杰克逊遗产

Question

似乎无法解决这个问题。我一直收到各种错误，所以我只用我从 Jackson 那里得到的当前错误来写这篇文章。

public class ResponseDetail {
    private Response response;
}

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME)
@JsonSubTypes({
    @JsonSubTypes.Type(value = ResponseTypeOne.class, name = "ResponseTypeOne"),
    @JsonSubTypes.Type(value = ResponseTypeTwo.class, name = "ResponseTypeTwo"),
    @JsonSubTypes.Type(value = ResponseTypeThree.class, name = "ResponseTypeThree")
})
@JsonIgnoreProperties(ignoreUnknown = true)
public abstract class Response {
}

在其他包中我有这三个：

public class ResponseTypeOne extends Response {
     private Integer status;
}
public class ResponseTypeTwo extends Response {
    private String message;
}
public class ResponseTypeThree extends Response {
     private String value;
}

错误：

Caused by: com.fasterxml.jackson.databind.exc.InvalidTypeIdException: Missing type id when trying to resolve subtype of [simple type, class com.services.models.Response]: missing type id property '@type' (for POJO property 'response')

我尝试了这个 @JsonTypeInfo 的各种迭代与各种 includes 和各种 property 也与 Id.CLASS 没有运气。

Answer 1

您需要声明如何识别类型。

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.EXISTING_PROPERTY, property = "@ttype")
@JsonSubTypes({
    @JsonSubTypes.Type(value = ResponseTypeOne.class, name = "ResponseTypeOne"),
    @JsonSubTypes.Type(value = ResponseTypeTwo.class, name = "ResponseTypeTwo"),
    @JsonSubTypes.Type(value = ResponseTypeThree.class, name = "ResponseTypeThree")
})

@JsonIgnoreProperties(ignoreUnknown = true)
public abstract class Response {
   @JsonProperty("@ttype")
   public abstract String getChildType();
}

并且在子类型中执行以下操作：

   @JsonTypeName("ResponseTypeOne")
    public class ResponseTypeOne extends Response {
        @Override
        public String getChildType() {
            return "ResponseTypeOne";
        }
    }

传入的 json 应该如下所示，以使 jackson 能够找到正确的子实现：

{
  //some attributes of child Response
  "@ttype": "ResponseTypeOne"
}

杰克逊遗产

Jackson Inheritance

jackson

spring-boot

jackson-databind