Scala映射Future[IOResult]到Future[Unit]

Question

我有一个函数必须返回一个FutureUnit。

在我的函数中，我编写了一个返回FutureIOResult的文件。未来和IOResult都有失败的状态。

我想在我的函数中检查未来和IOResult的成功和失败，但是从这个函数返回一个FailureUnit，这可能吗？

下面的代码是报告错误：

discarded non-Unit value
[error]           Future.successful(Unit)

discarded non-Unit value
[error]           Future.failed(e)

这是我的功能：

def create(filePath: String, fileStream: Source[ByteString, Any]): Future[Unit] = {

    val writeResultFuture: Future[IOResult] = fileStream.runWith(FileIO.toPath(filePath))

    writeResultFuture map { writeResult =>
      writeResult.status match {
        case Success(_) =>
          logger.info(s"Successfully uploaded: ${fileInfo.fileName} to: $filePath")
          Future.successful(Unit)
        case Failure(e) =>
          logger.error(s"Failed to upload: ${fileInfo.fileName} to: $filePath", e)
          Future.failed(e)
      }
    } recover {
      case e =>
        logger.error(s"Failed to upload: ${fileInfo.fileName} to: $filePath", e)
        Future.failed(e)
    }
  }

Answer 1

三件事。

1. Unit不是Unit类型的值(它是对scala库中定义的object scala.Unit的引用)。您希望()表示Unit的值。
2. 您给recover的函数应该返回实际结果，而不是Future (recoverWith与期货一起工作)。
3. 您给map的函数也返回实际结果。如果你真的需要返回一个flatMap，你会想要Future

所以，像这样的东西应该有效：

writeResultFuture.map { writeResult =>
    writeResult.status match {
        case Success(_) =>
          logger.info(s"Successfully uploaded: ${fileInfo.fileName} to: $filePath")
        case Failure(e) =>
          logger.error(s"Failed to upload: ${fileInfo.fileName} to: $filePath", e)
          throw e
    }
  }.recover { case e =>
        logger.error(s"Failed to upload: ${fileInfo.fileName} to: $filePath", e)
        throw e
  }

更好的方法是使用onFailure而不是recover -这样就不需要重新抛出了。只需执行.onFailure { e => logger.error(...) }，请注意，您正在以这种方式记录错误两次(一次在map中，另一次在recover/onFailure中).考虑一起删除recover部件。

  writeResultFuture.map(_.status).map { 
    case Success(_) => logger.info(s"Successfully uploaded: ${fileInfo.fileName} to: $filePath")
    case Failure(e) => logger.error(s"Failed to upload: ${fileInfo.fileName} to: $filePath", e)
        throw e
  }

Answer 2

这样做可能会奏效：

  writeResultFuture flatMap { writeResult =>
    writeResult.status match {
      case Success(_) =>
        logger.info(s"Successfully uploaded: ${fileInfo.fileName} to: $filePath")
        Future.successful()
      case Failure(e) =>
        Future.failed(e)
    }
  } recover {
    case e =>
      logger.error(s"Failed to upload: ${fileInfo.fileName} to: $filePath", e)
  }

使用flatMap将扁平match语句返回的嵌套Future，recover现在返回Unit，因此结果是Future[Unit]。recover将捕获生成的所有错误，因此不需要在内部Failure大小写中打印任何内容。

Answer 3

您正在返回未来的，所以您的最终结果将是类型未来[未来…]，在您的情况下，我们根本不需要嵌套，因此我们可以将代码更改为：

免责声明:我只在这里输入了代码，它还没有被输入，YMMV。没有保证。等

fileStream.runWith(FileIO.toPath(filePath)) transform {
  case Failure(e) => Failure(e)
  case Success(i: IOResult) => i.status
} andThen {
    case Success(_) =>
      logger.info(s"Successfully uploaded: ${fileInfo.fileName} to: $filePath")
    case Failure(e) =>
      logger.error(s"Failed to upload: ${fileInfo.fileName} to: $filePath", e)
  }
} map { _ => () }