Question

我有一个Play控制器，它接收来自用户界面的Model对象。此模型对象是我尝试插入数据库的用户。但在我在表中插入此用户之前，我会检查给定的电子邮件地址是否存在已存在的重复用户。如果是，我拒绝请求，如果不是，我插入。我正在使用Slick和Play框架，这是我的控制器代码：

  def registerNewUser(user: User) = {
    dbService.registerNewUser(User.toUserRow(user))
      .map(userID => Ok(Json.obj("status" -> "ok", "userId" -> userID)))
      .recover { case ex => InternalServerError(Json.obj("error" -> s"${ex.getMessage}")) }
  }

  def createUser() = Action.async(parse.tolerantJson) { request =>
    request.body.validate[User].fold(
      errors => Future.successful {
        BadRequest(Json.obj("status" -> "error", "message" -> JsError.toJson(errors)))
      },
      user => {
        val userExists: Future[Boolean] = dbService.userExists(user.email)
        userExists.map(value => {
          if (value) UnprocessableEntity(Json.obj("status" -> "error", "message" -> s"user with email ${user.email.toString} already exists"))
          else registerNewUser(user)
        }).recover {
          case ex => InternalServerError(Json.obj("error" -> s"${ex.getMessage}"))
        }
      } // compilation failure on this line [[ Line X ]]
    )
  }

我认为这种方法没有任何问题，但我的编译器并不满意。它在X线投诉：

Expression of type Future[Object] does not confirm to the expected type _X

这里究竟出现了什么问题？

Answer 1

顺便说一下，你不应该使用flatMap代替地图吗？

userExists.map(value => {
          if (value) UnprocessableEntity(Json.obj("status" -> "error", "message" -> s"user with email ${user.email.toString} already exists"))
          else registerNewUser(user)
        })

registerNewUser会返回Future吗？

也许是这样的：

        val userExists: Future[Boolean] = dbService.userExists(user.email)
        userExists.flatMap(value => {
          if (value) Future.successful(UnprocessableEntity(Json.obj("status" -> "error", "message" -> s"user with email ${user.email.toString} already exists")))
          else registerNewUser(user)
        }).recover {
          case ex => InternalServerError(Json.obj("error" -> s"${ex.getMessage}"))
        }

Play Framework Controller处理多个期货

1 个答案: