scala.concurrent.Future[play.api.mvc.Result] 要求：play.api.mvc.Result

Question

我想要一个 Action.async (1) 尝试从数据库中获取值。如果数据库不可用，它将尝试连接到另一个资源并 (2) 从那里获取值。因为我正在使用 return Future 的两个资源，所以我用 "recover" 关键字将它们分开。我不确定这是否是最好的方法.....但是recovery{}里面的语句有一个类型不匹配的错误：

  def show(url: String) = Action.async { implicit request: Request[AnyContent] =>
    println("url: " + url)

    val repositoryUrl = RepositoryUrl(url)
    val repositoryId = RepositoryId.createFromUrl(url)

    // Listing commits from the DB
    val f: Future[Seq[Commit]] = commit.listByRepository(repositoryId.toString())
    f.map { f: Seq[Commit] =>
      val json = JsObject(Seq(
        "project URL" -> JsString(url),
        "list of commits" -> Json.toJson(f)))
      Ok(json)
    }.recover {
      case e: scala.concurrent.TimeoutException =>
        // InternalServerError("timeout")
        // Listing commits from the Git CLI
        val github = rules.GitHub(repositoryUrl)
        val seq: Future[Seq[Commit]] = github.listCommits

        seq.map { seq: Seq[Commit] =>
          val json = JsObject(Seq(
            "project URL" -> JsString(url),
            "list of commits" -> Json.toJson(seq)))
          Ok(json)
        }
    }
  }

我在 seq.map { seq: Seq[Commit] => 行收到错误 type mismatch; found : scala.concurrent.Future[play.api.mvc.Result] required: play.api.mvc.Result。如果我的未来失败了，我怎么能return另一个结果？

谢谢！

Answer 1

Scala Future.recover 的签名是

def recover[U >: T](pf: PartialFunction[Throwable, U])(implicit executor: ExecutionContext): Future[U]

尝试改用recoverWith

def recoverWith[U >: T](pf: PartialFunction[Throwable, Future[U]])(implicit executor: ExecutionContext): Future[U]

Answer 2

recover 将普通结果包装在 Future 中（类似于 map），而 recoverWith 期望 Future 作为结果（类似于 flatMap）。 ()。所以，你应该使用 recoverWith:

def show(url: String): EssentialAction = Action.async { implicit request: Request[AnyContent] =>
  // This future will throw ArithmeticException because of division to zero
  val f: Future[Seq[Int]] = Future.successful(Seq[Int](1, 2, 3, 4 / 0))
  val fResult: Future[JsObject] = f.map { r =>
    JsObject(Seq(
      "project URL" -> JsString(url),
      "list of commits" -> Json.toJson(r)))
  }.recoverWith {
    case e: ArithmeticException =>
      val seq: Future[Seq[Int]] = Future.successful(Seq(1, 2, 3, 4))

      seq.map { seq: Seq[Int] =>
        JsObject(Seq(
          "project URL" -> JsString(url),
          "list of commits" -> Json.toJson(seq)))
      }
  }
  fResult.map { r =>
    Ok(r)
  }
}

scala.concurrent.Future[play.api.mvc.Result] 要求：play.api.mvc.Result

scala.concurrent.Future[play.api.mvc.Result] required: play.api.mvc.Result

scala

concurrent.futures

playframework-2.6