烧瓶插座向特定用户发出

Question

我看到有关此主题的问题，但未概述特定代码。说我只想向第一个客户发出。

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)

如何正确完成？

谢谢

Answer 1

我不确定我是否了解向第一个客户端发出消息的逻辑，但是无论如何，这是这样做的：

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，您可以request.sid在处理来自该客户端的事件时获得该ID
。因此，您要做的就是sid为所有客户存储此值，然后在emit呼叫中使用所需的房间名称。