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我有一些用于往返 scala 案例类的工作 jackson scala 模块代码。Jackson 非常适合扁平案例类，但是当我制作一个包含其他案例类列表的案例时，我似乎需要的代码量很多。考虑：

abstract class Message
case class CardDrawn(player: Long, card: Int, mType: String = "CardDrawn") extends Message
case class CardSet(cards: List[CardDrawn], mType: String = "CardSet") extends Message

为了让 CardSet 使用 jackson scala 模块往返于 json 之间，我使用了一个用 java 编写的自定义序列化器/反序列化器：

object ScrumGameMashaller {

  val mapper = new ObjectMapper() 
  val module = new SimpleModule("CustomSerializer")
  module.addSerializer(classOf[CardSet], new CardSetSerializer)
  module.addDeserializer(classOf[CardSet], new CardSetDeserializer)
  val scalaModule = DefaultScalaModule
  mapper.registerModule(scalaModule)
  mapper.registerModule(module)

  def jsonFrom(value: Any): String = {
    import java.io.StringWriter
    val writer = new StringWriter()
    mapper.writeValue(writer, value)
    writer.toString
  }

  private[this] def objectFrom[T: Manifest](value: String): T =
    mapper.readValue(value, typeReference[T])

  private[this] def typeReference[T: Manifest] = new TypeReference[T] {
    override def getType = typeFromManifest(manifest[T])
  }

  private[this] def typeFromManifest(m: Manifest[_]): Type = {
    if (m.typeArguments.isEmpty) { m.runtimeClass }
    else new ParameterizedType {
      def getRawType = m.runtimeClass
      def getActualTypeArguments = m.typeArguments.map(typeFromManifest).toArray
      def getOwnerType = null
    }
  }

使用序列化器：

public class CardSetSerializer extends JsonSerializer<CardSet> {
@Override
    public void serialize(CardSet cardSet, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException {
        jgen.writeStartObject();
        jgen.writeArrayFieldStart("cards");
        List<CardDrawn> cardsDrawn = cardSet.cards();
        scala.collection.Iterator<CardDrawn> iter = cardsDrawn.iterator();
        while(iter.hasNext()){
            CardDrawn cd = iter.next();
            cdSerialize(jgen,cd);
        }
        jgen.writeEndArray();
        jgen.writeStringField("mType", "CardSet");
        jgen.writeEndObject();      
    }

    private void cdSerialize(JsonGenerator jgen, CardDrawn cd) throws IOException, JsonProcessingException {
        jgen.writeStartObject();
        jgen.writeNumberField("player", cd.player());
        jgen.writeNumberField("card", cd.card());
        jgen.writeEndObject();
    }
}

和匹配的解串器： 

public class CardSetDeserializer extends JsonDeserializer<CardSet> {

    private static class CardDrawnTuple {
        Long player;
        Integer card;
    }

    @Override
    public CardSet deserialize(JsonParser jsonParser, DeserializationContext cxt) throws IOException, JsonProcessingException {
        ObjectCodec oc = jsonParser.getCodec();
        JsonNode root = oc.readTree(jsonParser);
        JsonNode cards = root.get("cards");
        Iterator<JsonNode> i = cards.elements();
        List<CardDrawn> cardObjects = new ArrayList<>();
        while( i.hasNext() ){
            CardDrawnTuple t = new CardDrawnTuple();
            ObjectNode c = (ObjectNode) i.next();
            Iterator<Entry<String, JsonNode>> fields = c.fields();
            while( fields.hasNext() ){
                Entry<String,JsonNode> f = fields.next();
                if( f.getKey().equals("player")) {
                    t.player = f.getValue().asLong();
                } else if( f.getKey().equals("card")){
                    t.card = f.getValue().asInt();
                } else { 
                    System.err.println(CardSetDeserializer.class.getCanonicalName()+ " : unknown field " + f.getKey());
                }
            }
            CardDrawn cd = new CardDrawn(t.player, t.card, "CardDrawn");
            cardObjects.add(cd);
        }

        return new CardSet(JavaConversions.asScalaBuffer(cardObjects).toList(), "CardSet");
    }

}

这似乎是很多代码来处理 scala 中相当普通的东西。这段代码可以改进吗（我错过了杰克逊必须让这件事变得简单的什么）？否则是否有一个库可以自动执行结构化案例类？jerkson 示例看起来很简单，但似乎已被放弃。

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2 回答 2

0

Argonaut 做得很好。Mark Hibbard帮助我完成了下面的示例。所需要的只是为这些类型创建一个编解码器，它将隐式添加asJson到您的对象中以将它们转换为字符串。它还将decodeOption[YourClass]在字符串中添加一个以提取对象。以下：

package argonaut.example

import argonaut._, Argonaut._

abstract class Message
case class CardDrawn(player: Long, card: Int, mType: String = "CardDrawn") extends Message
case class CardSet(cards: List[CardDrawn], mType: String = "CardSet") extends Message

object CardSetExample  {

  implicit lazy val CodecCardSet: CodecJson[CardSet] = casecodec2(CardSet.apply, CardSet.unapply)("cards","mType")
  implicit lazy val CodecCardDrawn: CodecJson[CardDrawn] = casecodec3(CardDrawn.apply, CardDrawn.unapply)("player", "card", "mType")

  def main(args: Array[String]): Unit = {
    val value = CardSet(List(CardDrawn(1L,2),CardDrawn(3L,4)))
    println(s"Got some good json ${value.asJson}")

    val jstring =
      """{
        | "cards":[
        |   {"player":"1","card":2,"mType":"CardDrawn"},
        |   {"player":"3","card":4,"mType":"CardDrawn"}
        | ],
        | "mType":"CardSet"
        | }""".stripMargin

    val parsed: Option[CardSet] =
      jstring.decodeOption[CardSet]

    println(s"Got a good object ${parsed.get}")
  }
}

输出：

Got some good json {"cards":[{"player":"1","card":2,"mType":"CardDrawn"},{"player":"3","card":4,"mType":"CardDrawn"}],"mType":"CardSet"}

Got a good object CardSet(List(CardDrawn(1,2,CardDrawn), CardDrawn(3,4,CardDrawn)),CardSet)

于 2013-09-07T17:18:27.077 回答
0

这个问题很老，但也许有人仍然会发现它有帮助。除了 Argonaut，Scala 还有几个 Json 库。在这里您可以找到更新到 2016 年初的列表（它仍然可以为您提供良好的整体情况）。它们中的大多数（可能全部）应该允许您提出自定义序列化器/解轨器的更干燥版本。我更喜欢json4s，它旨在为包括 Jackson 在内的多个库提供单个 AST（有点像 slf4j 为日志库所做的那样）。在这篇文章中，您可以找到一个使用 Json4s 和 Akka Http 的 Json 自定义序列化器/反序列化器的工作示例。

于 2018-07-25T09:39:00.197 回答