Question

我有一个处理Source并返回的方法。我正在尝试修改它，但似乎无法返回相同的内容：

原始

def originalMethod[as: AS, mat: MAT, ec: EC](checkType: String) 
: Flow[ByteString, MyValidation[MyClass], NotUsed]{
      collectStuff
      .map { ts =>
        val errors = MyEngine.checkAll(ts.code)
        (ts, errors)
      }
      .map { x =>
        x._2
          .leftMap(xs => {
            addInformation(x._1, xs.toList)
          })
          .toEither
      }
}

我正在使用其他来源进行修改，并将其结果传递给原始来源，但返回的内容相同：

def calculate[T: AS: MAT](source: Source[T, NotUsed]): Future[Seq[T]] = 
{
 source.runWith(Sink.seq)
}


def modifiedMethod[as: AS, mat: MAT, ec: EC](checkType: String, mySource: Source[LoanApplicationRegister, NotUsed]) 
: Flow[ByteString, MyValidation[MyClass], NotUsed]{
  for {
    calc <- calculate(mySource)
    orig <-  collectStuff
        .map { ts =>
          val errors = MyEngine.checkAll(ts.code, calc)
          (ts, errors)
        }
        .map { x =>
          x._2
            .leftMap(xs => {
              addInformation(x._1, xs.toList)
            })
            .toEither
        }
  }
  yield {
    orig
  }
}

但是我遇到编译错误Expression of type Future[Nothing] doesn't conform to existing type Flow[ByteString, MyValidation[MyClass]

如何像在Flow[ByteString, MyValidation[MyClass]之前一样在modifiedMethod中返回originalMethod

Answer 1

  for { calc <- calculate(mySource)}
  yield {
    collectStuff
        .map { ts =>
          val errors = MyEngine.checkAll(ts.code, calc)
          (ts, errors)
        }
        .map { x =>
          x._2
            .leftMap(xs => {
              addInformation(x._1, xs.toList)
            })
            .toEither
        }
  }

将给您一个Future[Flow[ByteString, MyValidation[MyClass], NotUsed]]而不是Future[Nothing] 但是如果您想删除Future，则需要在某个地方Await（在您调用calculate（然后不需要for）时或之后。通常，这不是使用期货的方法

如何将结果从一个源流传递到另一个

1 个答案: