Scala-如何以功能样式重构代码

2024-05-01 • 问答

我创建了两种类似的方法：

override def getUsers(organization: String,params: String): F[Either[CodecException,List[Users]]] = for {
    resp <- getUsersFromClient(organization,params)
    result <- Sync[F].delay(resp)
  } yield result

  override def getanimals(url: String,List[Animal]]] = for {
    resp <- getanimalsFromClient(url,params)
    result <- Sync[F].delay(resp)
  } yield result

我认为我应该以某种方式将它们重构为implicits仅具有一种方法或某种方法，但是我不知道应该如何以正确的方式完成功能样式的重构。你能帮我吗？

编辑：

def getUsersFromClient(param1: String,params: String): F[HttpResponse] = Hammock.getWithOpts(...)

def getanimals(param1: String,params: String): F[HttpResponse] = Hammock.getWithOpts(...)

此方法也相似，并返回F[HttpResponse]

def get[A](s: String,params: String,getFromClient: (String,String) => ...,decodeResponse: ... => ...): F[Either[CodecException,List[A]]] = for { resp <- getFromClient(s,params) result <- Sync[F].delay(decodeResponse(resp),CodecException.withMessage("Something goes wrong while decoding response.")) } yield result

Scala-如何以功能样式重构代码

wdwxhn 回答：Scala-如何以功能样式重构代码

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