Question

如何为列表中的每个JObject获取密钥scenario0和scenario1？

便便是：

poop: List[org.json4s.JsonAST.JValue] = List(JObject(List((street,JString(Bulevard)), (city,JString(Helsinki)))), JObject(List((street,JString(Bulevard)), (city,JString(Helsinki)))))

代码：

import org.json4s.jackson.JsonMethods._
import org.json4s._

implicit val formats = DefaultFormats

val json = parse("""
         {
            "address0": {
              "scenario0": {
                "street": "Bulevard",
                "city": "Helsinki"
              },
              "scenario1": {
                "street": "Bulevard",
                "city": "Helsinki"
              }
            },
            "address1": {
              "scenario0": {
                 "street": "Bulevard",
                 "city": "Helsinki"
               },
               "scenario1": {
                 "street": "Bulevard",
                 "city": "Helsinki"
               }
            }
         }""")

val poop = (json \ "address0").children
poop.foreach(p => {

})

Answer 1

你的意思是你想要这样，就像JObject每个人一样。

json.children.flatMap(_.children) 
res13: List[JValue] = List(
  JObject(List((street,JString(Bulevard)), (city,JString(Helsinki)))),
  JObject(List((street,JString(Bulevard)), (city,JString(Helsinki)))),
  JObject(List((street,JString(Bulevard)), (city,JString(Helsinki)))),
  JObject(List((street,JString(Bulevard)), (city,JString(Helsinki))))
)

但您可能会发现使用像circe，play-json或argonaut这样的库将它们解析为案例类会更有用。这样你就不必在任何地方继续解析json。

Answer 2

如果您需要值且没有固定架构： val poop: Map[String, Any] = json.asInstanceOf[JObject].values返回Map(address0 -> Map(scenario0 -> Map(street -> Bulevard, city -> Helsinki), scenario1 -> Map(street -> Bulevard, city -> Helsinki)), address1 -> Map(scenario0 -> Map(street -> Bulevard, city -> Helsinki), scenario1 -> Map(street -> Bulevard, city -> Helsinki)))
如果您需要值并且具有固定架构，那么extracting of case class可能会有用。
如果您需要AST：val poop = json.asInstanceOf[JObject].obj提供List[org.json4s.JsonAST.JField] = List((address0,JObject(List((scenario0,JObject(List((street,JString(Bulevard)), (city,JString(Helsinki))))), (scenario1,JObject(List((street,JString(Bulevard)), (city,JString(Helsinki)))))))), (address1,JObject(List((scenario0,JObject(List((street,JString(Bulevard)), (city,JString(Helsinki))))), (scenario1,JObject(List((street,JString(Bulevard)), (city,JString(Helsinki)))))))))

P.S。为简单起见，我省略了监护人检查。

使用json4s-jackson解析密钥和值的JObject

2 个答案: