如何使用Jackson将Scala值类序列化为字符串？

Question

I want to use Scala's value classes (or a normal case class) to give stronger types to some strings in my program. When I serialize instances of these classes using Jackson I want them to be strings.

For example:

case class Brand(name: String) extends AnyVal
val brands = Seq(Brand("Coke"), Brand("Disney"))
val brandCount = Map(Brand("Coke") -> 5, Brand("Disney") -> 10)

Since Brand is just a wrapper for a String, I want the corresponding JSON serialization for these variables to be:

brands:     ["Coke", "Disney"]
brandCount: {"Coke": 5, "Disney": 10}

By default I get:

import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.module.scala.DefaultScalaModule

val mapper = new ObjectMapper()
mapper.registerModule(DefaultScalaModule)

println(mapper.writeValueAsString(brands))
// ==> [{"name":"Coke"},{"name":"Disney"}]
println(mapper.writeValueAsString(brandCount))
// ==> {"Brand(Coke)":5,"Brand(Disney)":10}

The best I could come up with is to define a custom serializer and key serializer for Brand :

import com.fasterxml.jackson.databind.module.SimpleModule
import com.fasterxml.jackson.databind.JsonSerializer
import com.fasterxml.jackson.databind.SerializerProvider
import com.fasterxml.jackson.core.JsonGenerator

class BrandSerializer extends JsonSerializer[Brand] {
  override def serialize(
    b: Brand,
    json: JsonGenerator,
    provider: SerializerProvider
  ): Unit = {
    json.writeString(b.name)
  }
}

class BrandKeySerializer extends JsonSerializer[Brand] {
  override def serialize(
    b: Brand, 
    json: JsonGenerator,
    provider: SerializerProvider
  ): Unit = {
    json.writeFieldName(b.name)
  }
}

val serializers = new SimpleModule("Serializers");
serializers.addSerializer(classOf[Brand], new BrandSerializer())
serializers.addKeySerializer(classOf[Brand], new BrandKeySerializer());

val mapper = new ObjectMapper()
mapper.registerModule(DefaultScalaModule)
mapper.registerModule(serializers)

println(mapper.writeValueAsString(brands))
// ==> ["Coke","Disney"]
println(mapper.writeValueAsString(brandCount))
// ==> {"Coke":5,"Disney":10}

Is there a better (or less verbose) way to serialize these value classes (or any case class) as strings?

Answer 1

Well, I don't think, there is a way to make it do all you want, and nothing you don't want in one go ... The best thing you can do that comes to mind is replace Brand with Product1 , and b.name with b._1 (I also think, your key serializer is wrong - it should be writing out b.getClass.getSimpleName , not not b.name ).

That wouldn't make it any less verbose for writing out a single class, but will at least eliminate the need to create a separate serializer for every type like that.

The downside, of course, is that every one-field case class will end up written this way, which may be more than what you wished for ..