中断未来地图链并返回左

Question

如何在未来的地图链中间返回一个左，然后失败未来？我正在发送一个例子来澄清。

import scala.concurrent.Future
import scala.concurrent.ExecutionContext.Implicits.global
object TestApp extends App {
final case class Message(statusCode: Int, data: String)
final case class Success(data: String)
final case class Failure()
val msg = Message(500, "My data")
val future = Future { msg }
def getMessage(future: Future[Message]): Future[Either[Failure, Success]] = {
future.map { msg =>
// Evaluate msg here - if failure, make this feature fail and then return Left(Failure), otherwise, continue in the chain
msg
}.map(_.data).map(data => Right(Success(data)))
}
}

Answer 1

一种选择是抛出一个异常，该异常将使Future失败，然后recover到这样的Left

def getMessage(future: Future[Message]): Future[Either[Failure, Success]] =
future
.map(msg => if (msg.statusCode == 500) throw new RuntimeException("boom") else msg)
.map(_.data)
.map(data => Right(Success(data)))
.recover{case ex => Left(Failure())}

另一种选择是这样transform

def getMessage(future: Future[Message]): Future[Either[Failure, Success]] = {
future
.map(msg => if (msg.statusCode == 500) throw new RuntimeException("boom") else msg)
.map(_.data)
.transform(v => scala.util.Success {
v.toEither
.map(v => Success(v))
.left.map(e => Failure())
})
}

警告，您已经定义了自己的Success/Failure，它遮蔽了 Scala 的Success/Failure。