java - 如果 Json 属性不存在于 java 的构造函数中，则忽略 Json 属性

Question

我有一个类，如下所示:-

@Getter
@Setter
@NoArgsConstructor
public class Response extends ResponseMessage {

    @JsonProperty("ResponseDto")
    private myDTo myDto;

    @JsonProperty("revId")
    private Long revisionId;

    @JsonProperty("modelId")
    private Long model;

    public Response(HttpStatus status, String message) {
        super(status, message);
    }

    public Response(HttpStatus status, String message, Long revisionId) {
        super(status, message);
        this.revisionId = revisionId;
    }

    public Response(HttpStatus status, String message, Long revisionId, Long modelId) {
        super(status, message);
        this.revisionId = revisionId;
        this.model = modelId;
    }

    public Response(HttpStatus status, String message, myDTo myDto) {
        super(status, message);
        this.myDto = myDto;
    }
}

现在，当我返回 Response 类作为响应时，如下所示:-

return new Response(HttpStatus.OK,"done",123);

它返回响应 json 为:-

{
"revId":123,
"status":"OK",
"modelId":null,
"message":"done"
}

但希望响应取决于正在调用的构造函数。在这种情况下应该是:-

{
"revId":123,
"status":"OK",
"message":"done"
}

任何帮助将不胜感激!

Answer 1

您应该使用@JsonInclude(Include.NON_NULL)来忽略空字段。