烧瓶和烧瓶-插座IO集成和导入错误

Question

我正在尝试将Flask-SocketIO与我的Flask应用程序集成。

由于某种原因，我收到由 from .. import socketio我 events.py，我不知道为什么。

我将非常感谢所提供的任何帮助。

app.py

from factory import create_app
app = create_app()

factory.py

from flask import Flask
from flask.ext.socketio import SocketIO
from flask.ext.login import LoginManager
import os
from celery import Celery
lm = LoginManager()
socketio = SocketIO()
lm.login_view = 'main.login'
lm.session_protection = 'strong'

def create_app():
    app = Flask(__name__)
    app.config.from_object(os.environ.get('APP_CONFIG'))  # export APP_CONFIG=settings.Development
    lm.init_app(app)
    from project.main import main as main_blueprint
    app.register_blueprint(main_blueprint)
    socketio.init_app(app)
    return app

主/____init____.py

from flask import Blueprint
main = Blueprint('main', __name__)
import views, events

主要/事件.py

# from __future__ import absolute_import
from uuid import uuid4
from flask import current_app, session, request
from flask.ext.socketio import emit, disconnect
from .. import socketio
@socketio.on('status', namespace='/events')
def events_message(message):
    print 'socketio.on: status'
    emit('status', {'status': message['status']})

@socketio.on('disconnect request', namespace='/events')
def disconnect_request():
    print 'socketio.on: disconnect request'
    emit('status', {'status': 'Disconnected!'})
    disconnect()

@socketio.on('connect', namespace='/events')
def events_connect():
    print 'socketio.on: connect'
    websocket_id = str(uuid4())
    session['websocket_id'] = websocket_id
    current_app.clients[websocket_id] = request.namespace
    emit('websocket_id', {'websocket_id': websocket_id})

Answer 1

将main/____init____.py重命名为 main/__init__.py，您可以执行以下操作：

from . import socketio

如果socketio像你说的那样在__init__.py中定义。但是，它看起来像是在 factory.py 中定义的，在这种情况下，您可以执行以下操作：

from ..factory import socketio