JSON挑剔的Java

Question

我有一个简单的问题，假设我有这样的json

{
   "outer-field":{
      "outer-a":"something",
      "outer-b":12345678,
      "inner-field":{
         "inner-a":false,
         "inner-b":0.0,
         "inner-c":29.99
      }
   }
}

映射：

public class OuterObject {

    @JsonProperty("outer-a")
    public String outerA;
   
    @JsonProperty("outer-b")
    public Integer outerB;
   
    @JsonProperty("inner-field")
    public InnerObject innerField;

}

public class InnerObject{

    @JsonProperty("inner-a")
    public Boolean innerA;

    @JsonProperty("inner-b")
    public Double innerB;

    @JsonProperty("inner-c")
    public Double innerC;

我想知道是否可以通过使用一些自定义setter/entotation/shots或其他内容来保存内部对象中的外部字段：

public class InnerObject{

    @JsonProperty("inner-a")
    public Boolean innerA;

    @JsonProperty("inner-b")
    public Double innerB;

    @JsonProperty("inner-c")
    public Double innerC;

    //how to map this?
    @JsonProperty("outer-a")
    public String outerA;

ps：使用自定义由于JSON的复杂性，次要化是我的最后一项手段

Answer 1

如果有人问我如何解决问题，我决定通过使用内部的xxx方法使用自定义构建器将外部属性传递给创建期间的内部对象

@Jacksonized
@JsonDeserialize(builder = OuterObject.Builder.class)
@AllArgsConstructor
@Data
public class OuterObject {

    public String outerA;
    public Integer outerB;
    public InnerObject innerObject;

    public static class Builder {

        public String outerA;
        public Integer outerB;
        public InnerObject innerObject;

        Builder withInnerObject(InnerObject innerObject) {
            // set attributes here
            innerObject.outerA = this.outerA
            this.innerObject = innerObject
            return this;
        }

        public OuterObject build() {
            return new(outerA, outerB, innerObject)
        }

Hope someone might find this helpful