如何通过外键获取数据？

Posted 2023-02-24

【中文标题】如何通过外键获取数据？

【英文标题】：How to get data through Foreign key?

【发布时间】：2016-07-21 07:34:35

【问题描述】：

如何从外键（restaurant_id 和 channel_id）中获取内容？我的意思不是 FK 值，而是 restaurant_id 的餐厅名称和 channel_id 的频道名称。

表：

型号：

@Entity
@Table(name = "restaurant")
public class Restaurant 

    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name = "restaurant_id")
    private List<Booking> bookings;

    ...

@Entity
@Table(name = "channel")
public class Channel 

    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name = "channel_id")
    private List<Booking> bookings;

    ...

@Entity
@Table(name = "booking")
public class Booking 
    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column(name = "restaurant_id", unique = true)
    private Long restaurant_id;

    @Column(name = "channel_id", unique = true)
    private Long channel_id;

    ...

    @Override
    public String toString() 
        return "Booking [id=" + id + ", name=" + name + ", count=" + count + ", duration=" + duration + ", start="
                + start + ", phone=" + phone + ", comment=" + comment + ", updated=" + updated + ", created=" + created
                + ", active=" + active + ", restaurant_id=" + restaurant_id + ", channel_id=" + channel_id + "]";
    

    ...

DAO

@Repository
public class BookingDaoImpl implements BookingDao 

    private static final Logger logger = LoggerFactory.getLogger(BookingDaoImpl.class);

    @Resource(name = "localSessionFactoryBean")
    private SessionFactory sessionFactory;

    @Override
    public List<Booking> getBookings() 
        Session session = this.sessionFactory.getCurrentSession();
        List<Booking> bookingList = session.createCriteria(Booking.class).list();
        for (Booking booking : bookingList) 
            logger.info("Booking List::" + booking);
        
        return bookingList;
    

    ...

服务

@Service
public class BookingServiceImpl implements BookingService 

    @Autowired
    private BookingDao bookingDao;

    @Override
    @Transactional
    public List<Booking> getBookings() 
        return bookingDao.getBookings();
    

    ...

控制器

@Controller
public class MainController 

    @Autowired
    private BookingService bookingService;

    @RequestMapping(value = "bookings", method = RequestMethod.GET)
    public String bookings(Model model) 
        List<Booking> bookingList = bookingService.getBookings();
        model.addAttribute("bookings", bookingList);
        return "bookings";
    

    ...

JSP

<c:forEach items="$bookings" var="booking">
    <tr>
        <td>$booking</td>
    </tr>
</c:forEach>

toString() 结果：

【问题讨论】：

您希望从哪里获取信息？ JSP？控制器还是在 SQL 查询中？

JSP。现在我得到 restaurant_id=1，channel_id=1，但我想得到餐厅名称，或地址和频道名称。使用 id 我不能做太多事情。

【参考方案1】：

@Entity
@Table(name = "restaurant")
public class Restaurant 

    @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, mappedBy="restaurant")
    private List<Booking> bookings;



@Entity
@Table(name = "channel")
public class Channel 

    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, mappedBy="channel")
    private List<Booking> bookings;



@Entity
@Table(name = "booking")
public class Booking 

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "restaurant_id") 
    private Restaurant restaurant;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "channel_id") 
    private Channel channel;

您可以将 HQL 与 session.createQuery() 一起使用。

这让所有Booking 急切地获取关联。您也可以将Criteria 与 fetch 一起使用。

from Booking b left join fetch b.restaurant left join fetch b.channel 

这会获取名称，您可以将其处理为 List<Object[]> 或使用转换器到 DTO。

select booking.name, restaurant.name, channel.name
  from Booking booking left join booking.restaurant restaurant left join booking.channel channel

【讨论】：

我应该如何实现这个HQL？

@user3374117 欢迎您。您也可以使用list()，但在@ManyToOne 中不使用FetchType.LAZY。