汇总多个演员响应？ [英] Aggregate multiple actor responses?

问题描述

我正在尝试为我的应用程序收集健康信息，

I am trying to collect health information for my application as

class HealthMonitor extends Actor with ActorLogging {
  val statusReporter = new StatusReporter
  val versionInfo = context.actorOf(Props[VersionInfo], "versionInfo")
  val memoryInfo = context.actorOf(Props[MemoryInfo], "memoryInfo")

  def receive = LoggingReceive {
    case HealthReportRequest => log.debug("Generating Health Report")
      println("Generating Health Report")
      // todo (harit): should be concurrent calls and collect results
      versionInfo ! VersionInfoRequest
      memoryInfo ! MemoryInfoRequest
  }
}

我需要什么

我需要一种可以从 versionInfo ， memoryInfo 和其他一些方式收集响应的方法信息随后变成1个响应并将其发送到某个地方

What I need
I need a way wherein I can collect responses from versionInfo, memoryInfo, and some other info later into 1 response and send it somewhere

，而我不想按顺序发送或希望阻止通话，最好的方法是什么？

and I do not want a sequential or want to block the calls, what is the best way?

推荐答案

我会猜测您是要问而不是告诉，因为您是
在谈论响应，因此代码应该是

I'll make a guess you meant an ask instead of a tell because you're talking about responses, so the code should be

def receive = LoggingReceive {
  case HealthReportRequest => log.debug("Generating Health Report")
    println("Generating Health Report")
    versionInfo ? VersionInfoRequest
    memoryInfo ? MemoryInfoRequest
}

然后您可以输入期货。

def receive = LoggingReceive {
  case HealthReportRequest =>
    versionInfo ? VersionInfoRequest mapTo[VersionInfo]
    memoryInfo ? MemoryInfoRequest mapTo[VersionInfo]
}

然后合并

def receive = LoggingReceive {
  case HealthReportRequest =>
    val version = versionInfo ? VersionInfoRequest mapTo[VersionInfo]
    val memory = (memoryInfo ? MemoryInfoRequest mapTo[VersionInfo])
    version.flatMap(v =>
      memory.map(m =>
        fun(v, m)
    ))
}

或通过for（not

or via for (not sure on the syntax)

def receive = LoggingReceive {
  case HealthReportRequest =>
    val version = versionInfo ? VersionInfoRequest mapTo[VersionInfo]
    val memory = (memoryInfo ? MemoryInfoRequest mapTo[VersionInfo])
    for {
      v <- version
      m <- memory
    } yield fun(v, m)
}

这篇关于汇总多个演员响应？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！