I have a data class which has a property whose type is another data class, like this:
#Serializable
data class Vehicle (
val color: String,
val miles: Int,
val year: Int,
val garage: Garage
)
#Serializable
data class Garage (
val latitude: Float,
val longitude: Float,
val name: String
)
Upon serializing, it produces output like this:
{
"color" : "black" ,
"miles" : 35000 ,
"year" : 2017 ,
"garage" : { "latitude" : 43.478342 , "longitude" : -91.337000 , "name" : "Paul's Garage" }
}
However I would like garage to be a literal string of its JSON representation, not an actual JSON object. In other words, the desired output is:
{
"color" : "black" ,
"miles" : 35000 ,
"year" : 2017 ,
"garage" : "{ \"latitude\" : 43.478342 , \"longitude\" : -91.337000 , \"name\" : \"Paul's Garage\" }"
}
How can I accomplish this in Kotlin? Can it be done with just kotlinx.serialization or is Jackson/Gson absolutely necessary?
Note that this output is for a specific usage. I cannot overwrite the base serializer because I still need to serialize/deserialize from normal JSON (the first example). In other words, the best scenario would be to convert the first JSON sample to the second, not necessarily to have the data class produce the 2nd sample directly.
Thanks!
Create a custom SerializationStrategy for Vehicle as follows:
val vehicleStrategy = object : SerializationStrategy<Vehicle> {
override val descriptor: SerialDescriptor
get() = buildClassSerialDescriptor("Vehicle") {
element<String>("color")
element<Int>("miles")
element<Int>("year")
element<String>("garage")
}
override fun serialize(encoder: Encoder, value: Vehicle) {
encoder.encodeStructure(descriptor) {
encodeStringElement(descriptor, 0, value.color)
encodeIntElement(descriptor, 1, value.miles)
encodeIntElement(descriptor, 2, value.year)
encodeStringElement(descriptor, 3, Json.encodeToString(value.garage))
}
}
}
Then pass it to Json.encodeToString():
val string = Json.encodeToString(vehicleStrategy, vehicle)
Result:
{"color":"black","miles":35000,"year":2017,"garage":"{\"latitude\":43.47834,\"longitude\":-91.337,\"name\":\"Paul's Garage\"}"}
More info here
Here is a solution with a custom serializer for Garage and an additional class for Vehicle.
Garage to String serializer:
object GarageToStringSerializer : KSerializer<Garage> {
override val descriptor = PrimitiveSerialDescriptor("GarageToString", PrimitiveKind.STRING)
override fun serialize(encoder: Encoder, value: Garage) = encoder.encodeString(Json.encodeToString(value))
override fun deserialize(decoder: Decoder): Garage = Json.decodeFromString(decoder.decodeString())
}
Auxiliary class:
#Serializable
data class VehicleDto(
val color: String,
val miles: Int,
val year: Int,
#Serializable(GarageToStringSerializer::class)
val garage: Garage
) {
constructor(v: Vehicle) : this(v.color, v.miles, v.year, v.garage)
}
The demanded result can be received with:
Json.encodeToString(VehicleDto(vehicle))
Related
I am currently implementing an API client with Ktor. The API I am requesting does not return a consistent JSON format.
for Example:
sometimes the JSON looks like this:
{
"description": {
"lang": "en",
"value": "an English description..."
},
...
}
and sometimes like this:
{
"description": [
{
"lang": "en",
"value": "an English description..."
},
{
"lang": "fr",
"value": "a French description..."
}
],
...
}
Now my Question:
How can I implement a Custom Kotlinx Deserializer to Decode an Object of T or a List<T> to a List<T>
My classes look like this:
#Serializable
class ResourceResponse(
#SerialName("description")
val descriptions: List<Description>
) {
#Serializable
data class Description(
#SerialName("value")
val value: String,
#SerialName("lang")
val language: String,
)
}
I want that a Json with only one Description-Object will be deserialized to a List with one Object and not specifically for the description, but in general for classes.
I've found nothing really helpful in the Web.
One solution is to first deserialize it to JsonElement, introspect and then decide how to deserialize it further into ResourceResponse:
fun decode(s: String): ResourceResponse {
val json = Json.parseToJsonElement(s).jsonObject
return when (val desc = json["description"]) {
is JsonArray -> Json.decodeFromJsonElement(json)
is JsonObject -> {
val json2 = json.toMutableMap()
json2["description"] = JsonArray(listOf(desc))
Json.decodeFromJsonElement(JsonObject(json2))
}
else -> throw IllegalArgumentException("Invalid value for \"description\": $desc")
}
}
This solution is definitely not ideal. It may be potentially less performant as we need to deserialize the whole tree into the tree of JsonElement objects only to transform it to the final types (although, maybe the library does this internally anyway). It works only for json and it is tricky to use this solution if ResourceResponse is somewhere deep into the data structure.
