【发布时间】:2019-01-30 12:01:34
【问题描述】:
我需要一个住宿实体,因为当我单击此处时,我有位置列,如果我选择一个位置,它会显示许多位置,然后在这个实体中只有我有存储多个日期的列“save-dates”和“timeslotAvailable”列我需要存储多个时隙,例如“hh-mm-ss to hh-mm-ss”....我已经尝试了一些代码,请检查一下..
这是住宿类
@Entity
@Table(name="accommadation")
public class Accommadation {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
@Column(name="AccmdtnId")
private long AccmdtnId;
@DateTimeFormat(pattern = "yyyy-MM-dd")
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd")
@Column(name="saveDates")
private Date saveDates;
@DateTimeFormat(pattern = "yyyy-MM-dd")
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd")
@Column(name="available_Dates")
private Date availableDates;
@DateTimeFormat(pattern = "hh:mm:ss")
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "hh-mm-ss")
@Column(name="time_Slot_Available")
private Time timeSlotAvailable;
@JsonManagedReference
@OneToMany(mappedBy="accdtn",targetEntity=Location.class,cascade =
CascadeType.ALL)
private List<Location> locations=new ArrayList<Location>();
这是位置实体**
@Entity
public class Location {
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="location_Id")
private long locationId;
@Column(name="location")
private String location;
@JsonBackReference
@ManyToOne(cascade=CascadeType.ALL,fetch=FetchType.LAZY
,targetEntity=Accommadation.class)
@JoinColumn(name="AccmdtnId")
private Accommodation accdtn;
在这里,我采用了单独的实体,但它是否需要,我已经发布了数据,我在位置表下面得到了这样的输出,原因是 2 倍
{
"saveDates": "2019-02-02",
"availableDates": "2019-02-25",
"timeSlotAvailable": "02:00:00",
"locations": [
{
"locationId": 3,
"location": "banglore"
},
{
"locationId": 4,
"location": "ubbali"
}
],
"accmdtnId": 2,
"loactions": [
{
"locationId": 3,
"location": "banglore"
},
{
"locationId": 4,
"location": "ubbali"
}
]
}
【问题讨论】:
-
尝试更多地隔离您的问题,这是我们阅读大量代码的方式。你到底有什么问题?
-
在控制器中@GetMapping 代码如何在 DAO 中工作我有标准接口这是如何工作的
标签: spring-boot