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Following this article https://github.com/FasterXML/jackson-module-scala/wiki/Enumerations
The enumeration declaration is as
object UserStatus extends Enumeration {
type UserStatus = Value
val Active, Paused = Value
}
class UserStatusType extends TypeReference[UserStatus.type]
case class UserStatusHolder(@JsonScalaEnumeration(classOf[UserStatusType]) enum: UserStatus.UserStatus)
The DTO is declared as
class UserInfo(val emailAddress: String, val userStatus:UserStatusHolder) {
}
and the serialization code is
val mapper = new ObjectMapper()
mapper.registerModule(DefaultScalaModule)
def serialize(value: Any): String = {
import java.io.StringWriter
val writer = new StringWriter()
mapper.writeValue(writer, value)
writer.toString
}
The resulting JSON serialization is
{
"emailAddress":"[email protected]",
"userStatus":{"enum":"Active"}
}
Is it possible to get it the following form ?
{
"emailAddress":"[email protected]",
"userStatus":"Active"
}
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Note that Jackson does not use java. io. Serializable for anything: there is no real value for adding that. It gets ignored.
In order to serialize Enum, we take the help of ObjectMapper class. We use the writeValueAsString() method of ObjectMapper class for serializing Enum. If we serialize Enum by using the writeValueAsString() method, it will represent Java Enums as a simple string.
All you have to do is create a static method annotated with @JsonCreator in your enum. This should accept a parameter (the enum value) and return the corresponding enum. This method overrides the default mapping of Enum name to a json attribute .
Have you tried:
case class UserInfo(
emailAddress: String,
@JsonScalaEnumeration(classOf[UserStatusType]) userStatus: UserStatus.UserStatus
)
The jackson wiki's example is a little misleading. You don't need the holder class. Its just an example of a thing that has that element. The thing you need is the annotation
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