如何將未來[向量[UserLocation]]轉換爲未來[向量[User]]

Question

我有一個方法，我想返回Future[Vector[user]]。

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

其中用戶位置看起來像：

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


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

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

} 

我有一個基於用戶ID喜歡返回單個用戶的方法：

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

如何建立一個未來[矢量[用戶]鑑於上述？

Answer 1

如果map的Future[Vector[UserLocation]]，你可以很容易產生內Vector[Future[User]]，從包含的Vector[UserLocation]：

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

您可以使用Future.sequence反轉Vector[Future[User]]到Future[Vector[User]]：

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

或者使用Future.traverse：

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

這會給你一個Future[Future[Vector[User]]]，可以通過將map更改爲flatMap來修復。全部放在一起：

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

或者用換理解：

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