Flask-SocketIO Redis订阅 [英] Flask-SocketIO redis subscribe

问题描述

我正在使用 https://github.com/miguelgrinberg/Flask-SocketIO 实现WebSocket服务器.

I'm using https://github.com/miguelgrinberg/Flask-SocketIO to implement a WebSocket server.

我需要接收来自其他进程的消息(仅订阅)，并向特定房间中的客户端发送消息.

I need to receive a message from another process (only subscribe) and emit for clients in a specific room.

但是，当我尝试发送消息时，出现此错误:

But, when i try to send the message, i got this error:

无法将消息发送到家庭会议室:在请求上下文之外工作.

Could not send message to home room: Working outside of request context.

这是我的代码:

from flask import Flask, request
from flask_socketio import SocketIO, join_room, leave_room, send, rooms
import json
import eventlet
import logging
import redis
import threading

FORMAT = '%(asctime)-15s - %(message)s'
logging.basicConfig(format=FORMAT)
log = logging.getLogger(__name__)

app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret!'
socketio = SocketIO(app, async_mode='eventlet')

.
.
.

def _send_task_message():
    try:
        send(json.dumps({"type":"UPDATE_TASK"}), room='home')
    except Exception as e:
        log.error('Could not send message to home room: %s' % str(e)) 

class Listener(threading.Thread):
    def __init__(self, r, channels):
        threading.Thread.__init__(self)
        self.daemon = True
        self.redis = r
        self.pubsub = self.redis.pubsub()
        self.pubsub.psubscribe(channels)

    def work(self, item):
        if isinstance(item['data'], bytes):
            try:
                msg = item['data'].decode('utf-8')
                decode_msg = json.loads(msg)                
                if decode_msg['type'] == 'UPDATE_TASK':
                    _send_task_message()
            except ValueError as e:
                log.error("Error decoding msg to microservice: %s", str(e))

    def run(self):
        for item in self.pubsub.listen():
            self.work(item)


if __name__ == '__main__':

    r = redis.Redis()
    client = Listener(r, ['/bobguarana/socketio'])
    client.start()

    socketio.run(debug=True, app=app, port=8080)

推荐答案

我解决了将应用程序作为参数传递给类的问题，并按照错误说明的建议使用了它的上下文，但是名称空间也是必需的:

I Solved passing the app as an argument to the class and using it context as suggested by the error description, but the namespace is necessary too:

class Listener(threading.Thread):
    def __init__(self, r, channels, app):
    threading.Thread.__init__(self)
    self.daemon = True
    self.redis = r
    self.pubsub = self.redis.pubsub()
    self.pubsub.psubscribe(channels)
    self.app = app

    def work(self, item):
        with app.app_context():
            if isinstance(item['data'], bytes):
                try:
                    msg = item['data'].decode('utf-8')
                    decode_msg = json.loads(msg)                
                    if decode_msg['type'] == 'UPDATE_TASK':
                        send(json.dumps({"type":"UPDATE_TASK"}), room='home', namespace='/')
                    #_send_task_message()
                except ValueError as e:
                    log.error("Error decoding msg to microservice: %s", str(e))

    def run(self):
        for item in self.pubsub.listen():
            self.work(item)

if __name__ == '__main__':

    r = redis.Redis()
    client = Listener(r, ['/bobguarana/socketio'], app)
    client.start()

    socketio.run(debug=True, app=app, port=8080)

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