You can use a JsonContentPolymorphicSerializer to choose a deserializer based on the form of the JSON.
This one should work:
#Suppress("UNCHECKED_CAST")
class DescriptionsSerializer : JsonContentPolymorphicSerializer<List<ResourceResponse.Description>>(
List::class as KClass<List<ResourceResponse.Description>>
) {
// Here we check the form of the JSON we are decoding, and choose
// the serializer accordingly
override fun selectDeserializer(element: JsonElement): DeserializationStrategy<out List<ResourceResponse.Description>> {
return if (element is JsonArray)
ListSerializer(ResourceResponse.Description.serializer())
else
SingleDescriptionAsList()
}
class SingleDescriptionAsList : KSerializer<List<ResourceResponse.Description>> {
override val descriptor: SerialDescriptor
get() = ResourceResponse.Description.serializer().descriptor
override fun deserialize(decoder: Decoder): List<ResourceResponse.Description> {
return listOf(ResourceResponse.Description.serializer().deserialize(decoder))
}
override fun serialize(encoder: Encoder, value: List<ResourceResponse.Description>) {
throw Exception("Not in use")
}
}
}
You must also amend your original class to tell it to use this serializer:
#Serializable
class ResourceResponse(
#SerialName("description")
#Serializable(with = DescriptionsSerializer::class) val descriptions: List<Description>
) {
#Serializable
data class Description(
#SerialName("value")
val value: String,
#SerialName("lang")
val language: String,
)
}
Then you will be able to decode JSON objects with the single key "descriptions" using the ResourceResponse serializer.
For avoidance of doubt, if there are other keys in the JSON (it's not entirely clear from the question) then those should also be written into ResourceResponse definition.
After my research, I have now come up with a solution. For this you need a wrapper class. (here GenericResponse). I hope I can help others who have the same problem.
This is the Wrapper-Class
#Serializable(with = ListOrObjectSerializer::class)
class GenericResponse<T>(
val data: List<T> = emptyList()
) {
private var _isNothing : Boolean = false
val isNothing: Boolean
get() {
return this._isNothing
}
companion object {
fun <T> nothing(): GenericResponse<T> {
val o = GenericResponse(emptyList<T>())
o._isNothing = true
return o
}
}
}
And the Serializer looks like:
import kotlinx.serialization.KSerializer
import kotlinx.serialization.builtins.ListSerializer
import kotlinx.serialization.descriptors.SerialDescriptor
import kotlinx.serialization.encoding.Decoder
import kotlinx.serialization.encoding.Encoder
import kotlinx.serialization.json.*
class ListOrObjectSerializer<T : Any>(private val tSerializer: KSerializer<T>): KSerializer<GenericResponse<T>> {
override val descriptor: SerialDescriptor
get() = tSerializer.descriptor
override fun deserialize(decoder: Decoder): GenericResponse<T> {
val input = decoder as JsonDecoder
val jsonObj = input.decodeJsonElement()
return when(jsonObj) {
is JsonObject -> GenericResponse(listOf(Json.decodeFromJsonElement(tSerializer, jsonObj)))
is JsonArray -> GenericResponse(Json.decodeFromJsonElement(ListSerializer(tSerializer), jsonObj))
else -> return GenericResponse.nothing()
}
}
override fun serialize(encoder: Encoder, value: GenericResponse<T>) {
throw IllegalAccessError("serialize not supported")
}
}
My Data-Class look now like:
import kotlinx.serialization.SerialName
import kotlinx.serialization.Serializable
#Serializable
class ResourceResponse(
#SerialName("description")
val descriptions: GenericResponse<Description>? = null,
) {
#Serializable
data class Description(
#SerialName("value")
val value: String? = null,
#SerialName("lang")
val language: String? = null,
)
}
data class ResourceResponse(
#SerializedName("description") val descriptions: List<Description>,
)
data class Description(
#SerializedName("value") val value: String,
#SerializedName("lang") val language: String,
)
it should be like that
I have not very well structured JSON from BE (that I cannot change) which was formerly handled with moshi (custom adapter for the issue). Now I am trying to use pure kotlin serialization instead, but as said, JSON structure doesn't help me much.
