將 JSON object 轉換為列表<>

Question

我想將 JSON object 轉換為不同的格式。 我的 JSON object 看起來像。

{
    "schema": [
        {
            "name": "cust_id",
            "data_type": "String",
            "nullable": true
        },
        {
            "name": "source_type",
            "data_type": "String",
            "nullable": true
        }
    ]
}

我想要下面的 output：

val colType = List(("cust_id","String",true),("source_type","String",true))

我想要 Scala 中的轉換。

Answer 1

使用Circe怎么樣？

import io.circe._, io.circe.generic.auto._, io.circe.parser._, io.circe.syntax._
import cats.syntax.show._

case class Schema(schema: List[Container])
case class Container(name:String, data_type: String, nullable: Boolean)

val json = """
{
"schema": [{
        "name": "cust_id",
        "data_type": "String",
        "nullable": true
    },
    {
        "name": "source_type",
        "data_type": "String",
        "nullable": true
    }
]
}"""

decode[Schema](json) match {
  case Right(result) => println(result.schema.map{case Container(a,b,c) => (a,b,c)})
  case Left(failure) => println(failure.show)
    
}

Answer 2

您也可以使用 spray.json。

在 build.sbt 中添加依賴

libraryDependencies += "io.spray" %% "spray-json" % "1.3.6"

應用：

import spray.json._

//create your model classes
case class MyItem(name: String, data_type: String, nullable: Boolean)
case class MySchema(schema: Seq[MyItem])

//prepare JSON protocol
trait MyJsonProtocol extends DefaultJsonProtocol {
  //MyItem has 3 fields, so you have to use jsonFormat3
  implicit val myItemFormat: RootJsonFormat[MyItem] = jsonFormat3(MyItem)

  //MySchema has 1 field, so you have to use jsonFormat1
  implicit val mySchemaFormat: RootJsonFormat[MySchema] = jsonFormat1(MySchema)
}

//extends your application with your JSON protocol
object MyApplication extends App with MyJsonProtocol {
  val json =
    """{
      |"schema": [{
      |        "name": "cust_id",
      |        "data_type": "String",
      |        "nullable": true
      |    },
      |    {
      |        "name": "source_type",
      |        "data_type": "String",
      |        "nullable": true
      |    }
      |]
      |}""".stripMargin
  
  //execute parsing
  val parsed = json.parseJson.convertTo[MySchema]

  //true
  println(parsed.schema == List(MyItem("cust_id","String",true), MyItem("source_type","String",true)))
}

Answer 3

我將演示如何使用play-json來實現。 如果您想將其序列化為案例 class，如其他答案中所示，您可以定義類及其序列化程序：

case class Schema(name: String, data_type: String, nullable: Boolean)

object Schema {
  implicit val format: OFormat[Schema] = Json.format[Schema]
}

case class Schemas(schema: Seq[Schema])

object Schemas {
  implicit val format: OFormat[Schemas] = Json.format[Schemas]
}

然后只需調用：

println(Json.parse(jsonString).as[Schemas])

代碼在Scastie運行。

但如果你真的想要元組列表，你可以這樣做：

val path = JsPath \ "schema"

val schemas = path(Json.parse(jsonString)) match {
  case List(arr) =>
    arr match {
      case arr: JsArray =>
        arr.value.map(_.validate[JsObject] match {
          case JsSuccess(value, _) =>
            value.values.toList match {
              case List(a, b, c) =>
                (a, b, c)
              case _ => ???
            }
          case JsError(errors) =>
            ??? // Error handling
        }).toList
      case _ => ??? // Error handling
    }
  case _ => ??? // Error handling
}

在Scastie運行的代碼