1
我正在做Scala和Play的第一个应用程序!在我使用WebSocket的时候,就像他们在聊天室示例中一样,到目前为止,我可以向我的服务器发送消息,但我似乎无法理解我可以“处理”从客户那里获得的这些消息,我也想知道如果我可以从我的服务器发送JSON数组我的客户:服务器Scala与Play的响应
@Singleton
class HomeController @Inject()(cc: ControllerComponents)
(implicit actorSystem: ActorSystem,
mat: Materializer,
executionContext: ExecutionContext)
extends AbstractController(cc) {
private type WSMessage = String
private val logger = Logger(getClass)
private implicit val logging = Logging(actorSystem.eventStream, logger.underlyingLogger.getName)
// chat room many clients -> merge hub -> broadcasthub -> many clients
private val (chatSink, chatSource) = {
// Don't log MergeHub$ProducerFailed as error if the client disconnects.
// recoverWithRetries -1 is essentially "recoverWith"
val source = MergeHub.source[WSMessage]
.log("source")
.recoverWithRetries(-1, { case _: Exception ⇒ Source.empty })
val sink = BroadcastHub.sink[WSMessage]
source.toMat(sink)(Keep.both).run()
}
private val userFlow: Flow[WSMessage, WSMessage, _] = {
Flow.fromSinkAndSource(chatSink, chatSource)
}
def index: Action[AnyContent] = Action { implicit request: RequestHeader =>
val webSocketUrl = routes.HomeController.chat().webSocketURL()
logger.info(s"index: ")
Ok(views.html.index(webSocketUrl))
}
def chat(): WebSocket = {
WebSocket.acceptOrResult[WSMessage, WSMessage] {
case rh if sameOriginCheck(rh) =>
Future.successful(userFlow).map { flow =>
Right(flow)
}.recover {
case e: Exception =>
val msg = "Cannot create websocket"
logger.error(msg, e)
val result = InternalServerError(msg)
Left(result)
}
case rejected =>
logger.error(s"Request ${rejected} failed same origin check")
Future.successful {
Left(Forbidden("forbidden"))
}
}
}
}
顺便说一句,我通过jQuery函数发送从我的客户端的消息。
编辑我要处理这个消息的方法是将它们作为参数传递给函数,该函数将返回字符串或整数数组,我要返回给客户端