向除发件人 socket.io 以外的所有人广播

Question

i am building a chat in my app. I am using socket.io for this. When a user sends a message i send an api request to my server. The api stores the message in the database and only then emits, with a socket service to everyone in the room that there is a new message. I have a SocketService class with this method:

  private async broadcast({ type, data, chatId, senderId }: { type: string; data: any; chatId: string; senderId: string }) {
const excludedSocket = await this.getUserSocket(senderId);
if (chatId && excludedSocket) {
  excludedSocket.emit;
} else if (excludedSocket) {
  excludedSocket.emit(type, data);
} else if (room) {
  gIo.to(room).emit(type, data);
} else {
  gIo.emit(type, data);
}

}

The problem i have is that getUserSocket returns a RemoteSocket object that dont have the broadcast or to methods on it. So how can i achive this?

 private async getUserSocket(userId: string) {
const sockets = await this.getAllSockets();
const socket = sockets.find((s) => s.data.uid === userId);
return socket;

}

private async getAllSockets() {
const sockets = await this.io.fetchSockets();
return sockets

}

Answer 1

socket.broadcast.emit equates to a BroadcastOperator that sends to an empty Set of rooms (meaning "all rooms"), excluding the socket id "room" of the sender Socket.

The same BroadcastOperator can be created from the server instance with server.except()

const sockets = await this.io.except(socketIdString).emit('blah', x)