I have two files:
app.py
from flask import Flask
from flask_restful import Api
from celery import Celery
from resources.item import Item, ItemList, ItemInsert
from db import db
app = Flask(__name__)
app.config["SQLALCHEMY_DATABASE_URI"] = ""
app.config["SQLALCHEMY_TRACK_MODIFICATIONS"] = False
app.config['CELERY_BROKER_URL'] = ''
app.config['CELERY_RESULT_BACKEND'] = ''
celery = Celery(app.name, broker=app.config['CELERY_BROKER_URL'])
app.secret_key = ""
api = Api(app)
#app.before_first_request
def create_tables():
db.create_all()
api.add_resource(ItemList,"/items")
api.add_resource(Item,"/item/<int:id>")
api.add_resource(ItemInsert,"/items")
#celery.task
def update_row(data):
pass
if __name__ == "__main__":
db.init_app(app)
app.run(debug=True,port=5000)
item.py
from flask_restful import Resource, reqparse
from flask import request
from models.item import ItemModel
class ItemInsert(Resource):
def post(self):
file_task = update_row.apply_async((data,), countdown=3)
return item.json(), 201
As you can see in app.py I have imported classes from item.py, however now my celery (task) function call i.e update_row from item.py is left hanging, since I cannot import from app.py as it will result in a cyclic import. Is there any solution?
With simple project, you could implement the tasks inside app.py as you're doing for now. But with more complicated project, it's better to move the tasks definition into a separated package so that it could mitigate the cyclic import.
Like so:
App and celery configutation
**# app.py**
# App & celery
# ...
Tasks definitions
**# tasks.py**
from project_name.app import celery
#celery.task
def update_row(data):
pass
API
**# resources/item.py**
from project_name.tasks import update_row
# ...
Separate the tasks into another package (tasks package, which is auto discovered by Celery) could help you to prevent cyclic import and also good to maintain the code.
But if you're still want to use the current approach, to prevent cyclic import, you could import it dynamically when calling API:
**# resources/item.py**
# ...
class ItemInsert(Resource):
def post(self):
from project_name.app import update_row
file_task = update_row.apply_async((data,), countdown=3)
return item.json(), 201
celery_app.task is a decorator, so just a regular Python function. It works in the following way: it takes your function, registers it in the celery app and returns a wrapper object with methods delay, apply_async etc. And you can always get a registered task from celery_app.tasks dictionary by its name. Another trick to avoid circular imports is in storing celery_app reference as an attribute of flask_app, and inside request context you can always get current flask app from flask.current_app
app.py
from tasks import register_tasks
...
app = Flask(__name__)
...
app.celery = Celery(app.name, ...)
register_tasks(app.celery)
tasks.py
def update_row(data):
pass
def register_tasks(celery_app):
celery_app.task(update_row, name="update_row")
views.py
from flask import current_app
class ItemInsert(Resource):
def post(self):
update_row = current_app.celery.tasks["update_row"]
file_task = update_row.apply_async((data,), countdown=3)
return item.json(), 201
UPD: indeed the most canonical way is to use autodiscovery of tasks:
myapp/tasks.py
from celery import shared_task
#shared_task
def update_row(data):
pass
myapp/app.py
celery_app = Celery(...)
celery_app.set_default()
celery_app.autodiscover_tasks(["myapp"], force=True)
myapp/views.py
from .tasks import update_row
def index_view():
update_row.delay(...)
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 am running unittests in a Flask app and I keep getting 404 when views.py file is not imported even though it is not used. I have such tests.py package:
import unittest
from presence_analyzer import main, utils
from presence_analyzer import views
class PresenceAnalyzerViewsTestCase(unittest.TestCase):
def setUp(self):
self.client = main.app.test_client()
def test_mainpage(self):
resp = self.client.get('/')
self.assertEqual(resp.status_code, 302)
When I delete views import the described problem occurs. Views are organized in a similar way to this:
from presence_analyzer.main import app
#app.route('/')
def mainpage():
return redirect('/static/presence_weekday.html')
And the main.py file:
import os.path
from flask import Flask
app = Flask(__name__) # pylint: disable=invalid-name
app.config.update(
DEBUG=True,
)
I guess it's something similar to what happened in this case, so I'm trying to change the application so that I don't have to make this dumb imports while testing. I've been trying to make use of the answer from above, but still can't make it work and these docs don't seem helpful. What am I doing wrong? main.py:
from flask.blueprints import Blueprint
PROJECT_NAME = 'presence_analyzer'
blue_print = Blueprint(PROJECT_NAME, __name__)
def create_app():
app_to_create = Flask(__name__) # pylint: disable=invalid-name
app_to_create.register_blueprint(blue_print)
return app_to_create
app = create_app()
views.py:
from presence_analyzer.main import app, blue_print
#blue_print.route('/')
def mainpage():
return redirect('/static/presence_weekday.html')
And tests.py has remained unchanged.
