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.
这是方法：
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) – 这个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？我已经异步获取了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.
解决方法
Future.successful不会在提供的ExecutionContext上排队其他函数.它只是使用
Promise[T]来创建一个完整的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(_))
