Hi. I posted this at SO, may be a better place is here. I don't have a good understanding of connection management, and not sure how server can keep a handle of client connection and push data (instead of responding to API calls).
https://stackoverflow.com/questions/59726003/grpc-keep-a-reference-to-stream-to-send-data-to-multiple-clients I am starting with GRPC using go. I read the official docs, and few examples. In most examples, you don't identify the client, but use the stream to read / write data. I see there is API in Context to retrieve auth info, and can indentify the client for the ChatRequest. But, what if I want to keep a reference / index to the Stream based on the client id. For example, say I have 3 users in a chat room. I represent the rpc as (it could be server-streaming as well) rpc Chat(stream ChatRequest) returns (stream ChatResponse) {} say, one user send a message to the group, which need to send to other 2. So, if I need to send it through the Stream currently opened for those users, how safe is to keep a reference of the stream. The implementation will be like... type chatServiceServer struct { // keep a map of subscribers / users currently connected; protect with mutex } func (s *chatServiceServer) Chat(stream pb.ChatService_ChatServer) error { // md, ok := metadata.FromIncomingContext(stream.Context()) // p, ok := peer.FromContext(ctx) // ... identify client from above for { // save the message to DB // find other users in the chatroom is currently connected // if so, stream.Send(m) // else notify .... } } But, I see the warnings in API doc and wonder a better way. https://godoc.org/google.golang.org/grpc#ServerStream // It is safe to have a goroutine calling SendMsg and another goroutine // calling RecvMsg on the same stream at the same time, but it is not safe // to call SendMsg on the same stream in different goroutines. SendMsg(m interface{}) error Similar use case arise with any subscription (event, ....), and need to notify based on client id. Any example code, article would be great as well. Thank you -- You received this message because you are subscribed to the Google Groups "grpc.io" group. To unsubscribe from this group and stop receiving emails from it, send an email to grpc-io+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/grpc-io/6c6700f0-098e-4991-92a5-9f89b6ead7e0%40googlegroups.com.
