Question

在我的method1中，我需要异步调用另一个方法2，它返回Option（result1）。比如果result1为空，我需要异步调用另一个方法3，但如果result1不为空，我只需要返回它。

以下是方法：

  def signIn(username: String): Future[User] = {
    for {
      foundUser <- userService.findByUsername(username) // this method returns Future[Option[User]], 
      // foundUser is Option[User]
      user <- if (foundUser.isEmpty) {
        val newUser = User(username = "User123")
        userService.create(newUser).map(Some(_)) // this method returns Future[Option[User]]
      }
      else
        // Here I want to return just foundUser, of course, it is not possible. 
        // IS THIS APPROACH CORRECT?? DOES THIS LINE CREATE ASYNCHRONOUS CALL?          
        Future.successful(foundUser)
    } yield user 
  }

问题是：

Future.successful(foundUser) - 这种方法在上面的代码中是否正确？这行是否会创建异步调用？如果是这样，如何避免呢？我已经异步获取 foundUser ，并且我不想仅仅为了返回已经获取的值而进行额外的异步调用。

Answer 1

Future.successful不会在提供的ExecutionContext上排队其他功能。它只是使用Promise[T]来创建完整的Future[T]：

/** Creates an already completed Future with the specified result.
   *
   *  @tparam T       the type of the value in the future
   *  @param result   the given successful value
   *  @return         the newly created `Future` instance
   */
  def successful[T](result: T): Future[T] = Promise.successful(result).future

作为旁注，您可以使用Option.fold：

减少样板量

def signIn(username: String): Future[User] = 
  userService
    .findByUsername(username)
    .flatMap(_.fold(userService.create(User(username = "User123")))(Future.successful(_))

Answer 2

@Yuval Itzchakov回答了您的问题，但作为旁注，您可能希望在您的案例中直接使用flatMap模式匹配。我个人觉得它更具可读性：

def signIn(username: String): Future[User] =
  userService.findByUsername(username)
    .flatMap {
      case Some(user) => Future.successful(user)
      case _ => userService.create(User(username = "User123"))
    }

Scala Future - 为了理解，混合同步和异步

2 个答案: