开发case class
带有可选参数的 a 时,请使用Option
:
case class Character(name: String, location: Option[String])
Character("Tyrion Lannister", None)
然后你所要做的就是修改你的数据提取器,如果它没有找到数据,则传递一个 None 选项:
val tyrion = Map("location" -> "King's Landing", "name" -> "Cersei Lannister")
val cersei = Map("father" -> "Tywin Lannister?", "name" -> "Cersei Lannister")
val jaime = Map("father" -> "Tywin Lannister", "location" -> "Tower of the Hand")
val characters = List(tyrion, cersei, jaime)
case class Character(name: String, location: Option[String])
characters.map(x => Character(x.getOrElse("name", "A CHARACTER HAS NO NAME"), x.get("location")))
结果characters.map(...)
是这样的:
res0: List[Character] = List(
Character(Cersei Lannister,Some(King's Landing)),
Character(Cersei Lannister,None),
Character(A CHARACTER HAS NO NAME NAME,Some(Tower of the Hand)))
从 的源代码中RichSearchHit
,sourceAsMap
应该返回一个Map
对象:
def sourceAsMap: Map[String, AnyRef] = if (java.sourceAsMap == null) Map.empty else java.sourceAsMap.asScala.toMap
鉴于您使用的是Map
速记,您应该能够将代码转换为:
case class Character(name: String, location: Option[String])
implicit object CharacterHitAs extends HitAs[Character] {
override def as(hit: RichSearchHit): Character = {
Character(hit.sourceAsMap.getOrElse("name", "A CHARACTER HAS NO NAME"), hit.sourceAsMap.get("location")) }}