如何在Scala中打开选项[Map（A，B）]？

Question

我已经做了足够的Scala来了解丑陋的代码是什么样的。观察：

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

预期产出：

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

这是我的Tyler Perry代码，由M. Knight Shama Llama Yama执导：

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

然而，当sm为None时，这不起作用:-(。我得到一个错误，说没有什么没有＆＃34;过滤＆＃34;方法（它认为我们在线(k,v) <- m上进行过滤格拉西亚斯！

Answer 1

接受选项可迭代的事实

(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(", ")