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
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。