How can I get Gson to deserialize an interface type?

Question

I have an interface

public interace ABC {
}

Implementation of this is as follows:

public class XYZ implements ABC {
    private Map<String, String> mapValue;
    public void setMapValue( Map<String, String> mapValue) {
        this.mapValue = mapValue;
    }  

    public  Map<String, String> getMapValue() {
        return this.mapValue
    }
}

I want to deserialize a class using Gson which is implemented as

public class UVW {
    ABC abcObject;
}

when I try to deserialize it like gson.fromJson(jsonString, UVW.class); it returns me null. jsonString is UTF_8 String.

Is it because of interface used in UVW class? If yes, how do I deserialize such class?

Given only a field of type `ABC`, how can Gson know how to deserialize the JSON? — Sotirios Delimanolis, Aug 11 '15 at 20:59
I don't believe Gson has such built-in functionality (polymorphic deserialization). You'd need to write a custom deserializer that takes some kind of hint from the JSON for how to deserialize it. Look at Jackson. It has this feature built into it. — Sotirios Delimanolis, Aug 11 '15 at 21:03
@SotiriosDelimanolis That question is woefully out of date. He suggests using `RuntimeTypeAdapterFactory`, which is in the code base now (he says it isn't) and then suggests to use `JsonDeserializer`, which is basically deprecated now. — durron597, Aug 11 '15 at 21:39
You can provide an updated answer if you wish. No need to spread it across questions. — Sotirios Delimanolis, Aug 11 '15 at 21:40
@durron597 I don't mind reopening (you can too), I just don't want answers all over the place. There are numerous posts explaining how to solve this problem. — Sotirios Delimanolis, Aug 11 '15 at 21:57
@SotiriosDelimanolis I have a gold badge too, if I thought you were definitely wrong I would have just reopened; you reminded me of the right thing to do and I decided to leave it closed. Maybe I'll write a more canonical answer (on the other question) tomorrow. — durron597, Aug 11 '15 at 21:59

score 5 · Answer 1 · edited May 23 '17 at 11:57

You need to tell Gson to use XYZ when it deserializes ABC. You can do this using a TypeAdapterFactory.

Briefly, thus:

public class ABCAdapterFactory implements TypeAdapterFactory {
  private final Class<? extends ABC> implementationClass;

  public ABCAdapterFactory(Class<? extends ABC> implementationClass) {
     this.implementationClass = implementationClass;
  }

  @SuppressWarnings("unchecked")
  @Override
  public <T> TypeAdapter<T> create(Gson gson, TypeToken<T> type) {
    if (!ABC.class.equals(type.getRawType())) return null;

    return (TypeAdapter<T>) gson.getAdapter(implementationClass);
  }
}

Here is a complete working test harness that illustrates this example:

public class TypeAdapterFactoryExample {
  public static interface ABC {

  }

  public static class XYZ implements ABC {
    public String test = "hello";
  }

  public static class Foo {
    ABC something;
  }

  public static void main(String... args) {
    GsonBuilder builder = new GsonBuilder();
    builder.registerTypeAdapterFactory(new ABCAdapterFactory(XYZ.class));
    Gson g = builder.create();

    Foo foo = new Foo();
    foo.something = new XYZ();

    String json = g.toJson(foo);
    System.out.println(json);
    Foo f = g.fromJson(json, Foo.class);
    System.out.println(f.something.getClass());
  }
}

Output:

{"something":{"test":"hello"}}
class gson.TypeAdapterFactoryExample$XYZ

It's simple if you have a single subtype. In this case, you don't need to add any hints for the target type. Add another subtype and it becomes more complicated. — Sotirios Delimanolis, Aug 11 '15 at 21:34

How can I get Gson to deserialize an interface type?

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