使用与当前应用程序交互所需的功能是否意味着在 Flask 中创建视图？

Question

在我的 Flask 应用程序中 app/__init__.py 我包含了以下功能

def create_app(config_filename=None):
    _app = Flask(__name__, instance_relative_config=True)
    with _app.app_context():
        print "Creating web app",current_app.name
        cors = CORS(_app, resources={r'/*': {"origins": "*"}})
        sys.path.append(_app.instance_path)
        _app.config.from_pyfile(config_filename)
        from config import app_config
        config = app_config[_os.environ.get('APP_SETTINGS', app_config.development)]
        _app.config.from_object(config)
        register_blueprints(_app)
        # _app.app_context().push()
        return _app

现在我还有 /app/datastore/core.py 其中我有

import peewee
import os

from flask import g, current_app
cfg = current_app.config    
dbName = 'clinic_backend'


def connect_db():
    """Connects to the specific database."""
    return peewee.MySQLDatabase(dbName, 
    user=cfg['DB_USER'], 
    host=cfg['DB_HOST'], 
    port=3306, 
    password=cfg['DB_PWD']
    )

def get_db():
    """ Opens a new database connection if there is none yet for the
    current application context.
    """
    if not hasattr(g, 'db'):
        g.db = connect_db()
    return g.db

当我创建我的运行我的应用程序 start.py 时，它会创建应用程序对象，但是当我尝试在浏览器中访问 URL 时，我收到错误提示

 File "/Users/ciasto/Development/python/backend/app/datastore/core.py",
 line 31, in get_db
     g.db = connect_db()   File "/Users/ciasto/Development/python/backend/venv/lib/python2.7/site-packages/werkzeug/local.py",
 line 364, in <lambda>
     __setattr__ = lambda x, n, v: setattr(x._get_current_object(), n, v)   File
 "/Users/ciasto/Development/python/backend/venv/lib/python2.7/site-packages/werkzeug/local.py",
 line 306, in _get_current_object
     return self.__local()   File "/Users/ciasto/Development/python/backend/venv/lib/python2.7/site-packages/flask/globals.py",
line 44, in _lookup_app_object
    raise RuntimeError(_app_ctx_err_msg) 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.

我做错了什么？

Answer 1

what am I doing wrong here?

几乎所有内容。

Peewee 已经将数据库连接公开为本地线程，因此您正在做的事情毫无意义。特别是通过阅读您的评论，您只是在尝试添加连接挂钩。

peewee 文档非常清晰：http://docs.peewee-orm.com/en/latest/peewee/database.html#flask

顺便说一句，阅读该死的文档。您已经发布了多少可以通过阅读文档轻松回答的问题？

使用与当前应用程序交互所需的功能是否意味着在 Flask 中创建视图？

Does using functionality needed to interface with current app means creating view in flask?

python

flask

peewee