I'm having some trouble understanding how to incorporate Flask-Pymongo. My app is initiated from my rrapp.py Inside of this file, I have
rrapp.py
#
# Imports up here
#
app = Flask(__name__)
mongo = PyMongo(app)
# Code down here
Now, to use this, I simply do mongo.db.users.find(). This works fine.
Now, say I have another file called userservice.py that I call methods from one of my endpoints within rrapp.py. How do I incorporate PyMongo(app) in my userservice.py file if I don't have access to the app object? Or am I missing something obvious here?
you should first define mongo oustside create_app to have access to it from inside other files.
then init_app with that like the following:
from flask import Flask, current_app
from flask_pymongo import PyMongo
mongo = PyMongo()
def create_app(config_name):
app = Flask(__name__, instance_relative_config=False)
app.config.from_object(app_config[config_name])
# INIT EXTENSIONS ----------------------
mongo.init_app(app)
return app
then in any file you can import mongo from above file. for example:
from ../factory import mongo
Related
I am trying to define a mongodb object inside main flask app. And I want to send that object to one of the blueprints that I created. I may have to create more database objects in main app and import them in different blueprints. I tried to do it this way.
from flask import Flask, render_template
import pymongo
from admin_component.bp1 import bp_1
def init_db1():
try:
mongo = pymongo.MongoClient(
host='mongodb+srv://<username>:<passwrd>#cluster0.bslkwxdx.mongodb.net/?retryWrites=true&w=majority',
serverSelectionTimeoutMS = 1000
)
db1 = mongo.test_db1.test_collection1
mongo.server_info() #this is the line that triggers exception.
return db1
except:
print('Cannot connect to db!!')
app = Flask(__name__)
app.register_blueprint(bp_1, url_prefix='/admin') #only if we see /admin in url we gonna extend things in bp_1
with app.app_context():
db1 = init_db1()
#app.route('/')
def test():
return '<h1>This is a Test</h1>'
if __name__ == '__main__':
app.run(port=10001, debug=True)
And this is the blueprint and I tried to import the init_db1 using current_app.
from flask import Blueprint, render_template, Response, request, current_app
import pymongo
from bson.objectid import ObjectId
import json
bp_1 = Blueprint('bp1', __name__, static_folder='static', template_folder='templates')
print(current_app.config)
db = current_app.config['db1']
But it gives this error without specifying more details into deep.
raise RuntimeError(unbound_message) from None
RuntimeError: Working outside of application context.
This typically means that you attempted to use functionality that needed
the current application. To solve this, set up an application context
with app.app_context(). See the documentation for more information.
Can someone point out what am I doing wrong here??
The idea you are attempting is correct; however it just needs to be done a little differently.
First, start by declaring your mongo object in your application factory:
In your app/__init__.py:
import pymongo
from flask import Flask
mongo = pymongo.MongoClient(
host='mongodb+srv://<username>:<passwrd>#cluster0.bslkwxdx.mongodb.net/?retryWrites=true&w=majority',
serverSelectionTimeoutMS = 1000
)
# Mongo is declared outside of function
def create_app(app):
app = Flask(__name__)
return app
And then in your other blueprint, you would call:
from app import mongo # This right here will get you the mongo object
from flask import Blueprint
bp_1 = Blueprint('bp1', __name__, static_folder='static', template_folder='templates')
db = mongo
I am trying to use SQLAlchemy not in a view function (I was doing something like this with Flask-APSheduler).
I know that there were already a lot of topics related to this theme, but none of them were helpful to me.
So, first of all I will show my code:
./run.py
from app import create_app
from flask_config import DevConfig, ProdConfig
app = create_app(DevConfig)
if __name__ == '__main__':
app.run(host='0.0.0.0', port=80)
./app/__init__.py
from flask import Flask
from .node import node
from .models import db
def create_app(app_config=None):
app = Flask(__name__, instance_relative_config=False)
app.config.from_object(app_config)
db.init_app(app)
app.register_blueprint(node)
return app
./app/models.py
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
class Users(BaseFuncs, db.Model):
...
./app/node.py
from flask import Blueprint, request
from .bot import bot, secret
import telebot
node = Blueprint('node', __name__)
#node.route('/{}'.format(secret), methods=['POST'])
def handler():
bot.process_new_updates([telebot.types.Update.de_json(request.get_data().decode('utf-8'))])
return 'ok', 200
./app/bot.py
from flask import current_app as app
...
#bot.message_handler(commands=['test'])
def cmd_test(message):
with app.app_context():
print(Users.query.filter_by(id=0).first())
So when I am trying to call cmd_test from my application I am getting this error:
RuntimeError: Working outside of application context.
This typically means that you attempted to use functionality that needed
to interface with the current application object in some way. To solve
this, set up an application context with app.app_context(). See the
documentation for more information.
I tried to use g variable and before_request methods, because every time before calling the database there is a call to the route 'handler', but this also doesn't work.
I also tried to use db.get_app(), but there was no effect.
So my question is how to call database right outside the views?
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 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 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?
You can import the object app into database.py by putting:
from index import app
in database.py.
Edited answer after comment:
Simply use
from index import app
in database.py.
Alternatively, with the
import index
statement, use index.app instead of app only.
This should help you get out of python's import hell.
Btw: Not sure which IDE you are using. I like pycharm a lot. Using it you can refactor code and issues such as above are prevented automagically.