Question

我想要一个Action.async，它（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行上收到错误seq.map { seq: Seq[Commit] =>。如果我未来失败，我该如何返回另一个结果呢？

谢谢！

Answer 1

recover在Future中包含普通结果（类似于map），而recoverWith期望Future作为结果（类似于flatMap）。（https://stackoverflow.com/a/36585703/5794617）。因此，您应该使用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)
  }
}

Answer 2

Scala Future.recover的签名为

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]

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

2 个答案: