访问由Source.actorRef创建的akka​​流Source的基础ActorRef [英] Accessing the underlying ActorRef of an akka stream Source created by Source.actorRef

问题描述

我正在尝试使用 Source.actorRef 方法来创建 akka.stream.scaladsl.Source 对象。格式如下

I'm trying to use the Source.actorRef method to create an akka.stream.scaladsl.Source object. Something of the form

import akka.stream.OverflowStrategy.fail
import akka.stream.scaladsl.Source

case class Weather(zip : String, temp : Double, raining : Boolean)

val weatherSource = Source.actorRef[Weather](Int.MaxValue, fail)

val sunnySource = weatherSource.filter(!_.raining)
...

我的问题是：如何将数据发送到基于ActorRef的Source对象？

我以为向源发送消息是某种形式

I assumed sending messages to the Source was something of the form

//does not compile
weatherSource ! Weather("90210", 72.0, false)
weatherSource ! Weather("02139", 32.0, true)

但是 weatherSource 没有！运算符或 tell 方法。

But weatherSource doesn't have a ! operator or tell method.

文档不太说明如何使用Source.actorRef，它只是说您可以...

The documentation isn't too descriptive on how to use Source.actorRef, it just says you can...

在此先感谢您的评论和答复。 p>

Thank you in advance for your review and response.

推荐答案

您需要流程：

  import akka.stream.OverflowStrategy.fail
  import akka.stream.scaladsl.Source
  import akka.stream.scaladsl.{Sink, Flow}

  case class Weather(zip : String, temp : Double, raining : Boolean)

  val weatherSource = Source.actorRef[Weather](Int.MaxValue, fail)

  val sunnySource = weatherSource.filter(!_.raining)

  val ref = Flow[Weather]
    .to(Sink.ignore)
    .runWith(sunnySource)

  ref ! Weather("02139", 32.0, true)

请记住，这都是实验性的，可能会改变！

Remember this is all experimental and may change!

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