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，我收到了由。触发的导入错误我在events.py中有from .. import socketio，我不知道为什么。

我非常感谢您提供的任何帮助。

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/ ____的init ____。PY

from flask import Blueprint

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

主/ events.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中定义的__init__.py就像你说的那样。但是，看起来它在factory.py中定义，在这种情况下你会这样做：

from ..factory import socketio

Flask和Flask-SocketIO集成和导入错误

1 个答案: