Flask RESTful create_app使用芹菜问题

Question

I am new to Python Flask RESTful API. Now the project that I'm working on was pre built from the previous developers. I am able to put more logic into this project. Now as requirements goes I need to use CELERY for extensive calculations. I went through different articles in the WEB (eg https://blog.miguelgrinberg.com/post/celery-and-the-flask-application-factory-pattern ) and other articles here but still no luck on this.

[repo: tracker/]
   __init.py__
   app.py
   config.py
   models.py
   celery.py
   tasks.py
   /resources/locate.py
   /resources/create.py

init .py - has the following contents:

from tracker import app

app.py - has the following contents:

from tracker.resources.locate import Locate
from tracker.resources.create import Create

from .celery import create_celery
from .redis_repo import redis_store
from .config import app_config
from .models import db

import collections

def create_app(config_name):

    app = Flask(__name__)
    app.config.from_object(app_config[config_name])

    api = Api(app, catch_all_404s=True)
    api.add_resource(Locate, '/api/v1/locate/<string:ud>', methods=['GET'])
    api.add_resource(Create, '/api/v1/create/<string:ud>', methods=['GET','POST'])

    redis_store.init_app(app)
    db.init_app(app)

       return app

    app = create_app('development')

    if __name__ == '__main__':
        app.run(threaded=True, debug=False)

config.py - has the following contents:

import tempfile

class DevelopmentConfig(Config):
    DEBUG = True
    SQLALCHEMY_POOL_SIZE = 10
    SQLALCHEMY_DATABASE_URI = 'postgres://xxx'
    REDIS_URL = "redis://127.0.0.1:32769/0" 
    CELERY_BROKER_URL = "redis://127.0.0.1:32769/0"
    CELERY_BACKEND = "db+postgresql://xxx" 

app_config = {'development': DevelopmentConfig}

models.py - has the following contents:

from flask_sqlalchemy import SQLAlchemy
from sqlalchemy import func, exc, or_
import datetime
from sqlalchemy_utils import UUIDType

import tracker

db = SQLAlchemy()

class T_Logs(db.Model)
      ....
      ....
      ....    

locate.py - has the following contents:

from flask import request, current_app, jsonify, after_this_request
from flask_restful import Resource
from sqlalchemy import exc

from tracker import models, config, redis_repo, utility
from datetime import datetime, timedelta

import uuid, json

class Locate(Resource):

    def get(self, ud):
    ....
    ....    

Using those documentations or tutorials, it's quite easy to run and understand how these works. But putting it altogether in our implementations it doesnt seem to work.

with celery.py

from celery import Celery

def create_celery(app=None):
    application = app or create_app
    celery = Celery(app.import_name,
                    broker=app.config['CELERY_BROKER_URL'])
    celery.conf.update(app.config)
    TaskBase = celery.Task

    class ContextTask(TaskBase):
        abstract = True

        def __call__(self, *args, **kwargs):
            with application.app_context():
                return TaskBase.__call__(self, *args, **kwargs)

    celery.Task = ContextTask
    return celery

And sample tasks.py

from app import create_celery

@celery.task(name="tasks.add")
def add(a,b):
    return a+ b

So when I run these via visual studio code, I dont see any errors or warnings. But when i run this on the other command prompt to start the worker:

celery -A app.tasks.add worker -l info -P eventlet

It saying from tracker.resources.locate import Locate ModuleNotFoundError: No module named 'tracker'

It never get executed. I'm not yet trying to call the tasks on the resources yet, But I cant proceed yet to understand how to do this.

If anyone could possibly enlighten me with this it would be greatly appreciated.

Answer 1

I have manage to solve the issue, it seems that the problem is on the absolute path.

so made some relevant changes like:

try:
   from tracker.resources.locate import Locate
except ImportError:
   from resources.locate import Locate

And so with the other import statements.

Now my next problem is, How can I call the task let say inside Locate Route in GET method.

class Locate(Resource):

    def get(self, ud):

Let me know if any has an idea on this.