在Json4s中使用案例类将(非常)复杂的JSON对象转换为Scala对象 [英] Converting (Very) Complicated JSON objects to Scala Objects With Case Classes In Json4s
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问题描述
我有一个非常复杂的JSON文件,如下所示:
I have a very complicated JSON file that looks like this:
{
"Animals": [
[
100,
"Mammals",
[
1,
"Cat",
50,
45,
57,
-1
],
[
2,
"Dog",
31,
44,
18,
-1
]
],
[
159,
"Reptiles",
[
1,
"Lizard",
11,
12,
9,
-1
]
]
]
}
我正在尝试解析此结构,并以某种方式从中获取scala对象.
I am attempting to parse this structure and somehow get scala objects out of it.
这是我的尝试:
case class Facts(number: Int, subTypeOfAnimal: String, data: List[Int])
case class Animaltype(value: Int, typeOfAnimal: String, characteristics: List[Facts])
case class Animal(rows: List[Animaltype])
这当然不能转换数据.它返回一个JNothing.我想知道如何在这种JArrays中正确表达复杂的JArray.
This, of course, fails to convert the data. It returns a JNothing. I am wondering how I can express complex JArrays within JArrays of this kind properly.
任何帮助都是有用的
谢谢!
推荐答案
您可以为Facts
和AnimalType
定义CustomSerializer
.
import scala.util.Try
import org.json4s._
import org.json4s.JsonDSL._
import org.json4s.native.JsonMethods._
import org.json4s.native.Serialization
case class Facts(number: Int, subTypeOfAnimal: String, data: List[Int])
case class AnimalType(value: Int, typeOfAnimal: String, characteristics: List[Facts])
case class Animal(Animals: List[AnimalType])
implicit val formats = Serialization.formats(NoTypeHints) +
new AnimalTypeSerializer + new FactsSerializer
class FactsSerializer extends CustomSerializer[Facts](format => ( {
case JArray(JInt(nr) :: JString(subType) :: data) =>
Facts(nr.toInt, subType, data.collect{ case JInt(i) => i.toInt})
}, { case _ => throw new RuntimeException("No serializing")}))
class AnimalTypeSerializer extends CustomSerializer[AnimalType](format => ( {
case JArray(JInt(value) :: JString(typeAnimal) :: factsArrays) =>
val facts = factsArrays.collect { case facts: JArray =>
Try(facts.extract[Facts]).toOption
}.flatten
AnimalType(value.toInt, typeAnimal, facts)
}, { case _ => throw new RuntimeException("No serializing")}))
如果将输入json作为值json
,则可以使用:
If you take your input json as the value json
you can deserialize it with :
parse(json).extract[Animal]
// Animal = Animal(List(
// AnimalType(100,Mammals,List(
// Facts(1,Cat,List(50, 45, 57, -1)), Facts(2,Dog,List(31, 44, 18, -1))
// )),
// AnimalType(159,Reptiles,List(
// Facts(1,Lizard,List(11, 12, 9, -1))
// ))
// ))
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