scala - Playframework : Type mismatch found scala.并发.Future[play.api.mvc.Result] 需要 : play. api.mvc.Result

Question

我的 PlayFramework Controller 中有以下代码:

  def auth = Action.async(parse.json) { request =>
    {

      val authRequest = request.body.validate[AuthRequest]
      authRequest.fold(
        errors => Future(BadRequest),
        auth => {
          credentialsManager.checkEmailPassword(auth.email, auth.password).map {

            case Some(credential: Credentials) => {

              sessionManager.createSession(credential.authAccountId).map { //Throws an error
                case Some(authResponse: AuthResponse) => Ok(Json.toJson(authResponse))
                case None => InternalServerError

              }

            }

            case (None) => Unauthorized
          }

        })
    }
  }

我在上面带有错误注释的行收到以下错误:

Type Mismatch:
[error]  found   : scala.concurrent.Future[play.api.mvc.Result]
[error]  required: play.api.mvc.Result
[error]               sessionManager.createSession(credential.authAccountId).map {

那里的 createSession 调用返回一个 Future[Option[Object]] 但我不知道如何解决这个问题。

任何帮助将不胜感激。

Answer 1

简短回答:将 credentialsManager.checkEmailPassword(auth.email, auth.password).map 行中的 .map 更改为 .flatMap 和 case (None ) => 未经授权 至 case None => future (未经授权)

说明:

credentialsManager.checkEmailPassword(auth.email, auth.password) 返回一个 Future[Option[Credentials]] 并且对其进行映射将始终返回一个 Future 并在其中 sessionManager.createSession(credential.authAccountId) 还返回一个 Future 因此， credentialsManager.checkEmailPassword(auth.email, auth) 的最终结果.password) 是 Future[Future[something]] 为了避免这种情况，您可以展平它，然后映射它并可以通过 flatmap

一步完成