oauth2在Play scala web-socket中的实现

Question

我使用Play2.3 scala开发了web套接字，我正在尝试使用oauth-provider在Play scala web套接字中实现oauth2。

web-socket代码片段:

object LoginWS {
  def props(out: ActorRef) = Props(new LoginWS(out))
}
class LoginWS(out: ActorRef) extends Actor {
  def receive = {
    case json_req: JsObject =>
        val login_status : String = "Success"
        out ! Json.toJson(JsObject(Seq("login_status" -> JsString(login_status))))
  }
}
object WebSocketController extends Controller {
  def login = WebSocket.acceptWithActor[JsValue, JsValue] { request =>
    out =>
      LoginWS.props(out)
  }
}

oauth2的REST服务代码片段:

我使用了"scala-oauth2-provider-slick"示例来实现oauth2。

trait MyOAuth extends OAuth2Provider {
  override val tokenEndpoint: TokenEndpoint = MyTokenEndpoint
}
object OAuth2Controller extends Controller with MyOAuth {
  val cats = TableQuery[CatsTable] 
  implicit val catFormat = Json.format[Cat]
  def accessToken = Action.async { implicit request =>
    issueAccessToken(new Oauth2DataHandler())
  }
  def list = Action.async { implicit request =>
    authorize(new Oauth2DataHandler()) { authInfo =>
      DBAction { implicit rs =>
        Ok(toJson(cats.list))
      }.apply(request)
    }
  }
}

我如何实现oauth2在上面的web套接字代码?

有人知道，如何实现oauth2以上的web套接字?

更新:

我试图实现oauth和web套接字:

代码片段:

  def login = WebSocket.tryAcceptWithActor[JsValue, JsValue] { request =>
      authorize(new Oauth2DataHandler()) { authInfo =>
      val user = authInfo.user // User is defined on your system
      println(user.toString())
    }
  }

得到以下问题:

1) polymorphic expression cannot be instantiated to expected type; found : [A]scala.concurrent.Future[play.api.mvc.Result] required: Either[play.api.mvc.Result,play.api.mvc.WebSocket.HandlerProps] (which expands to) Either[play.api.mvc.Result,akka.actor.ActorRef ⇒ akka.actor.Props]
2) type mismatch; found : Unit required: scala.concurrent.Future[play.api.mvc.Result]

Answer 1

OAuth2Provider提供了protectedResource实例，所以你可以直接调用handleRequest方法在你的WebSocket控制器中进行授权。客户端通过HTTP头或查询参数发送访问令牌。

我建议你使用tryAcceptWithActor而不是acceptWithActor。因为如果用户发送无效的访问令牌，您需要返回错误的请求。

object LoginWS {
  def props(user: User)(out: ActorRef) = Props(new LoginWS(user, out))
}
object WebSocketController extends Controller extends MyOAuth {
  def login = WebSocket.tryAcceptWithActor[JsValue, JsValue] { request =>
    Future.successful {
      (protectedResource.handleRequest(request, new Oauth2DataHandler())).map {
        case Left(e) => e match {
          case e: OAuthError if e.statusCode == 400 => Left(BadRequest.withHeaders(responseOAuthErrorHeader(e)))
          case e: OAuthError if e.statusCode == 401 => Left(Unauthorized.withHeaders(responseOAuthErrorHeader(e)))
        }
        case Right(authInfo) => Right(LoginWS.props(authInfo.user) _)
      }
    }
  }
}

我不确定这段代码是否可以编译。但是我已经使用这个库编写了类似的代码。