Question

我有一个方法，我想返回Future[Vector[user]]。

方法userLocationService.getUserLocationsInList将返回Future[Vector[UserLocation]]。

其中UserLocation类似于：

case class UserLocation(id: Int, locationId: Int, userId: Int)


def getUsersInLocation(locationIdList: Set[Int]): Future[Vector[User]] = {

   userLocationService.getUserLocationsInList(locationIdList).map{
      userLocations =>
          // ????????????
   }

}

我有一个基于UserId返回单个用户的方法，如：

userService.getById(userId: Int): Future[User]

如何根据上述内容构建Future [Vector [User]]？

Answer 1

如果您map Future[Vector[UserLocation]]，则可以从包含的Vector[Future[User]]内轻松生成Vector[UserLocation]：

userLocations.map(location => userService.getById(location.userId))

您可以使用Future.sequence将Vector[Future[User]]转换为Future[Vector[User]]：

Future.sequence(userLocations.map(location => userService.getById(location.userId)))

或使用Future.traverse：

Future.traverse(userLocations) { location => userService.getById(location.userId) }

这将为您留下Future[Future[Vector[User]]]，可以通过将map更改为flatMap来解决此问题。把它们放在一起：

def getUsersInLocation(locationIdList: Set[Int]): Future[Vector[User]] = {
  userLocationService.getUserLocationsInList(locationIdList).flatMap { locations =>
    Future.traverse(locations) { location =>
      userService.getById(location.userId)
    }
  }
}

或者为了理解：

def getUsersInLocation(locationIdList: Set[Int]): Future[Vector[User]] = {
  for {
    locations <- userLocationService.getUserLocationsInList(locationIdList)
    users <- Future.traverse(locations) { location =>
      userService.getById(location.userId)
    }
  } yield users
}

如何将Future [Vector [UserLocation]]转换为Future [Vector [User]]

1 个答案: