Custom JSON deserialization only if certain field is present (using Jackson)

问题

When deserializing MyEntity (which is an interface) I either have

the following input:
```
{ "id": 123 }
```
in which case I would like to deserialize it into a
```
new MyEntityRef(123)
```
or I have the following input:
```
{
    "id": 123,
    "message": "Hello world",
    "otherEntity": {
        "field": "value",
        ...
    }
}
```
in which case I would like to deserialize it as
```
new MyEntityImpl(123, "Hello world", otherEntity);
```
where otherEntity is deserialized the same way as if it was found outside the context of MyEntity.

I've figured out how to register my own custom deserializer through a SimpleModule but I don't know how to

Choose a custom deserializer based on the presense of some field (such as message above).
Fallback on the "default" serializer for certain fields (such as otherEntity above).

回答1:

Finally solved it by configuring my ObjectMapper as follows:

ObjectMapper mapper = new ObjectMapper();

SimpleModule idAsRefModule = new SimpleModule("ID-to-ref",
                                              new Version(1, 0, 0, null));

idAsRefModule.addDeserializer(TestEntity.class,
                              new JsonDeserializer<TestEntity>() {
    @Override
    public TestEntity deserialize(JsonParser jp, DeserializationContext dc)
            throws IOException, JsonProcessingException {

        ObjectCodec codec = jp.getCodec();
        JsonNode node = codec.readTree(jp);
        boolean isFullImpl = node.has("message");
        Class<? extends TestEntity> cls = isFullImpl ? TestEntityImpl.class
                                                     : TestEntityRef.class;
        return codec.treeToValue(node, cls);
    }
});

mapper.registerModule(idAsRefModule);

return mapper;

来源：https://stackoverflow.com/questions/19463141/custom-json-deserialization-only-if-certain-field-is-present-using-jackson

标签

json

jackson

deserialization