Playframework：发现类型不匹配 scala.concurrent.Future[play.api.mvc.Result] 必需：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 调用返回 aFuture[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：更改.map为.flatMapin linecredentialsManager.checkEmailPassword(auth.email, auth.password).map和case (None) => Unauthorizedtocase None => Future(Unauthorized)

解释：

credentialsManager.checkEmailPassword(auth.email, auth.password)返回 aFuture[Option[Credentials]]并且其上的映射将始终返回 aFuture并且在其内部sessionManager.createSession(credential.authAccountId)也返回 aFuture所以，最终结果是credentialsManager.checkEmailPassword(auth.email, auth.password)为了Future[Future[something]]避免这种情况，您可以改为flatten它，然后map它可以通过以下步骤完成flatmap