如何使用 Scala 中的任何库将通用的潜在嵌套映射 Map[String, Any] 转换为 case 类？

Question

I've not had much joy with reflection, this answer using shapeless works for some cases (but seems to have many edge cases) Shapeless code to convert Map[String, Any] to case class cannot handle optional substructures

Does anyone know of a nice library that does this in just a few LOCs?

Answer 1

I've found a reasonably neat way to do it using Spray Json

First we define a way to get to a JsObject from a Map[String, Any]

def mapToJsObject(map: Map[String, Any]): JsObject =
  JsObject(fields = map.mapValues(anyToJsValue))

def anyToJsValue(any: Any): JsValue = any match {
  case n: Int => JsNumber(n)
  case n: Long => JsNumber(n)
  case n: Double => JsNumber(n)
  case s: String => JsString(s)
  case true => JsTrue
  case false => JsFalse
  case null | None => JsNull
  case list: List[_] => JsArray(list.map(anyToJsValue).toVector)
  case Some(any) => anyToJsValue(any)
  case map: Map[String, Any] => mapToJsObject(map)
}

Then we can just use convertTo provided we have the implicit JsonFormat in scope

case class Address(street: String, zip: Int)
case class Person(name: String, address: Address)

implicit val addressFormat = jsonFormat2(Address.apply)
implicit val personFormat = jsonFormat2(Person.apply)

"Convert Person example map to Person JsObject" in {
  JsonUtils.mapToJsObject(
    Map(
      "name" -> "Tom",
      "address" -> Map("street" -> "Jefferson st", "zip" -> 10000)
    )
  ).convertTo[Person] must_=== Person("Tom", Address("Jefferson st", 10000))
}

CAVEATs

Spray json only has out of box jsonFormat up to 22 fields!

Can not handle any custom types, eg java.sql.Timestamp , since this isn't a JSON type.

Answer 2

Using jackson:

libraryDependencies += "com.fasterxml.jackson.core" % "jackson-databind" % "2.9.8"
libraryDependencies += "com.fasterxml.jackson.module" %% "jackson-module-scala" % "2.9.8"

case class Foo(a: List[Int], b: Option[Double])
case class Bar(c: Int, d: String, e: Foo)

val mapper = new ObjectMapper().registerModule(DefaultScalaModule)
println(mapper.convertValue(Map(
  "c" -> 3, 
  "d" -> "foo", 
  "e" -> Map("a" -> List(1, 2))), classOf[Bar]))

Output: Bar(3,foo,Foo(List(1, 2),None))

Answer 3

We can use circe

import io.circe._
import io.circe.generic.auto._
import io.circe.parser._
import io.circe.syntax._


def mapToJson(map: Map[String, Any]): Json =
    map.mapValues(anyToJson).asJson

  def anyToJson(any: Any): Json = any match {
    case n: Int => n.asJson
    case n: Long => n.asJson
    case n: Double => n.asJson
    case s: String => s.asJson
    case true => true.asJson
    case false => false.asJson
    case null | None => None.asJson
    case list: List[_] => list.map(anyToJson).asJson
    case list: Vector[_] => list.map(anyToJson).asJson
    case Some(any) => anyToJson(any)
    case map: Map[String, Any] => mapToJson(map)
  }

def mapToCaseClass[T : Decoder](map: Map[String, Any]): T = mapToJson(map).as[T].right.get

Then, if we have any types that are not primitive, we just need to add these to our anyToJson along with an encoder/decoder pair that can encode/decode this type as something primitive.

Eg we can represent java.sql.Timestamp with Long , then

import cats.syntax.either._

  import io.circe.Decoder
  import io.circe.Encoder

  implicit val decodeTimestamp: Decoder[Timestamp] = Decoder.decodeLong.emap(long =>
    Either.catchNonFatal(new Timestamp(long)).leftMap(_ => "Timestamp")
  )

implicit val encodeTimestamp: Encoder[Timestamp] = Encoder.encodeLong.contramap[Timestamp](_.getTime)

and we need to add the line to anyToJson

case n: Timestamp => n.asJson