How to map complex object to simple fields - json

Question

I have a JSON document similar to the following:
{
"aaa": [
{
"value": "ewfwefew"
}
],
"bbb": [
{
"value": "ewfewfe"
}
]
}
I need to deserialize this into something more clean such as:
public class MyEntity{
private String aaa;
private String bbb;
}
What's the best way to unwrap each array and extract the "value" field on deserialization?

Answer 1

#Tim Mac's response is correct, but you can make it more elegant by writing a custom deserializer for your MyEntity class.
It should be something like this:
private class MyEntityDeserializer implements JsonDeserializer<MyEntity> {
public MyEntity deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context)
throws JsonParseException {
JsonObject rootObj = json.getAsJsonObject();
String nid = rootObj
.get("nid")
.getAsJsonArray()
.get(0)
.getAsJsonObject()
.get("value")
.getAsString();
String uuid = rootObj
.get("uuid")
.getAsJsonArray()
.get(0)
.getAsJsonObject()
.get("value")
.getAsString();
MyEntity entity = new MyEntity(nid, uuid);
return entity;
}
}
Then you have to register the TypeAdapter with:
Gson gson = new GsonBuilder().registerTypeAdapter(MyEntity.class, new MyEntityDeserializer()).create();
And finally you just have to parse your JSON as usual, with:
MyEntity entity = gson.fromJson(yourJsonString, MyEntity.class);
Gson will automatically use your custom deserializer to parse your JSON into your MyEntity class.

Answer 2

If you can't change the json that you're getting, you might consider deserializing it the way it is, and then converting it to something more manageable?
public class TmpEntity {
public Value[] nid {get;set;}
public Value[] uuid {get;set;}
}
public class Value {
public string value {get;set;}
}
public class MyEntity {
public string nid {get;set;}
public string uuid {get;set;}
}
var tmp = ...; //deserialize using javascriptserializer
var converted = tmp.Select(a => new MyEntity()
{
nid = a.nid.First().value,
uuid = a.uuid.First().value
}