问题
My JSON File containes below details { "category":"age, gender,post_code" }
My scala code is below one
val filename = args.head
println(s"Reading ${args.head} ...")
val json = Source.fromFile(filename)
val mapper = new ObjectMapper() with ScalaObjectMapper
mapper.registerModule(DefaultScalaModule)
val parsedJson = mapper.readValue[Map[String, Any]](json.reader())
val data = parsedJson.get("category").toSeq
It's returning Seq(Any) = example List(age, gender,post_code) but I need Seq(String) output please if any has an idea about this please help me.
回答1:
The idea in scala is to be typesafe whenever possible which you are giving away using Map[String, Any].
So, I recommend using a data class that represents your JSON data.
Example,
define a mapper,
scala> import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.databind.ObjectMapper
scala> import com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper
import com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper
scala> import com.fasterxml.jackson.module.scala.DefaultScalaModule
import com.fasterxml.jackson.module.scala.DefaultScalaModule
scala> val mapper = new ObjectMapper() with ScalaObjectMapper
mapper: com.fasterxml.jackson.databind.ObjectMapper with com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper = $anon$1@d486a4d
scala> mapper.registerModule(DefaultScalaModule)
res0: com.fasterxml.jackson.databind.ObjectMapper = $anon$1@d486a4d
Now, when you deserialise to Map[K, V] you can not specify all the nested data-structures,
scala> val jsonString = """{"category": ["metal", "metalcore"], "age": 10, "gender": "M", "postCode": "98109"}"""
jsonString: String = {"category": ["metal", "metalcore"], "age": 10, "gender": "M", "postCode": "98109"}
scala> mapper.readValue[Map[String, Any]](jsonString)
res2: Map[String,Any] = Map(category -> List(metal, metalcore), age -> 10, gender -> M, postCode -> 98109)
Following is a solution casting some key to desired data-structure but I personally don not recommend.
scala> mapper.readValue[Map[String, Any]](jsonString).get("category").map(_.asInstanceOf[List[String]]).getOrElse(List.empty[String])
res3: List[String] = List(metal, metalcore)
Best solution is to define a data class which I'm calling SomeData in following example and deserialize to it. SomeData is defined based on your JSON data-structure.
scala> final case class SomeData(category: List[String], age: Int, gender: String, postCode: String)
defined class SomeData
scala> mapper.readValue[SomeData](jsonString)
res4: SomeData = SomeData(List(metal, metalcore),10,M,98109)
scala> mapper.readValue[SomeData](jsonString).category
res5: List[String] = List(metal, metalcore)
回答2:
Just read the JSON as a JsonNode, and access the property directly:
val jsonNode = objectMapper.readTree(json.reader())
val parsedJson = jsonNode.get("category").asText
来源：https://stackoverflow.com/questions/51130430/scala-deserialize-json-to-collection