{
"foo": {
"version":1,
"mask": [
{
"values": [
{ "bar": 1, ... }
]
},
{
"values": [
"important text i guess"
]
},
]
}
}
Now as you can see my issue is with values that can contains both object as well as string. All this should be parsed into kotlin where values looks like this:
data class Values(
val bar: Int? = null,
val text: String? = null,
...
)
Theoretically I can change local implementation, e.g. split the values class or something.
I've already tried to apply Polymorphic deserialization, but as I understand it was not able to recognise difference between two descendants of values without classDiscriminator set in JSON.
Any good advice?
Update
The polymorphic version I tried (in case I just made some error)
#Polymorphic
#Serializable
sealed class Values{
#Serializable
data class ObjectValues(
val bar: Int? = null,
...
)
#Serializable
data class TextValues(
val text: String? = null
)
}
and use it:
Json {
ignoreUnknownKeys = true
serializersModule = SerializersModule {
polymorphic(Values::class) {
subclass(Values.ObjectValues::class, Values.ObjectValues.serializer())
subclass(Values.TextValues::class, Values.TextValues.serializer())
}
}
error:
Polymorphic serializer was not found for missing class discriminator ('null')
JSON input: {"bar":42,...}
Disclaimer
I know that this would fix it all :)
"values": [ { "text":"important text i guess" } ]
Having classDiscriminator is not a mandatory requirement for polymorphic deserialization. You can use content-based polymorphic deserialization in this case.
All you need is to define a JsonContentPolymorphicSerializer for Values class with some logic for exact subclass serializer selection:
object ValuesSerializer : JsonContentPolymorphicSerializer<Values>(Values::class) {
override fun selectDeserializer(element: JsonElement) = when {
element.jsonObject["values"]!!.jsonArray.first() is JsonObject -> ObjectValues.serializer()
else -> TextValues.serializer()
}
}
and wire it as a serializer for Values class:
#Serializable(with = ValuesSerializer::class)
sealed class Values
#Serializable
data class TextValues(val values: List<String>) : Values()
#Serializable
data class ObjectValues(val values: List<Bar>) : Values()
#Serializable
data class Bar(val bar: Int)
#Serializable
data class MyJson(val foo: Foo)
#Serializable
data class Foo(val version: Int, val mask: List<Values>)
No need for serializersModule:
val result = Json.decodeFromString<MyJson>(json)
I'm looking for a solution to this but for json spray and my searches and attempts to get this working with json spray have failed thus far.
If I have the following json:
{
"personalData": {
"person": {
"first": "first_name",
"last": "last_name"
},
"contact": {
"phone": "1111111",
"address": {
"line": "123 Main St",
"city": "New York"
}
},
"xx": "yy", // unknown in advanced
"zz": { // unknown in advanced
"aa": "aa",
"bb": "bb",
"cc": {
"dd": "dd",
"ee": "ee"
}
}
}
}
We know for sure that the json will contain person and contact but we don't know what other fields may be created upstream from us that we don't care about/use.
I want to serialize this JSON into a case class containing person and contact but on the other hand, I don't want to lose the other fields (save them in a map so the class will be deserialized to the same json as received).
This is how far I've made it:
case class Address(
line: String,
city: String,
postalCode: Option[String]
)
case class Contact(
phone: String,
address: Address
)
case class Person(
first: String,
last: String
)
case class PersonalData(
person: Person,
contact: Contact,
extra: Map[String, JsValue]
)
implicit val personFormat = jsonFormat2(Person)
implicit val addressFormat = jsonFormat3(Address)
implicit val contactFormat = jsonFormat2(Contact)
implicit val personalDataFormat = new RootJsonFormat[PersonalData] {
def write(personalData: PersonalData): JsValue = {
JsObject(
"person" -> personalData.person.toJson,
"contact" -> personalData.contact.toJson,
// NOT SURE HOW TO REPRESENT extra input
)
}
def read(value: JsValue): CAERequestBEP = ???
}
Can someone help me do this with spray.json instead of play? I've spent such a long time trying to do this and can't seem to make it work.
