将一组将来列表平化为一个将来列表

Question

I'm trying to call persistenceService.find() returning a Future[List[Scan]] foreach email passed to my method findByEmails which should return the merged Future[List[Scan]] of every calls to persistenceService.find()

So currently, I have this:

def findByEmails(emails: Set[String]): Future[List[Scan]] = {
    val results: Set[Future[List[Scan]]] = emails.map(email => persistenceService.find("report.commitsStats.contributors." + email -> BSONDocument("$exists" -> true)))
}

I can't figure out how to merge each Future[List[Scan]] in the results Set so my method return one Future[List[Scan]] .

Any thoughts?

Answer 1

You can use Future.sequence to combine a list of futures, then a flatten on the list should be enough, the code might look similar to this not tested code:

def f(str: String): Future[List[Scan]] = ???

def g(emails: List[String]): Future[List[Scan] = {
  val futures: List[Future[list[Scan]]] = emails.map(f)
  val futureListList: Future[List[List[Scan]] = Future.sequence(futures)
  futureListList.map(_.flatten)
}