java - fastjson解析jpa，怎么返回了状态？

Question

new: false ?????

我没有这个字段，这个是jpa的创建状态吧，false不是新建对象

怎么删掉？

@Entity
@Table(name = "tb_user")
@DynamicUpdate
@Getter
@Setter
@AllArgsConstructor
@NoArgsConstructor
@Access(AccessType.FIELD) //强制hibernate全部访问字段注解
@Accessors(chain = true)
public class User extends BaseUUIDEntity implements Serializable {

    private static final long serialVersionUID = 1L;

    public enum Sex {
        @MetaData("男")
        MAN, //男
        @MetaData("女")
        WOMAN //女
    }

    @Column(name = "user_name", length = 20, nullable = false)
    @MetaData("用户名")
    private String userName;

    @Column(name = "pass_word", length = 32)
    @MetaData("密码")
    private String passWord;

    @Column(length = 12)
    @MetaData("手机号")
    private String phone;

    @Column(name = "head_photo")
    @MetaData("头像")
    private String headPhoto;

    @Enumerated(EnumType.STRING)
    @Column(length = 5, columnDefinition = "varchar(5) default 'MAN'")
    @MetaData("性别")
    private Sex sex;

    @Temporal(TemporalType.DATE)
    @MetaData("生日")
    private Date birthday;

    @JSONField(serialize = false)
    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.MERGE)
    @JoinColumn(name = "merchant_id")
    private Merchant merchant;
}

    @ResponseBody
    
    return ResponseEntity.ok(new Msg(HttpStatus.OK.value(), Constant.MESSAGE_SUCCESS, user.setPassWord("you don't want to know")));