Question

我想解构从服务器

获得的HttpResponse

Success(HttpResponse(200 OK,List(Server: akka-http/2.3.12, Date: Tue, 04 Aug 2015 22:20:21 GMT),HttpEntity.Default(application/json,69,akka.stream.scaladsl.Source@52ffaba9),HttpProtocol(HTTP/1.1)))

我看了documentation，但我不确定如何解构它

我正在尝试的方式是

 val responseFuture: Future[HttpResponse] =
        Http().singleRequest(HttpRequest(uri = "http://localhost:8080"))

      responseFuture onComplete {
        case Success(_) =>
            log.info("response received {}", response)
            log.info("notified about EC Failure")
        }

但我不确定模式匹配的有效方法是什么

Answer 1

像Scala中的其他monadic类型一样处理期货...... map，flatMap，forEach和朋友：

val responseFuture: Future[HttpResponse] =
  Http().singleRequest(HttpRequest(uri = "http://localhost:8080"))

responseFuture forEach { response =>
  log.info("response received {}", response)
  log.info("notified about EC Failure")
}

// or

for(response <- responseFuture) {
  log.info("response received {}", response)
  log.info("notified about EC Failure")
}

Answer 2

像这样：

responseFuture onComplete {
  case (Success(HttpResponse(statusCode, header, entity, protocol)), _) =>
    log.info("Request returned status code {} with entity {}", statusCode, entity)
  case (Failure(response), _) =>
    log.info("Request failed with response {}", response)
}

您可以通过与akka.http.scaladsl.model.StatusCodes匹配并添加更多案例来处理特定的状态代码。

Answer 3

您可以捕获Success内的HttpResponse类型，并将其分配给您可以在模式匹配块中使用的变量。如果Success包含HttpResponse，则会将其分配给变量hr，然后我们可以使用HttpResponse访问类hr上的方法。

    case Success(hr:HttpResponse) =>
        log.info("HTTP code: " + hr.status)
    }

如何解构HttpResponse？

3 个答案: