Jackson JSON Serialization without field name from one string row

Question

I have this JSON to deserialize:

"data": {
  "type": 18,        
  "msg": "firstName,lastName,15" 
  "timestamp": 1551770400000 
}

I want to get this data in my model:

class DataDto(
    type: String,
    timestamp: Long,
    msg: DataMsgDto?
) {
    @JsonFormat(shape = JsonFormat.Shape.STRING)
    @JsonPropertyOrder("firstName", "lastName", "age")
    class DataMsgDto(
        firstName: String,
        lastName: String,
        age: Long
    )
}

I use this code to get data:

DataBuffer payload //this is what I get from server
val jsonNode = objectMapper.readTree(payload.toString(StandardCharsets.UTF_8))
objectMapper.treeToValue(jsonNode, DataDto::class.java)

But this does not work because in msg I don't have fields. So, how can I do this?

Answer 1

You have a JSON property that contains comma-separated values. The JSON message is valid, and can be deserialized by Jackson.

It's possible to write custom Jackson code that can handle this case automatically

• How to deserialize a string separated by comma to list with Jackson commonly?
• Jackson json property (string) to instance

But the most practical way to solve this is unlikely to be digging around trying to make Jackson understand msg, or writing a custom deserializer. Instead, allow msg to be a String

class DataDto(
    val type: String,
    val timestamp: Long,
    // just use a normal String for msg
    val msg: String?
) 

And then manually decode it in Kotlin

fun parseDataDtoMsg(dto: DataDto): DataMsgDto? {
  val msgElements = dto.msg?.split(",") ?: return null

  // parse each message - you can chose whether to 
  // throw an exception if the `msg` does not match what you expect, 
  // or ignore invalid data and return `null`

  val firstName = ...
  val lastName = ...
  val age = ...

  return DataDtoMsg(
      firstName = firstName,
      lastName = lastName,
      age = age,
  )
}

data class DataMsgDto(
    val firstName: String
    val lastName: String
    val age: Long,
)