scala 在ZIO中，如何创建一个在满足多个条件时重复尝试的时间表，分别计算每个条件的尝试次数

r8uurelv 于 5个月前发布在 Scala

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我正在使用ZIO，我的任务是用这样的逻辑（简化）重试对某些服务的请求：
1.如果响应包含一些不正确的数据-重试N次，如果情况没有改变-保留结果
1.如果响应包含的数据不足-重试M次，如果情况未发生变化-保留结果
我计划使用ZIO的.repeat函数，但不知道如何编写将为每个条件单独计算重试次数的计划。

package com.mycompany

import zio.Schedule.WithState
import zio.{ExitCode, Schedule, Task, URIO, ZIO, ZIOAppDefault}

import scala.util.Random

object Main3 extends ZIOAppDefault {
  def makeRequest: Task[Float] = ZIO.attempt {
    new Random().nextFloat() * 10
  }

  def makeSchedule(): WithState[(Unit, Unit), Any, Float, Float] = {
    val firstCondition =
      Schedule.recurUntil[Float](float => float > 3)
        // TODO: make only 3 tries for this condition
        .onDecision {
          case (input, value, decision) => ZIO.log(s"FirstCondition: ${input}, ${value}, ${decision}")
        }

    val secondCondition =
      Schedule.recurUntil[Float](float => float < 9)
        // TODO: make only 5 tries for this condition
        .onDecision {
          case (input, value, decision) => ZIO.log(s"SecondCondition: ${input}, ${value}, ${decision}")
        }

    firstCondition.||(secondCondition).map(_._1)
  }

  override def run: URIO[Any, ExitCode] = {
    makeRequest
      .repeat(makeSchedule())
      .tap(res => ZIO.log(f"RESULT IS: $res"))
      .exitCode
  }
}

字符串

scala

来源：https://stackoverflow.com/questions/77645441/in-zio-how-to-make-a-schedule-which-repeats-tries-while-multiple-conditions-are

1条答案

按热度按时间

6ovsh4lw1#

我找到了解决办法：

.repetitions
.whileOutput(repetition => repetition < 3)
.passthrough[Float]

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调用passthrough[Float]允许对齐类型并从repeat中获取结果。最后的代码片段是

package com.mycompany

import zio.Schedule.WithState
import zio.{ExitCode, Schedule, Task, URIO, ZIO, ZIOAppDefault}

import scala.util.Random

object Main3 extends ZIOAppDefault {
 def makeRequest: Task[Float] = ZIO.attempt {
   new Random().nextFloat() * 10
 }

 def makeSchedule() = {
   val firstCondition =
     Schedule.recurUntil[Float](float => float > 8)
       .repetitions
       .whileOutput(repetition => repetition < 3)
       .passthrough[Float]
       .onDecision {
         case (input, value, decision) => ZIO.log(s"FirstCondition: ${input}, ${value}, ${decision}")
       }

   val secondCondition: WithState[(Unit, Long), Any, Float, Float] =
     Schedule.recurUntil[Float](float => float < 9)
       .repetitions
       .whileOutput(repetition => repetition < 5)
       .passthrough[Float]
       .onDecision {
         case (input, value, decision) => ZIO.log(s"SecondCondition: ${input}, ${value}, ${decision}")
       }

   firstCondition.||(secondCondition).map(_._1)
 }

 override def run: URIO[Any, ExitCode] = {
   makeRequest
     .repeat(makeSchedule())
     .tap(res => ZIO.log(f"RESULT IS: $res"))
     .exitCode
 }
}

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scala 在ZIO中，如何创建一个在满足多个条件时重复尝试的时间表，分别计算每个条件的尝试次数

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