汇总多个演员响应？

本文介绍了汇总多个演员响应？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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) }