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

问题描述 投票:0回答:2

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。因为我使用的两个资源返回Future,我将它们与“recover”关键字分开。我不确定它是否是最好的方法.....但是恢复{}中的语句有一个类型不匹配错误:

  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type mismatch; found : scala.concurrent.Future[play.api.mvc.Result] required: play.api.mvc.Result线上得到错误seq.map { seq: Seq[Commit] =>。如果我未来失败,我该如何返回另一个结果?

谢谢!

scala concurrent.futures playframework-2.6
2个回答
2
投票

recover为你包装了Future的简单结果(地图的类比),而recoverWith期望Future作为结果(flatMap的模拟)。 (https://stackoverflow.com/a/36585703/5794617)。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)
  }
}

0
投票

Scala Future.recover有一个签名

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

尝试使用recoverWith代替

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