scala - 如何使用 Akka Streams 和 HTTP 将 HTTP 资源下载到文件？

Question

在过去的几天里，我一直在尝试找出使用 Akka Streams 和 HTTP 将 HTTP 资源下载到文件的最佳方法。

最初我从 Future-Based Variant 开始看起来像这样:

def downloadViaFutures(uri: Uri, file: File): Future[Long] = {
  val request = Get(uri)
  val responseFuture = Http().singleRequest(request)
  responseFuture.flatMap { response =>
    val source = response.entity.dataBytes
    source.runWith(FileIO.toFile(file))
  }
}

那还好，但是一旦我了解了更多关于纯 Akka Streams 的知识，我想尝试使用 Flow-Based Variant创建从 Source[HttpRequest] 开始的流.起初这完全让我难倒，直到我偶然发现了 flatMapConcat流量转换。这最终变得更加冗长:

def responseOrFail[T](in: (Try[HttpResponse], T)): (HttpResponse, T) = in match {
  case (responseTry, context) => (responseTry.get, context)
}

def responseToByteSource[T](in: (HttpResponse, T)): Source[ByteString, Any] = in match {
  case (response, _) => response.entity.dataBytes
}

def downloadViaFlow(uri: Uri, file: File): Future[Long] = {
  val request = Get(uri)
  val source = Source.single((request, ()))
  val requestResponseFlow = Http().superPool[Unit]()
  source.
    via(requestResponseFlow).
    map(responseOrFail).
    flatMapConcat(responseToByteSource).
    runWith(FileIO.toFile(file))
}

然后我想变得有点棘手并使用 Content-Disposition标题。

回到基于 future 的变体:

def destinationFile(downloadDir: File, response: HttpResponse): File = {
  val fileName = response.header[ContentDisposition].get.value
  val file = new File(downloadDir, fileName)
  file.createNewFile()
  file
}

def downloadViaFutures2(uri: Uri, downloadDir: File): Future[Long] = {
  val request = Get(uri)
  val responseFuture = Http().singleRequest(request)
  responseFuture.flatMap { response =>
    val file = destinationFile(downloadDir, response)
    val source = response.entity.dataBytes
    source.runWith(FileIO.toFile(file))
  }
}

但是现在我不知道如何使用基于 future 的变体来做到这一点。这是我得到的:

def responseToByteSourceWithDest[T](in: (HttpResponse, T), downloadDir: File): Source[(ByteString, File), Any] = in match {
  case (response, _) =>
    val source = responseToByteSource(in)
    val file = destinationFile(downloadDir, response)
    source.map((_, file))
}

def downloadViaFlow2(uri: Uri, downloadDir: File): Future[Long] = {
  val request = Get(uri)
  val source = Source.single((request, ()))
  val requestResponseFlow = Http().superPool[Unit]()
  val sourceWithDest: Source[(ByteString, File), Unit] = source.
    via(requestResponseFlow).
    map(responseOrFail).
    flatMapConcat(responseToByteSourceWithDest(_, downloadDir))
  sourceWithDest.runWith(???)
}

所以现在我有一个 Source将发出一个或多个 (ByteString, File)每个元素 File (我说每个 File 因为没有理由原始 Source 必须是单个 HttpRequest )。

有没有办法把这些和路由到一个动态 Sink ?

我在想类似 flatMapConcat ， 如:

def runWithMap[T, Mat2](f: T => Graph[SinkShape[Out], Mat2])(implicit materializer: Materializer): Mat2 = ???

这样我才能完成 downloadViaFlow2和:

def destToSink(destination: File): Sink[(ByteString, File), Future[Long]] = {
  val sink = FileIO.toFile(destination, true)
  Flow[(ByteString, File)].map(_._1).toMat(sink)(Keep.right)
}
sourceWithDest.runWithMap {
  case (_, file) => destToSink(file)
}

Answer 1

该解决方案不需要 flatMapConcat。如果您不需要文件写入的任何返回值，那么您可以使用 Sink.foreach :

def writeFile(downloadDir : File)(httpResponse : HttpResponse) : Future[Long] = {
  val file = destinationFile(downloadDir, httpResponse)
  httpResponse.entity.dataBytes.runWith(FileIO.toFile(file))
}

def downloadViaFlow2(uri: Uri, downloadDir: File) : Future[Unit] = {
  val request = HttpRequest(uri=uri)
  val source = Source.single((request, ()))
  val requestResponseFlow = Http().superPool[Unit]()

  source.via(requestResponseFlow)
        .map(responseOrFail)
        .map(_._1)
        .runWith(Sink.foreach(writeFile(downloadDir)))
}

请注意 Sink.foreach创建 Futures来自 writeFile功能。因此没有太多的背压。 writeFile 可能会被硬盘驱动器减慢，但流将继续生成 Futures。要控制它，您可以使用 Flow.mapAsyncUnordered (或 Flow.mapAsync):

val parallelism = 10

source.via(requestResponseFlow)
      .map(responseOrFail)
      .map(_._1)
      .mapAsyncUnordered(parallelism)(writeFile(downloadDir))
      .runWith(Sink.ignore)

如果要累积总计数的 Long 值，则需要结合 Sink.fold :

source.via(requestResponseFlow)
      .map(responseOrFail)
      .map(_._1)
      .mapAsyncUnordered(parallelism)(writeFile(downloadDir))
      .runWith(Sink.fold(0L)(_ + _))

当请求源枯竭时，折叠将保持运行总和并发出最终值。