I use this class to store data
public class Item(var name:String,
var description:String?=null){
}
And use it in ArrayList
public var itemList = ArrayList<Item>()
Use this code to serialize the object
val gs=Gson()
val itemListJsonString = gs.toJson(itemList)
And deserialize
itemList = gs.fromJson<ArrayList<Item>>(itemListJsonString, ArrayList::class.java)
But this method will give me LinkedTreeMap
, not Item
, I cannot cast LinkedTreeMap to Item
What is correct way to deserialize to json in Kotlin?
Deserialization in the context of Gson means converting a JSON string to an equivalent Java object. In order to do the deserialization, we need a Gson object and call the function fromJson() and pass two parameters i.e. JSON string and expected java type after parsing is finished. Program output.
Both Gson and Jackson are good options for serializing/deserializing JSON data, simple to use and well documented. Advantages of Gson: Simplicity of toJson/fromJson in the simple cases. For deserialization, do not need access to the Java entities.
Try this code for deserialize list
val gson = Gson()
val itemType = object : TypeToken<List<Item>>() {}.type
itemList = gson.fromJson<List<Item>>(itemListJsonString, itemType)
You can define a inline reified extension function like:
internal inline fun <reified T> Gson.fromJson(json: String) =
fromJson<T>(json, object : TypeToken<T>() {}.type)
And use it like:
val itemList: List<Item> = gson.fromJson(itemListJsonString)
By default, types are erased at runtime, so Gson cannot know which kind of List
it has to deserialize. However, when you declare the type as reified
you preserve it at runtime. So now Gson has enough information to deserialize the List
(or any other generic Object).
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