Flask：无法从 socket.io 监听器中访问 current_app答案

【问题标题】：Flask: Unable to access current_app from within socket.io listenerFlask：无法从 socket.io 监听器中访问 current_app
【发布时间】：2013-03-24 12:13:20
【问题描述】：

我正在尝试从侦听器中访问 current_app，以便我可以使用应用配置值来订阅哪个频道。但是我收到“RuntimeError：在应用程序上下文之外工作”。

这里是有问题的代码：

from flask import Blueprint, Response, request, current_app
from socketio import socketio_manage
from socketio.namespace import BaseNamespace
from redis import StrictRedis
import pprint

socketapp = Blueprint('socketapp', __name__)


class MessageNamespace(BaseNamespace):
    def listener(self):
        pprint.pprint(current_app.config)
        r = StrictRedis()
        p = r.pubsub()
        p.subscribe('message-channel')

        messages = r.lrange('message', 0, -1)
        self.emit('message-message', ''.join(messages))

        for m in p.listen():
            if m['type'] == 'message':
                self.emit('message-message', m['data'])

    def on_subscribe(self):
        self.spawn(self.listener)


@socketapp.route('/socket.io/<path:remaining>')
def socketio(remaining):
    try:
        socketio_manage(request.environ, {'/messages': MessageNamespace}, request)
    except BaseException:
        pass
    return Response()


@socketapp.route('/message', methods=['GET'])
def say():
    msg = request.args.get('msg', None)
    if msg:
        r = StrictRedis(host=current_app.config['REDIS_HOST'])
        r.rpush('message', msg)
        r.publish('message-channel', msg)

        return Response('Message sent!')
    else:
        return Response('Please specify your message in the "msg" parameter')

【问题讨论】：

你见过这个sn-p - flask.pocoo.org/snippets/105
@SeanVieira，您的 sn-p 在命名空间构造函数中使用 current_app，但它不可用。我根据您的代码编写了我的 hack 版本。请查看。

标签： python redis socket.io flask gevent

【解决方案1】：

current_app 仅在处理 HTTP 请求时有效（它是暂时指向实际应用程序对象的代理）。您需要从此模块访问实际的应用程序对象，或者通过current_app._get_current_object() 窃取对它的引用。

【讨论】：

【解决方案2】：

Sean 的 sn-p 对我不起作用。我基于它编写了以下 hack：

@bp.route('/<path:remaining>')
def socketio(remaining):
    app = current_app._get_current_object()
    try:
        # Hack: set app instead of request to make it available in the namespace.
        socketio_manage(request.environ, {'': ChatNamespace}, app)
    except:
        app.logger.error("Exception while handling socket.io connection", exc_info=True)
    return Response()


class ChatNamespace(BaseNamespace, RoomsMixin, BroadcastMixin):
    def __init__(self, environ, ns_name, request=None):
        self.context = None
        if request:
            # Hack: initialize context with app that was set instead of request. Then miss it in parent constructor call.
            app = request
            self.context = app.request_context(environ)
            self.context.push()
            app.preprocess_request()
        super(ChatNamespace, self).__init__(environ, ns_name)

def log(self, message):
        # Now we can access app and other goodies.
        self.context.app.logger.info("[{0}] {1}".format(self.socket.sessid, message))

def disconnect(self, *args, **kwargs):
        if self.context:
            self.context.pop()
        super(ChatNamespace, self).disconnect(*args, **kwargs)

【讨论】：