CODEWITHSUNDEEP
javajackson
KIC
KIC

Reputation: 6121

Jackson: treat object as primitive

I have a class which is more or less a wrapping class around a double. When I serialize my class via jackson I will receive something like: { "value" : 123.0 }. What I basically would like to happen is, that jackson gives me just 123.0. My problem would be solved if I could extend Number, but since I am already extending another class this is not an option.

Class:

@JsonIgnoreProperties(ignoreUnknown = true)
@SuppressWarnings("unused")
public class TestValue {
    @JsonProperty
    private final Double d;

    public TestValue(Double d) {
        this.d = d;
    }
}

Results in: 

{
  "d" : 123.0
}

What would work like expected:

public class TestValue extends Number {
    private final Double d;

    public TestValue(Double d) {
        this.d = d;
    }

    public double doubleValue() {
        return d;
    }

    public float floatValue() {
        return d.floatValue();
    }

    public int intValue() {
        return d.intValue();
    }

    public long longValue() {
        return d.longValue();
    }

    public String toString() {
        return d.toString();
    }
}

.. which results in: 123.0

But - you know - I am already extending an other abstract class. How can I get the expteced result?

Upvotes: 4

Views: 2043

Answers (3)

Hui Zheng
Hui Zheng

Reputation: 10224

First, create a custom serializer for type TestValue.

public class TestValueSerializer extends JsonSerializer<TestValue> {
@Override
public void serialize(TestValue value, JsonGenerator gen, SerializerProvider provider) throws IOException,
        JsonProcessingException {
    gen.writeString(value.toString());
}

}

Then add annotation @JsonSerialize(using=TestValueSerializer.class) before TestValue type.

BTW, even if TestValue had not extended other class, it's better to avoid extending it from Number. Always use implementation inheritance in caution.

Upvotes: 1

KIC
KIC

Reputation: 6121

Since I am sure somene can reuse this, I will answer my own question (with thanks to Gavin showing me the way):

public class TestValue {
    private final Double d;

    public TestValue(Double d) {
        this.d = d;
    }

    @JsonValue
    public Double getValue() {
        return d;
    }
}

Upvotes: 5

Gavin Xiong
Gavin Xiong

Reputation: 977

It seems that you want to customize the serialization, see here.

Upvotes: 2

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