Question

在Scala 2.10.6中，使用json4s 3.2.11，我现在受约束，使用以下代码：

object Thing1 extends Enumeration {
    type Thing1 = Value
    val A = Value(0, "A")
    val B = Value(1, "B")
}

object Thing2 extends Enumeration {
    type Thing2 = Value
    val A = Value(0, "A")
    val B = Value(1, "B")
    val C = Value(2, "C")
}

case class ThingHolder(thing1: Thing1, thing2: Thing2)

class ThingSpec extends FlatSpec with Matchers {
    "desrialized" should "match original" in {
        import org.json4s.native.Serialization.writePretty
        implicit val formats = DefaultFormats + new EnumNameSerializer(Thing1) + new EnumNameSerializer(Thing2)

        val original = ThingHolder(Thing1.A, Thing2.C)
        println(original)

        val serialized = writePretty(original)
        println(serialized)

        val jValue = JsonParser.parse(serialized)
        val deserialized = jValue.camelizeKeys.extract[ThingHolder]
        println(deserialized)
        println(deserialized.thing1)

        deserialized should be(original)
    }
}

结果：

ThingHolder(A,C)
{
  "thing1":"A",
  "thing2":"C"
}
ThingHolder(A,C)
A

ThingHolder(A,C) was not equal to ThingHolder(A,C)

这与Enumeration有关，但不是其他类型，例如String。如果case类只有一个Enumeration，它可以正常工作。这是为什么？我可以通过使用字符串反序列化为case类，然后按照我希望的方式映射到case类来纠正这个问题。有没有办法哄骗json4s直接反序列化，以便反序列化的对象与原始对象匹配？

编辑：

这是我提到的用于映射的黑客攻击：

case class ThingHolderSerialized(thing1: String, thing2: String)

...

val deserialized = jValue.camelizeKeys.extract[ThingHolderSerialized]
val reconstituted = ThingHolder(Thing1.withName(deserialized.thing1), Thing2.withName(deserialized.thing2))

编辑：

实际上，不需要单独的序列化程序类：

val deserialized = jValue.camelizeKeys.extract[ThingHolder]
val reconstituted = ThingHolder(Thing1.withName(deserialized.thing1.toString), Thing2.withName(deserialized.thing2.toString))

编辑：

似乎没有正确反序列化的东西，因为这也构建了匹配的重构对象：

val reconstituted = deserialized.copy(thing1 = Thing1.withName(deserialized.thing1.toString))

但不是这样：

val reconstituted = deserialized.copy(thing2 = Thing2.withName(deserialized.thing2.toString))

Answer 1

我把它作为一个解决方案，用于需要在多个Enumerations存在的情况下支持json4s案例类反序列化的人。这不是我正在寻找的答案，但在此期间有效：

object Thing1 extends Enumeration {
    type Thing1 = Value
    val A = Value(0, "A")
    val B = Value(1, "B")
}

object Thing2 extends Enumeration {
    type Thing2 = Value
    val A = Value(0, "A")
    val B = Value(1, "B")
    val C = Value(2, "C")
}

case class ThingHolder(thing1: Thing1, thing2: Thing2)

class ThingSpec extends FlatSpec with Matchers {
    "reconstituted" should "match original" in {
        import org.json4s.native.Serialization.writePretty
        implicit val formats = DefaultFormats + new EnumNameSerializer(Thing1) + new EnumNameSerializer(Thing2)

        val original = ThingHolder(Thing1.A, Thing2.C)
        println(original)

        val serialized = writePretty(original)
        println(serialized)

        val jValue = JsonParser.parse(serialized)
        val deserialized = jValue.camelizeKeys.extract[ThingHolder]
        println(deserialized)

        val reconstituted = deserialized.copy(thing1 = Thing1.withName(deserialized.thing1.toString))
        reconstituted should be(original)
    }
}

关键是我们正在构建reconstituted的行。我们制作副本，并在此过程中，将thing1转换为字符串并返回枚举。幸运的是，toString在没有正确反序列化字段的情况下工作。

json4s反序列化对象与原始对象不匹配

1 个答案: