Jackson: treat object as primitive

Question

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?

score 5 · Accepted Answer · answered Jan 06 '13 at 10:19

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;
    }
}

score 2 · Answer 2 · answered Jan 06 '13 at 09:54

2

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

answered Jan 06 '13 at 09:54

Gavin Xiong

957
6
9

Thanks, this led me into the right direction. Actually it leds me to another question: http://stackoverflow.com/questions/7161638/how-do-i-use-a-custom-serializer-with-jackson So the solution is much more simple using the @JsonValue annontation is enough for me - thanks! – KIC Jan 06 '13 at 10:16

Hui Zheng · Answer 3 · 2013-01-06T10:24:13.773

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.

Jackson: treat object as primitive

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