烧瓶socketio向特定用户发出

时间:2016-09-10 07:19:27

标签: python flask flask-socketio

我看到有关于此主题的问题,但具体代码未概述。假设我只想发射给第一个客户端。

例如(在events.py中):

clients = []

@socketio.on('joined', namespace='/chat')
def joined(message):
    """Sent by clients when they enter a room.
    A status message is broadcast to all people in the room."""
    #Add client to client list
    clients.append([session.get('name'), request.namespace])
    room = session.get('room')
    join_room(room)
    emit('status', {'msg': session.get('name') + ' has entered the room.'}, room=room)
    #I want to do something like this, emit message to the first client
    clients[0].emit('status', {'msg': session.get('name') + ' has entered the room.'}, room=room)

这是如何正确完成的?

由于

1 个答案:

答案 0 :(得分:6)

我不确定我理解向第一个客户端发出的逻辑,但无论如何,这是如何做到的:

clients = []

@socketio.on('joined', namespace='/chat')
def joined(message):
    """Sent by clients when they enter a room.
    A status message is broadcast to all people in the room."""
    # Add client to client list
    clients.append(request.sid)

    room = session.get('room')
    join_room(room)

    # emit to the first client that joined the room
    emit('status', {'msg': session.get('name') + ' has entered the room.'}, room=clients[0])

如您所见,每个客户都有自己的空间。该房间的名称是Socket.IO会话ID,当您从该客户端处理事件时,您可以获得request.sid。因此,您只需为所有客户存储此sid值,然后在emit调用中将所需的值用作房间名称。