问题
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?
回答1:
Try this code for deserialize list
val gson = Gson()
val itemType = object : TypeToken<List<Item>>() {}.type
itemList = gson.fromJson<List<Item>>(itemListJsonString, itemType)
回答2:
In my code I just use:
import com.google.gson.Gson
Gson().fromJson(string_var, Array<Item>::class.java).toList() as ArrayList<Type>
I give here a complet example.
First the type and the list array:
class Item(var name:String,
var description:String?=null)
var itemList = ArrayList<Item>()
The main code:
itemList.add( Item("Ball","round stuff"))
itemList.add(Item("Box","parallelepiped stuff"))
val striJSON = Gson().toJson(itemList) // To JSON
val backList = Gson().fromJson( // Back to another variable
striJSON, Array<Item>::class.java).toList() as ArrayList<Item>
val striJSONBack = Gson().toJson(backList) // To JSON again
if (striJSON==striJSONBack) println("***ok***")
The exit:
***OK***
来源:https://stackoverflow.com/questions/51376954/how-to-use-gson-deserialize-to-arraylist-in-kotlin