchemy()
def create_app(config_class=Config):
app = Flask(__name__)
app.config.from_object(config_class)
db.init_app(app)
...
return app
When I set FLASK_APP=myapi_app.py and run flask run from the app_dir directory, the flask service starts. However, when I make a request, I get the following error: flask.cli.NoAppException: Could not import "myapi_app". Where am I going wrong?
Your problem is that you are setting $FLASK_APP to the file in which the app variable is stored, you should instead set it to the python object path, e.g.
FLASK_APP=myapi_app:app
However, this is not necessary, as you could also just do:
FLASK_APP=myapi
as Flask will look for a create_app function in the package on its own.
Following this tutorial on how to structure a Flask app, I have:
project/
__init__.py
app.py
models/
__init__.py
base.py
base.py
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
models/__init__.py
from .base import db
def init_app(app):
db.init_app(app)
project/__init__.py
from flask import Flask
def create_app()
from . import models, routes, services
app = Flask(__name__)
models.init_app(app)
# routes.init_app(app)
# services.init_app(app)
return app
finally, in app.py, I try to run it:
from . import create_app
app = create_app()
if __name__ == '__main__':
app.run(use_reloader=True, threaded=True, debug=True)
but I'm getting the error:
from . import create_app
ValueError: Attempted relative import in non-package
Am I building it right, what am I doing wrong?
I guess you are running your program by:
python project/app.py
In this case, you are not treat your "project" as a python package, which will raise the error you got. Instead, you can run your project with:
FLASK_APP=project.app flask run
I've a question related to Flask-migrate.
I'm creating a set of web services with Flask. I've split each web service in his own package in a python application.
The application structure look like this:
MyApp
WS1
models.py
WS2
models.py
CommonPackage
models.py
How can I import all the modules and init the db? I've tried to import them all manually but is not working.
I know that it works if I import the "app" from WS1 or Ws2 separately, but I would like to do this in a single operation, is this possible?
Here you can find the code for Flask-migrate:
#!virtualenv/bin/python
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from flask_migrate import Migrate, MigrateCommand
from flask_script import Manager
from config import SQLALCHEMY_DATABASE_URI
app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = SQLALCHEMY_DATABASE_URI
db = SQLAlchemy(app)
migrate = Migrate(app, db)
manager = Manager(app)
manager.add_command('db', MigrateCommand)
from WS1.models import Class1, Class2, Class3 <--- This is not importing
from WS2.models import Class4, Class5, Class6 <--- This is not importing
if __name__=='__main__':
manager.run()
This is the modules where I initialize SQLAlchemy session:
from sqlalchemy import create_engine
from sqlalchemy.orm import scoped_session, sessionmaker
from sqlalchemy.ext.declarative import declarative_base
from config import SQLALCHEMY_DATABASE_URI
engine = create_engine(SQLALCHEMY_DATABASE_URI, convert_unicode = True)
db_session = scoped_session(sessionmaker(autocommit=False,
autoflush=False,
bind=engine))
Base = declarative_base()
Base.query = db_session.query_property()
All the models import this module and inherit from Base.
Thanks a lot,
Enrico
When you are defining a custom declarative base you're not really using Flask-SQLAlchemy anymore—and Flask-Migrate reads the models from Flask-SQLAlchemy's internal declarative base. (That is why you have to initialize Migrate with both app and db).
Flask has an answer to the import cycle problem: Most Flask extensions have an .init_app() method to defer initializing the extension where necessary.
In the module where you currently create Base by hand, just do db = SQLAlchemy() and then in your models import db and use db.Models as your declarative base instead of Base. In your app.py, do the following:
from dbpackagename import db
app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = SQLALCHEMY_DATABASE_URI
db.init_app(app)
for example:
from flask import Flask
from flask.ext.admin import Admin, BaseView, expose
class MyView(BaseView):
#expose('/')
def index(self):
return self.render('index.html')
app = Flask(__name__)
admin = Admin(app)
admin.add_view(MyView(name='Hello'))
app.run()
but, if I need a new file, called 'views.py', how can I add a view into views.py to admin?