You must import views, or the route will not be registered. No, you are not executing the views directly, but importing executes code all module-level code. Executing code calls route. route registers the view function. You cannot get around needing to import a module in order to use the module.
I'm currently using the Flask Application Factory pattern with Blueprints. The issue that I'm having is how do I access the app.config object outside of the application factory?
I don't need all the configuration options from the Flask app. I just need 6 keys. So the current way I do this is when the create_app(application factory) is called, I basically create a global_config dictionary object and I just set the global_config dictionary to have the 6 keys that I need.
Then, the other modules that need those configuration options, they just import global_config dictionary.
I'm thinking, there has to be a better way to do this right?
So, on to the code
My current init.py file:
def set_global_config(app_config):
global_config['CUPS_SAFETY'] = app_config['CUPS_SAFETY']
global_config['CUPS_SERVERS'] = app_config['CUPS_SERVERS']
global_config['API_SAFE_MODE'] = app_config['API_SAFE_MODE']
global_config['XSS_SAFETY'] = app_config['XSS_SAFETY']
global_config['ALLOWED_HOSTS'] = app_config['ALLOWED_HOSTS']
global_config['SQLALCHEMY_DATABASE_URI'] = app_config['SQLALCHEMY_DATABASE_URI']
def create_app(config_file):
app = Flask(__name__, instance_relative_config=True)
try:
app.config.from_pyfile(config_file)
except IOError:
app.config.from_pyfile('default.py')
cel.conf.update(app.config)
set_global_config(app.config)
else:
cel.conf.update(app.config)
set_global_config(app.config)
CORS(app, resources=r'/*')
Compress(app)
# Initialize app with SQLAlchemy
db.init_app(app)
with app.app_context():
db.Model.metadata.reflect(db.engine)
db.create_all()
from authenication.auth import auth
from club.view import club
from tms.view import tms
from reports.view import reports
from conveyor.view import conveyor
# Register blueprints
app.register_blueprint(auth)
app.register_blueprint(club)
app.register_blueprint(tms)
app.register_blueprint(reports)
app.register_blueprint(conveyor)
return app
An example of a module that needs access to those global_config options:
from package import global_config as config
club = Blueprint('club', __name__)
#club.route('/get_printers', methods=['GET', 'POST'])
def getListOfPrinters():
dict = {}
for eachPrinter in config['CUPS_SERVERS']:
dict[eachPrinter] = {
'code': eachPrinter,
'name': eachPrinter
}
outDict = {'printers': dict, 'success': True}
return jsonify(outDict)
There has to be a better way then passing a global dictionary around the application correct?
There is no need to use global names here, that defeats the purpose of using an app factory in the first place.
Within views, such as in your example, current_app is bound to the app handling the current app/request context.
from flask import current_app
#bp.route('/')
def example():
servers = current_app.config['CUPS_SERVERS']
...
If you need access to the app while setting up a blueprint, the record decorator marks functions that are called with the state the blueprint is being registered with.
#bp.record
def setup(state):
servers = state.app.config['CUPS_SERVERS']
...
In order to simplify the __init__.py main module, I want to push helper functionality to a different file/class. This requires passing many flask extensions instances when initializing the class, which seems inelegant. My current structure is as follows:
__init__.py:
from flask import Flask, render_template,request
from flask.ext.sqlalchemy import SQLAlchemy
from flask_mail import Mail
from FEUtils import FEUtils
# .. and more imports of various extensions ..
db = SQLAlchemy()
app = Flask(__name__)
db.init_app(app)
mail = Mail(app)
fe_utils = FEUtils(db,mail,app.config)
# Flask code..
if __name__ == '__main__':
app.run()
and FEUtils.py:
from models import User
class FEUtils(object):
def __init__(self,db,mail,config):
self.session = db.session # to access database
self.mail = mail # to send emails
self.config = config # to access app config dictionary
def count_users(self): # example helper method
return self.session.query(User).count()
This all works fine, but seems cumbersome. I'd like the helper class to inherit the various extension instances from the main module, and be able to access the flask config parameters from within the helper class, without passing each when the helper class is instantiated.
Asked differently, is there a way to have the helper class behave as if each of its methods was defined in the main module in an elegant way?