[英]Scala convert Map[A, Seq[Future[Option[B]]]] to Future[Map[A, Seq[B]]]?
[英]How to open an Option[Map(A,B)] in Scala?
我已經做了足夠的Scala來了解丑陋的代碼。 觀察:
val sm Option[Map[String,String]] = Some(Map("Foo" -> "won", "Bar" -> "too", "Baz" -> "tree"))
預期產量:
: String = Foo=won,Bar=too,Baz=tree
這是我的M. Knight Shama Llama Yama指揮的Tyler Perry代碼:
val result = (
for {
m <- sm.toSeq;
(k,v) <- m
} yield s"$k=$v"
).mkString(",")
但是,當sm為None時,這是行不通的:-(。我收到一條錯誤消息,說Nothing沒有“過濾”方法(它認為我們在(k,v) <- m
行上進行過濾)Gracias!
(for {
map <- sm.iterator
(k, v) <- map.iterator
} yield s"$k=$v").mkString(",")
res1: String = "Foo=won,Bar=too,Baz=tree"
scala> val sm: Option[Map[String, String]] = None
sm: Option[Map[String, String]] = None
scala> (for {
map <- sm.iterator
(k, v) <- map.iterator
} yield s"$k=$v").mkString(",")
res44: String = ""
scala> val sm: Option[Map[String,String]] = Some(Map("Foo" -> "won", "Bar" -> "too", "Baz" -> "tree"))
sm: Option[Map[String,String]] = Some(Map(Foo -> won, Bar -> too, Baz -> tree))
scala> val yourString = sm.getOrElse(Map[String, String]()).toList.map({
case (key, value) => s"$key=$value"
}).mkString(", ")
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