杰克逊：将对象视为原始对象 [英] Jackson: treat object as primitive

问题描述

我有一个类，它或多或少是一个double的包装类。当我通过jackson序列化我的课程时，我将收到类似：{value：123.0}的内容。我基本上想要发生的是，杰克逊给了我123.0。如果我可以扩展Number，我的问题就会解决，但由于我已经扩展了另一个类，所以这不是一个选项。

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;
    }
}

结果：

{
  "d" : 123.0
}

预期会有什么效果：

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();
    }
}

..导致：123.0

.. which results in: 123.0

但是 - 你知道 - 我已经扩展了另一个抽象类。我怎样才能得到明确的结果？

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

推荐答案

既然我确定somene可以重用这个，我会回答我自己的问题（与感谢Gavin向我展示的方式）：

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;
    }
}

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