Question

在我有一个Future[Something]：a的情况下，成功需要触发第二个Future[Unit]：b。我想将a和b链接在一起，但我只关心a成功。如果b失败，我可以记录一个错误并将其留在那：

到目前为止，我有：

def updateSomething(something: Something): Future[Something] = {
    val eventual: Future[Something] = repository.update(something)
    eventual.onSuccess({
      case updated =>
        repository.audit(updated.id, "Update successful")
          .onFailure({
            case throwable: Throwable => Logger.error("Audit failed", throwable)
          })
        Logger.info("Update Complete")
    })
    eventual
}

但这不会将update和audit的生命周期联系在一起。例如，当我Await.result(service.updateSomething(...), duration)不能保证repository.audit的未来已经完成时。

Answer 1

select player, team, min(timeid), max(timeid) from (select lp.*, row_number() over (partition by player, team order by timeid) as seqnum, cast(cast(timeid + 1 as varchar(255)) as datetime) as yyyymm from logplayer lp ) lp group by player, team, dateadd(month, - seqnum, yyyymm) order by player, team, min(timeid);是您的朋友。您可以使用理解+恢复：

flatMap

Answer 2

您还可以使用更友好的形式：

execudeA().flatMap(a =>
  executeB().recover { case e => println(e); () }.map(_ => a)
)

Answer 3

此外，您可以只使用val

val = a <- executeA()
a.andThen{ case _ => executeB(b).recover{case e => println(e)} }
a //and return a

Scala：链接期货，返回第一个

3 个答案: