如何 return 在 Scala 中赋值或抛出错误

Question

我有一个电子邮件列表，对于每个电子邮件，我都会在电子邮件 table 中查找它以查看该电子邮件是否存在。如果是这样，什么都不做我会抛出错误。这是我的代码；

def lookupEmailStatus(email: EmailAddress, requestId: RequestId)(
      implicit ec: ExecutionContext): HttpServiceResult[List[EmailStatusDTO]] = {
    emailDatabase
      .getEmailStatusByEmail(email, requestId)
      .map(
        l =>
        if (l.isEmpty) {
          logger.error(
            LoggingMessage(
              requestId,
              s"Email status not found by ${email.email} failed"))
          EntityNotFound(s"${email.email}", requestId)
        } else {
          l
        }
      )
      .leftMap[HttpError] {
        case e =>
          logger.error(
            LoggingMessage(
              requestId,
              s"Retrieve email status by ${email.email} failed"))
          DatabaseError(e.message, requestId)
      }
  }

当我运行代码出错时：

Error:(57, 27) type mismatch;
 found   : cats.data.EitherT[model.HttpError,Product with Serializable]
 required: model.HttpServiceResult[List[EmailStatusDTO]]
    (which expands to)  cats.data.EitherT[model.HttpError,List[EmailStatusDTO]]
      .leftMap[HttpError] {

如果我删除 .map(..) 方法它会工作正常，但这不是我想要的：

def lookupEmailStatus(email: EmailAddress, requestId: RequestId)(
          implicit ec: ExecutionContext): HttpServiceResult[List[EmailStatusDTO]] = {
        emailDatabase
          .getEmailStatusByEmail(email, requestId)
          .leftMap[HttpError] {
            case e =>
              logger.error(
                LoggingMessage(
                  requestId,
                  s"Retrieve email status by ${email.email} failed"))
              DatabaseError(e.message, requestId)
          }
      }

这里是类型定义：

type HttpServiceResult[A] = ServiceResult[HttpError, A]
type ServiceResult[Err, A] = EitherT[Future, Err, A]

Answer 1

如果 if returns 一个列表的一个分支和其他 returns 一个错误，则完全 if returns Product with Serializable.

尝试用 flatMap 替换 map 并用 EitherT

包装分支结果

def lookupEmailStatus(email: EmailAddress, requestId: RequestId)(
    implicit ec: ExecutionContext): HttpServiceResult[List[EmailStatusDTO]] = 
    emailDatabase
      .getEmailStatusByEmail(email, requestId)
      .leftMap[HttpError] {
        case e =>
          logger.error(
            LoggingMessage(
              requestId,
              s"Retrieve email status by ${email.email} failed"))
          DatabaseError(e.message, requestId)
      }.flatMap[HttpError, List[EmailStatusDTO]](
        /*(*/l/*: List[EmailStatusDTO])*/ =>
          if (l.isEmpty) {
            logger.error(
              LoggingMessage(
                requestId,
                s"Email status not found by ${email.email} failed"))
            EitherT.leftT/*[Future, List[EmailStatusDTO]]*/(EntityNotFound(s"${email.email}", requestId))
          } else {
            EitherT.rightT/*[Future, HttpError]*/(l)
          }
      )

Answer 2

正如@dmytro-mitin 已经提到的，问题的根本原因是您在条件的两个分支中没有 returning 相同的类型。解决此问题的一种方法是确保您 return 使用正确的类型。

另一种在我看来更好的方法是使用 cats.data.EitherT.ensure 进行 list.isEmpty 检查。这样你就可以明确你关心的事情（即如果列表为空，return 错误）并且不必手动处理满意的情况。

您的代码将变为：

def lookupEmailStatus(email: EmailAddress, requestId: RequestId)(
      implicit ec: ExecutionContext): HttpServiceResult[List[EmailStatusDTO]] = {
    emailDatabase
      .getEmailStatusByEmail(email, requestId)
      .ensure({
         logger.error(LoggingMessage(requestId, s"Email status not found by ${email.email} failed"))}
         EntityNotFound(s"${email.email}", requestId)
      })(!_.isEmpty)
      .leftMap[HttpError] {
        case e =>
          logger.error(
            LoggingMessage(
              requestId,
              s"Retrieve email status by ${email.email} failed"))
          DatabaseError(e.message, requestId)
      }
  }

如何 return 在 Scala 中赋值或抛出错误

how to return either value or throw error in Scala

scala

scala-cats