尝试 return Future[Result] 时在 playframework 中获取控制器 return 类型错误

Question

我的控制器流程非常复杂，这就是它的使用方式：

- get the request 

- translate it to my model
  case it succeed: 
      insert some record to my db
         case it succeed:
            perform some api call
               case it succeed:
                  match its returned value 
               case failed:
         recover with proper result
         case failed:
            recover with proper result
  case failed:
     recover with proper result

但是在我的代码中我得到一些错误，我在某处没有正确返回结果的未来...

错误：

expression of type unit doesnt conform to expected type Future[Result]

这是代码：

def myApiMethod ...

    val requestAsModel = request.body.extractOpt[MyModel]

    requestAsModel match {
      case Some(req) =>
        // returning Future
        dbService.insert.onComplete {
          case Success(x) =>
            myApiService.doSomething(req).map {
              case good: GoodRes => Ok(Extraction.decompose(good))
              case bad: BadRes => BadRequest(Extraction.decompose(bad))
            } recover {
              case ex =>
                InternalServerError
            }
          case Failure(ex) => Future {InternalServerError}
        }
      case None =>
        Future.successful(BadRequest("not a good body content"))
    }
}

所以你看到这里的问题了吗？

以这种方式编写控制器看起来是个好习惯吗？

谢谢

Answer 1

未来 onComplete 方法 returns 单元 documentation。 您需要将 onComplete 替换为 flatMap 调用才能将结果 future 转换为 Future[Result]:

    dbService.insert.flatMap{
        case Success(x) => 
              myApiService.doSomething(req).map {
                  case good: GoodRes => Ok(Extraction.decompose(good))
                  case bad: BadRes => BadRequest(Extraction.decompose(bad))
             } recover {
              case ex =>
                InternalServerError
            }
        case Failure(ex) => Future {InternalServerError}
    }

我建议阅读这篇文章Scala Futures guide。特别是关于组成期货的部分。