Playframework：发现类型不匹配 scala.concurrent.Future[play.api.mvc.Result] 需要：play.api.mvc.Result

Question

我在 PlayFramework 的控制器中有以下代码：

  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 returns 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

不确定，但这应该有效：

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).flatMap { //flatMap

        case Some(credential: Credentials) => {

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

          }

        }

        case None => Future(Unauthorized) //Wrap it
      }

    })
}

}

这是对您的代码的简化，带有一些注释。我希望这足以抓住这个想法：

 Future(Option("validCredentials")).flatMap {
   case Some(credential) => Future("OK")
   case None => Future("Unauthorized")
 }
 //Future[Option[String]].flatMap(Option[String] => Future[String])
 //Future[A].flatMap(A => Future[B]) //where A =:= Option[String] and B =:= String

Answer 2

简答：在 credentialsManager.checkEmailPassword(auth.email, auth.password).map 行中将 .map 更改为 .flatMap，将 case (None) => Unauthorized 更改为 case None => Future(Unauthorized)

解释：

credentialsManager.checkEmailPassword(auth.email, auth.password) returns a Future[Option[Credentials]] 并且在其上的映射将始终 return a Future 并且在其内部 sessionManager.createSession(credential.authAccountId) 也 returns a Future 所以，credentialsManager.checkEmailPassword(auth.email, auth.password) 的最终结果是 Future[Future[something]] 为了避免这种情况，你可以改为 flatten 然后 map 它可以完成通过 flatmap

一步

Answer 3

Future("Unauthorized") 这无效请使用 Future.successful(Ok("OK")) Future.successful(BadRequest(未授权))

Playframework：发现类型不匹配 scala.concurrent.Future[play.api.mvc.Result] 需要：play.api.mvc.Result

Playframework: Type mismatch found scala.concurrent.Future[play.api.mvc.Result] required: play.api.mvc.Result

scala

playframework-2.5