带create_app，SQLAlchemy和Celery的flask

Question

我真的很难为Flask，SQLAlchemy和Celery设置正确的设置。我进行了广泛的搜索并尝试了不同的方法，但似乎没有任何效果。我错过了应用程序上下文，或者无法运行工作程序，或者还有其他问题。该结构非常通用，因此我可以构建一个更大的应用程序。

我正在使用：Flask 0.10.1，SQLAlchemy 1.0，Celery 3.1.13，我当前的设置如下：

app / init.py

#Empty
app / config.py

import os
basedir = os.path.abspath(os.path.dirname(__file__))

class Config:

    @staticmethod
    def init_app(app):
        pass

class LocalConfig(Config):
    DEBUG = True
    SQLALCHEMY_DATABASE_URI = r"sqlite:///" + os.path.join(basedir, 
                                 "data-dev.sqlite")
    CELERY_BROKER_URL = 'amqp://guest:guest@localhost:5672//'


config = {
    "local": LocalConfig}
app / exstensions.py

from flask.ext.sqlalchemy import SQLAlchemy
from celery import Celery

db = SQLAlchemy()
celery = Celery()

app / factory.py

from extensions import db, celery
from flask import Flask
from flask import g
from config import config

def create_before_request(app):
    def before_request():
        g.db = db
    return before_request


def create_app(config_name):
    app = Flask(__name__)
    app.config.from_object(config[config_name])

    db.init_app(app)
    celery.config_from_object(config)

    # Register the blueprints

    # Add the before request handler
    app.before_request(create_before_request(app))
    return app

app / manage.py

from factory import create_app

app = create_app("local")

from flask import render_template
from flask import request

@app.route('/test', methods=['POST'])
def task_simple():
    import tasks
    tasks.do_some_stuff.delay()
    return ""

if __name__ == "__main__":
    app.run()

app / models.py

from extensions import db

class User(db.Model):
    __tablename__ = "user"

    id = db.Column(db.Integer, primary_key=True)
    username = db.Column(db.String(128), unique=True, nullable=False)

app / tasks.py

from extensions import celery
from celery.signals import task_prerun
from flask import g, current_app


@task_prerun.connect
def close_session(*args, **kwargs):
    with current_app.app_context():
       # use g.db
       print g

@celery.task()
def do_some_stuff():
    with current_app.app_context():
       # use g.db
       print g

在文件夹应用中：

• 通过以下方式启动开发Web服务器： python.exe manage.py
• 使工人开始： celery.exe worker -A tasks

我收到对我没有任何意义的导入错误。我应该以不同的方式构造应用程序吗？最后，我认为我需要一个非常基本的设置，例如，使用带有工厂模式的Flask，能够使用Flask-SQLAlchmey扩展，并需要一些工作人员来访问数据库。

非常感谢你的帮助。

启动celery工作者时执行回溯。

Traceback (most recent call last):

  File "[PATH]\scripts\celery-script.py", line 9, in <module>
    load_entry_point('celery==3.1.13', 'console_scripts', 'celery')()

  File "[PATH]\lib\site-packages\celery\__main__.py", line 30, in main
    main()

  File "[PATH]\lib\site-packages\celery\bin\celery.py", line 81, in main
    cmd.execute_from_commandline(argv)

  File "[PATH]\lib\site-packages\celery\bin\celery.py", line 769, in execute_from_commandline
    super(CeleryCommand, self).execute_from_commandline(argv)))

  File "[PATH]\lib\site-packages\celery\bin\base.py", line 305, in execute_from_commandline
    argv = self.setup_app_from_commandline(argv)

  File "[PATH]\lib\site-packages\celery\bin\base.py", line 473, in setup_app_from_commandline
    user_preload = tuple(self.app.user_options['preload'] or ())
AttributeError: 'Flask' object has no attribute 'user_options'

更新我根据注释中的建议更改了代码。该工作人员现在可以启动，但是在对它的get请求进行测试时http://127.0.0.1:5000/test。我得到以下回溯：

Traceback (most recent call last):
  File "[PATH]\lib\site-packages\celery\app\trace.py", line 230, in trace_task
    args=args, kwargs=kwargs)

  File "[PATH]\lib\site-packages\celery\utils\dispatch\signal.py", line 166, in send
    response = receiver(signal=self, sender=sender, \**named)

  File "[PATH]\app\stackoverflow\tasks.py", line 7, in close_session
    with current_app.app_context():

  File "[PATH]\lib\site-packages\werkzeug\local.py", line 338, in __getattr__
    return getattr(self._get_current_object(), name)

  File "[PATH]\lib\site-packages\werkzeug\local.py", line 297, in _get_current_object
    return self.__local()

  File "[PATH]\lib\site-packages\flask\globals.py", line 34, in _find_app
    raise RuntimeError('working outside of application context')
RuntimeError: working outside of application context exc, exc_info.traceback)))

Answer 1

我不喜欢current_app的建议。

你的celery对象需要访问应用程序上下文。我在网上找到了一些有关使用工厂功能创建Celery对象的信息。以下示例在没有消息代理的情况下进行了测试。

#factory.py
from celery import Celery
from config import config

def create_celery_app(app=None):
    app = app or create_app(config)
    celery = Celery(__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 app.app_context():
                return TaskBase.__call__(self, *args, **kwargs)

    celery.Task = ContextTask
    return celery

在tasks.py中：

#tasks.py
from factory import create_celery_app
from celery.signals import task_prerun
from flask import g

celery = create_celery_app()

@task_prerun.connect
def celery_prerun(*args, **kwargs):
    #print g
    with celery.app.app_context():
    #   # use g.db
       print g

@celery.task()
def do_some_stuff():
    with celery.app.app_context():
        # use g.db
        g.user = "test"
        print g.user