Can jackson determine root object type to deserialize to when json includes type property?

Question

Assume serialization to json includes the class name of the actual object, using this annotation on the Class:

@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "@type")
class MyClass {
    String foo;
}

So json is for example:

{"@type": "com.example.MyClass", "foo": "bar"}

Can this be deserialized without specifying the type? And I mean not even the super type. Just something like:

objectMapper.readValue(value, Object.class);

which doesn't actually work, it brings back a Map.

See also https://stackoverflow.com/q/30362446/873282 – koppor Jan 05 '23 at 21:58 — koppor, Jan 05 '23 at 21:58

score 10 · Accepted Answer · answered Feb 07 '15 at 16:18

Well, it is certainly possible to do that although I have personally never used Jackson that way. You can deserialize it to a JsonNode object and then convert it to the proper type.

final ObjectMapper objectMapper = new ObjectMapper();
final MyClass myClass = new MyClass();
myClass.foo = "bar";

// Serialize
final String json = objectMapper.writeValueAsString(myClass);

// Deserialize
final JsonNode jsonNode = objectMapper.readTree(json);

// Get the @type
final String type = jsonNode.get("@type").asText();

// Create a Class-object
final Class<?> cls = Class.forName(type);

// And convert it
final Object o = objectMapper.convertValue(jsonNode, cls);

System.out.println(o.getClass());

The output is:

MyClass

Oh, yeah, I guess that works. I was wondering if there was a more built-in way but I guess that wouldn't make any sense as Jackson could not possibly know how I have configured the JsonTypeInfo, without knowing the class. Thanks a lot. — Nazaret K., Feb 07 '15 at 16:42

score 3 · Answer 2 · answered Feb 10 '15 at 00:24

3

    ObjectMapper mapper = new ObjectMapper();
    mapper.enableDefaultTyping(ObjectMapper.DefaultTyping.NON_FINAL);

    MyClass original = new MyClass();
    original.foo = "hello";

    String json = mapper.writeValueAsString(original);
    MyClass foo = (MyClass) mapper.readValue(json, MyClass.class);

This should work and is very convenient.

answered Feb 10 '15 at 00:24

sydraz

653
6
12

It works if you know the JSON represents `MyClass` at deserialization time. I took the question to refer to a scenario where you don't know the type of the object. That's the situation I'm currently trying to handle. – Rob Fletcher Jun 17 '15 at 16:47
If you don't know the type, you can just assign to an object. The object.getClass() will tell you what type it is. – sydraz Jun 19 '15 at 22:01

score 2 · Answer 3 · answered Feb 09 '15 at 22:45

Yes, but there is a caveat: type you give MUST be something that includes @JsonTypeInfo you specify. Object.class will not have, unless you use "mix-in annotations" to associate it.

However, if you need to add type information for properties of (declared type of) java.lang.Object, you probably want to enable default typing: see ObjectMapper.enableDefaultTyping(...) for more information. That will actually enable inclusion (and use) of type information for larger categories of classes, without need to add annotations.

score 1 · Answer 4 · answered Nov 08 '17 at 18:44

1

What you want to use is Jackson Polymorphic Deserialization.

answered Nov 08 '17 at 18:44

Farrukh Najmi

5,055
3
35
54

Can jackson determine root object type to deserialize to when json includes type property?

4 Answers4

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