Iam使用hibernate实体类并与另一个模型具有ManyToOne关系,如下所示
@ManyToOne
@JoinColumn(name ="`brand-id`", referencedColumnName="`id`", nullable=false, insertable = false, updatable = false)
private HotelBrand brand;
我的模态如下
@Entity
@Table(name = "`hotel`" })
public class Hotel implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@Column(name="`id`", unique = true, nullable = false)
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
@NotEmpty
@Column(name="`brand-id`", nullable=false)
private Integer brandid;
@NotEmpty
@Column(name="`hotel-code`", nullable=false)
private String hotelid;
@NotEmpty
@Column(name="`hotel-name`", nullable=false)
private String hotelname;
@NotEmpty
@Column(name="`have-reports`", nullable=false)
private String havereports;
@ManyToOne
@JoinColumn(name ="`brand-id`", referencedColumnName="`id`", nullable=false, insertable = false, updatable = false)
private HotelBrand brand;
public Hotel() {}
public Hotel(Integer id, Integer brandid, String hotelid, String hotelname, HotelBrand brand ) {
super();
this.id = id;
this.brandid = brandid;
this.hotelid = hotelid;
this.hotelname = hotelname;
this.brand = brand;
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public Integer getBrandid() {
return brandid;
}
public void setBrandid(Integer brandid) {
this.brandid = brandid;
}
public String getHotelid() {
return hotelid;
}
public void setHotelid(String hotelid) {
this.hotelid = hotelid;
}
public HotelBrand getBrand() {
return brand;
}
public void setBrand(HotelBrand brand) {
this.brand = brand;
}
public String getHotelname() {
return hotelname;
}
public void setHotelname(String hotelname) {
this.hotelname = hotelname;
}
}
因此,在获取数据时,我将连接列作为单独的json对象,但我只需要来自该连接模型的单个列。
我得到的结果如下
{
"id": 115,
"brandid": 7,
"hotelid": "ABC",
"hotelname": "sample1",
"brand": {
"id": 7,
"brandname": "brand1"
}
}
但我希望如此
{
"id": 115,
"brandid": 7,
"hotelid": "ABC",
"hotelname": "sample1",
"brandname": "brand1"
}
任何帮助都会非常感激。
答案 0 :(得分:0)
您必须使用注释@JsonIgnore
注释字段品牌。
比你不会在json回复中收到它