How to access default jackson serialization in a custom serializer

Question

I want to create a custom serializer which does a tiny bit of work and then leaves the rest for default serialization.

For example:

@JsonSerialize(u         


        

Answer 1

@JsonSerialize(using = MyClassSerializer.class)
public class MyClass {
...
}

public class MyClassSerializer extends JsonSerializer<MyClass> {
    @Override
     public void serialize(MyClass myClass, JsonGenerator generator, 
                      SerializerProvider provider) 
        throws JsonGenerationException, IOException {
        if (myClass.getSomeProperty() == someCalculationResult) {
            provider.setAttribute("special", true);
        } 
        provider.defaultSerializeValue(myClass, generator);

    }  
}

if you are just writing an object as normal use the above

Answer 2

You can use @JsonGetter instead of using a custom serializer if that's the only change you want to make.

public class MyClass{

    @JsonGetter("special")
    protected boolean getSpecialForJackson() {
        return myClass.getSomeProperty() == someCalculationResult;
    }

}

Answer 3

A BeanSerializerModifier will provide you access to the default serialization.

Inject a default serializer into the custom serializer

public class MyClassSerializer extends JsonSerializer<MyClass> {
    private final JsonSerializer<Object> defaultSerializer;

    public MyClassSerializer(JsonSerializer<Object> defaultSerializer) {
        this.defaultSerializer = checkNotNull(defaultSerializer);
    }

    @Override
    public void serialize(MyClass myclass, JsonGenerator gen, SerializerProvider provider) throws IOException {
        if (myclass.getSomeProperty() == true) {
            provider.setAttribute("special", true);
        }
        defaultSerializer.serialize(myclass, gen, provider);
    }
}

Create a BeanSerializerModifier for MyClass

public class MyClassSerializerModifier extends BeanSerializerModifier {
    @Override
    public JsonSerializer<?> modifySerializer(SerializationConfig config, BeanDescription beanDesc, JsonSerializer<?> serializer) {
        if (beanDesc.getBeanClass() == MySpecificClass.class) {
            return new MyClassSerializer((JsonSerializer<Object>) serializer);
        }
        return serializer;
    }
}

Register the serializer modifier

ObjectMapper om = new ObjectMapper()
        .registerModule(new SimpleModule()
                .setSerializerModifier(new MyClassSerializerModifier()));

Answer 4

To add to the chosen answer, the serializer implementation may also have to implement ContextualSerializer and ResolvableSerializer interfaces. Please take a look at a related issue here https://github.com/FasterXML/jackson-dataformat-xml/issues/259

public class MyClassSerializer extends JsonSerializer<MyClass>
    implements ContextualSerializer, ResolvableSerializer {
private final JsonSerializer<Object> defaultSerializer;

public MyClassSerializer(JsonSerializer<Object> defaultSerializer) {
    this.defaultSerializer = checkNotNull(defaultSerializer);
}

@Override
public void serialize(MyClass myclass, JsonGenerator gen, SerializerProvider provider)
        throws IOException {
    if (myclass.getSomeProperty() == true) {
        provider.setAttribute("special", true);
    }
    defaultSerializer.serialize(myclass, gen, provider);
}

@Override
public JsonSerializer<?> createContextual(SerializerProvider prov, BeanProperty property)
        throws JsonMappingException {
    if (defaultSerializer instanceof ContextualSerializer) {
        JsonSerializer<?> contextual = ((ContextualSerializer)defaultSerializer).createContextual(prov, property);
        return new MyClassSerializer((JsonSerializer<Object>)contextual);
    }
    return new MyClassSerializer(defaultSerializer);
}

@Override
public void resolve(SerializerProvider provider) throws JsonMappingException {
    if (defaultSerializer instanceof ResolvableSerializer) {
        ((ResolvableSerializer)defaultSerializer).resolve(provider);
    }
}

}