Do I need to use a blueprint?
For my project I actually made a child class of Blueprint that supports flask admin:
from flask import Blueprint
from flask_admin.contrib.sqla import ModelView
from flask_admin import Admin
class AdminBlueprint(Blueprint):
views=None
def __init__(self,*args, **kargs):
self.views = []
return super(AdminBlueprint, self).__init__('admin2', __name__,url_prefix='/admin2',static_folder='static', static_url_path='/static/admin')
def add_view(self, view):
self.views.append(view)
def register(self,app, options, first_registration=False):
admin = Admin(app, name='microblog', template_mode='adminlte')
for v in self.views:
admin.add_view(v)
return super(AdminBlueprint, self).register(app, options, first_registration)
For details you may like to read my blog here: http://blog.sadafnoor.me/blog/how-to-add-flask-admin-to-a-blueprint/
I am very late for this question, but anyway... My guess is that you want to use the Application Factory pattern and use the Flask-Admin. There is a nice discussion about the underlying problems. I used a very ugly solution, instantiating the Flask-Admin in the init.py file:
from flask_admin.contrib.sqla import ModelView
class UserModelView(ModelView):
create_modal = True
edit_modal = True
can_export = True
def create_app(config_name):
app = Flask(__name__)
app.config.from_object(config[config_name])
db.init_app(app)
# import models here because otherwise it will throw errors
from models import User, Sector, Article
admin.init_app(app)
admin.add_view(UserModelView(User, db.session))
# attach blueprints
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')
return app
You don't need a blueprint for that. In views.py add an import for the admin object you defined in your main project:
from projectmodule import admin
from flask.ext.admin import BaseView, expose
class MyView(BaseView):
#expose('/')
def index(self):
return self.render('index.html')
admin.add_view(MyView(name='Hello'))
and in your main projectmodule file use:
from flask import Flask
from flask.ext.admin import Admin
app = Flask(__name__)
admin = Admin(app)
# import the views
import views
app.run()
e.g. you add import views after the line that sets admin = Admin(app).
I have an flask app with one blueprint (and login/logout to admin).
This is the best solution I found to implement flask admin with some custom features.
My structure as follows:
my_app
main
__init__.py
routes.py
static
templates
__init__.py
config.py
models.py
run.py
Customized admin index view from models.py
from flask_admin import AdminIndexView
class MyAdminIndexView(AdminIndexView):
def is_accessible(self):
return current_user.is_authenticated
def inaccessible_callback(self, name, **kwargs):
return redirect(url_for('main.home'))
Main init.py as follows:
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from flask_admin import Admin
from flask_admin.menu import MenuLink
from my_app.config import Config
# create extensions
db = SQLAlchemy()
admin = Admin()
def create_app(config_class=Config): # default configutation
app = Flask(__name__)
app.config.from_object(Config)
#initialize extensions
db.init_app(app)
...
# customized admin views
from my_app.models import MyAdminIndexView
admin.init_app(app,index_view=MyAdminIndexView())
admin.add_link(MenuLink(name='Home', url='/'))
admin.add_link(MenuLink(name='Logout', url='/logout'))
#blueprint
from my_app.main.routes import main
app.register_blueprint(main)
return app
I think this is the most elegant solution I came up so far.
In order to keep clean the __init__.py root file:
# app/__init__.py
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from app.config import Config
from flask_admin import Admin
db = SQLAlchemy()
admin = Admin(template_mode="bootstrap3")
def create_app(config_class=Config):
app = Flask(__name__)
app.config.from_object(config_class)
db.init_app(app)
admin.init_app(app)
from app.admin import bp
app.register_blueprint(bp, url_prefix='/admin')
return app
Then in the __init__.py of the Blueprint admin app
# app/admin/__init__.py
from flask import Blueprint
from app import admin, db
from app.models import User
from flask_admin.contrib.sqla import ModelView
bp = Blueprint('admin_app', __name__)
admin.name = 'Admin panel'
admin.add_view(ModelView(User, db.session))
# all the rest stuff