In order to do that, you need to write your own formatter for PersonalDataFormat:
case class Person(first: String, last: String)
case class Address(line: String, city: String)
case class Contact(phone: String, address: Address)
case class PersonalData(person: Person, contact: Contact, extra: Map[String, JsValue])
case class Entity(personalData: PersonalData)
implicit val personFormat = jsonFormat2(Person)
implicit val addressFormat = jsonFormat2(Address)
implicit val contactFormat = jsonFormat2(Contact)
implicit object PersonalDataFormat extends RootJsonFormat[PersonalData] {
override def read(json: JsValue): PersonalData = {
val fields = json.asJsObject.fields
val person = fields.get("person").map(_.convertTo[Person]).getOrElse(???) // Do error handling instead of ???
val contact = fields.get("contact").map(_.convertTo[Contact]).getOrElse(???) // Do error handling instead of ???
PersonalData(person, contact, fields - "person" - "contact")
}
override def write(personalData: PersonalData): JsValue = {
JsObject(personalData.extra ++ ("person" -> personalData.person.toJson, "contact" -> personalData.contact.toJson))
}
}
implicit val entityFormat = jsonFormat1(Entity)
val jsonResult = jsonString.parseJson.convertTo[Entity]
The result is:
Entity(PersonalData(Person(first_name,last_name),Contact(1111111,Address(123 Main St,New York)),Map(xx -> "yy", zz -> {"aa":"aa","bb":"bb","cc":{}})))
(Assuming the json is not exactly the json above, but a valid similar one)
Code run in Scastie
I've created the classes of Eqs and Service, got the service objects but can't get the list of eqs. Can anyone help me with this?
This the Eqs class
data class Eqs(
val name: String,
val imageUrl: String,
val description: String?,
val responsible: String
)
That's the Service class which gets its values
data class Service(
val title: String,
val servings: Int,
val eqs: List<Eqs>
) {
companion object {
fun getServicesFromFile(filename: String, context: Context): ArrayList<Service> {
val serviceList = ArrayList<Service>()
try {
// Load data
val jsonString = loadJsonFromAsset("services.json", context)
val json = JSONObject(jsonString)
val services = json.getJSONArray("services")
(0 until services.length()).mapTo(serviceList) {
Service(services.getJSONObject(it).getString("title"),
services.getJSONObject(it).getInt("servings"),
}
} catch (e: JSONException) {
e.printStackTrace()
}
return serviceList
}
I can't get the List of Eqs in my getServicesFromFile function. How to parse and get it correctly?
I recommend you to use Jackson library. It's simple and saves you a lot of time. You can find it's documentation here: https://www.baeldung.com/jackson-kotlin
You also can use some websites to generate the data class needed for Jackson like https://app.quicktype.io/
Use Json to Kotlin plugin
In tool bar of android studio Code >> Generate and copy & paste you API into it and give the class name
[
{
"id": 1,
"name" : "Madoldoowa",
"description": "Madol Doova (මඩොල් දූව) is a children's novel and coming-of-age story written by Sri Lankan writer
Martin Wickramasinghe and first published in 1947",
"language" : "Sinhala",
"isbn" : "ISBN232673434",
"file_size" : 300,
"no_of_pages" : 500,
"price" : 970,
"ratings" : "5.1K",
"cover_page" : "https://upload.wikimedia.org/wikipedia/en/5/5c/MadolDoova.jpg",
"author" : {
"name" : "Martin Wickramasinghe"
}
]
data class Model(
val author: Author,
val cover_page: String,
val description: String,
val file_size: Int,
val id: Int,
val isbn: String,
val language: String,
val name: String,
val no_of_pages: Int,
val price: Int,
val ratings: String
)
data class Author(
val name: String
)
Suppose I've multiple DTOs, like:
data class ActionDetailDTO(
#JsonProperty("priority")
val priority: String,
#JsonProperty("reason")
val reason: String
)
data class IntroDTO(
#JsonProperty("name")
val name: String,
#JsonProperty("number")
val number: String
)
and I've a json of these dtos stored as strings,
when I parse them doing something like this:
fun parseStringBasedOnType(action: SomeDTOType) :Any{
val obj = when (action.actionType){
"CREATED" -> objectMapper.readValue(action.actionDetails, ActionDetailDTO::class.java)
"INTRO" -> objectMapper.readValue(action.actionDetails, IntroDTO::class.java)
else -> "hh"
}
return obj
}
so:
val nn = parseStringBasedOnType(SomeActionObject) //type: CREATED
if(nn.actionType == "CREATED"){
println(nn.reason)
}
This obviously does not work, how can this be handled?
You could either define a common interface with an actionType method or cast the values:
if(nn is ActionDetailDTO) {
println(nn.reason)
}
Or use when if you plan to do something for the other type too:
when(nn) {
is ActionDetailDTO -> println(nn.reason)
is IntroDTO -> println(nn.number)
}