如何在Scala中打开Option [Map（A，B）]？

Question

我已经做了足够的Scala来了解丑陋的代码。 观察：

 val sm Option[Map[String,String]] = Some(Map("Foo" -> "won", "Bar" -> "too", "Baz" -> "tree"))

预期产量：

 : String = Foo=won,Bar=too,Baz=tree

这是我的M. Knight Shama Llama Yama指挥的Tyler Perry代码：

 val result = (
     for { 
         m <- sm.toSeq; 
         (k,v) <- m
     } yield s"$k=$v"
 ).mkString(",")

但是，当sm为None时，这是行不通的:-(。我收到一条错误消息，说Nothing没有“过滤”方法（它认为我们在(k,v) <- m行上进行过滤）Gracias！

Answer 1

拥抱选项是可迭代的事实

(for {
   map <- sm.iterator
   (k, v) <- map.iterator
  } yield s"$k=$v").mkString(",")

res1: String = "Foo=won,Bar=too,Baz=tree"

无抵抗

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 = ""

Answer 2

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(", ")