GSON. workaround for enum and int - json

Question

I get a json-answer from server. And i'm parsing it with GSON-library.
A key within json has an integer value. Is it somehow possible without to change the server answer (it is the external server interface, we have no influence on it) to cast the integer value to an enumeration?
Thank you.
UPD:
The json-Response. NOTE: we can't change it
"testObject":{
"id":123,
"type":42
}
The enumeration:
public enum ObjectTypeEnum
{
UNKNOWN_TYPE(0),
SIMPLE_TYPE(11),
COMPLEX_TYPE(42);
private int value;
private ObjectTypeEnum(int value)
{
this.value = value;
}
public static ObjectTypeEnum findByAbbr(int value)
{
for (ObjectTypeEnum currEnum : ObjectTypeEnum.values())
{
if (currEnum.value == value)
{
return currEnum;
}
}
return null;
}
public int getValue()
{
return value;
}
}
And the object class
public class TestObject
{
publuc int id;
public ObjectTypeEnum type;
}

Answer 1

You can just use the #SerializedName annotation to determine what value gets serialized to/from the wire. Then you don't need to write a custom TypeAdapter.
import com.google.gson.annotations.SerializedName;
public enum ObjectTypeEnum {
#SerializedName("0")
UNKNOWN_TYPE(0),
#SerializedName("11")
SIMPLE_TYPE(11),
#SerialziedName("42")
COMPLEX_TYPE(42);
private int value;
private ObjectTypeEnum(int value) {
this.value = value;
}
public int getValue() {
return value;
}
}
If you don't have a need for getting the wire value in your code you can eliminate the "value" field and related code.
public enum ObjectTypeEnum {
#SerializedName("0")
UNKNOWN_TYPE,
#SerializedName("11")
SIMPLE_TYPE,
#SerialziedName("42")
COMPLEX_TYPE;
}

Answer 2

Using an answer from Chin and help from my workmate I get following solution.
I wrote an inner class in the parser class.
private static class ObjectTypeDeserializer implements
JsonDeserializer<ObjectTypeEnum>
{
#Override
public PreconditioningStatusEnum deserialize(JsonElement json,
Type typeOfT, JsonDeserializationContext ctx)
throws JsonParseException
{
int typeInt = json.getAsInt();
return ObjectTypeEnum
.findByAbbr(typeInt);
}
}
and created GSON-Object on following way:
GsonBuilder gsonBuilder = new GsonBuilder();
gsonBuilder.registerTypeAdapter(ObjectTypeEnum.class, new ObjectTypeDeserializer() );
Gson gson = gsonBuilder.create();
http://sites.google.com/site/gson/gson-user-guide#TOC-Custom-Serialization-and-Deserializ

Answer 3

From kjones' answer, here's a Kotlin translation:
enum class ObjectTypeEnum(val value:Int) {
#SerializedName("0")
UNKNOWN_TYPE(0),
#SerializedName("11")
SIMPLE_TYPE(11),
#SerializedName("42")
COMPLEX_TYPE(42)
}
Or, without needing the Int values:
enum class ObjectTypeEnum {
#SerializedName("0")
UNKNOWN_TYPE,
#SerializedName("11")
SIMPLE_TYPE,
#SerializedName("42")
COMPLEX_TYPE
}

Answer 4

public enum Color {
GREEN(1), BLUE(2), RED(3);
private int key;
private Color(int key) {
this.key = key;
}
public static Color findByAbbr(int key) {
for (Color c : values()) {
if (c.key == key) {
return c;
}
}
return null;
}
}

Answer 5

I'm new to SO so I don't know how to add to Mur Votema's answer above, but just a small correction;
GsonBuilder gsonBuilder = new GsonBuilder();
gsonBuilder.registerTypeAdapter(ObjectTypeEnum.class, new ObjectTypeDeserializer() );
Gson gson = gsonBuilder.create();
Note; you need brackets to provide an instance of the class for the gsonBuilder.
Apart from that, great answer! Did exactly what I was looking for.