Jackson将无嵌套JSON转嵌套对象

例如，我有以下JSON

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{age: 18, lastname: "Liu", firstname: "Ryan"}

我需要将这个JSON装成Person对象

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class Person {
    private int age;
    private Name name;
}
class Name {
    private String lastname;
    private String firstname;
}

只需要使用Jackson的注解@JsonUnwrapped

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class Person {
    private int age;
    @JsonUnwrapped
    private Name name;
}

还有另外一种情况。

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class Person {
    private Id id;
    private int age;
}
class Id {
    private int val;
    
    public int val() {
        return this.val;
    }
}

转成

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{id: 1, age: 18}

这是可使用@JsonValue注解。

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class Id {
    private int val;
    
    @JsonValue
    public int val() {
        return this.val;
    }
}