JPA的Hibernate映射超过2个实体

Question

我想我的实体以下列方式映射：

我有一个用户，他买了车（具体品牌，型号和变体）。

为此，我创建了以下类。

Car.java

@Entity 
@Table(name="CARS") 
public class Car { 

    @Id 
    @GeneratedValue(strategy=GenerationType.AUTO) 
    @Column(name="CAR_ID") 
    private int id; 

    @Column(name="CAR_MAKER") 
    private String maker; 

    @OneToMany(mappedBy="car") 
    private Collection<CarModel> models =new ArrayList<CarModel>(); 

    ...... 
    Getters and Setters go here 
    ...... 
} 

CarModel.java

@Entity 
@Table(name="CAR_MODELS") 
public class CarModel { 


    @Id 
    @GeneratedValue(strategy=GenerationType.AUTO) 
    @Column(name="MODEL_ID") 
    private int modelId; 

    @Column(name="MODEL_NAME") 
    private String modelName; 

    @ManyToOne 
    @JoinColumn(name="CAR_ID") 
    private Car car; 

    @OneToMany(mappedBy="model") 
    private Collection<CarVariant> variants = new ArrayList<CarVariant>(); 

    ...... 
    Getters and Setters go here 
    ...... 
} 

CarVariant.java

@Entity 
@Table(name="CAR_VARIANT") 
public class CarVariant { 

    @Id 
    @GeneratedValue(strategy=GenerationType.AUTO) 
    @Column (name="VARIANT_ID") 
    private int variantId; 

    @Column(name="VARIANT_NAME") 
    private String variantName; 

    @Column(name="FUEL_TYPE") 
    private String fuelType; 

    @ManyToOne 
    @JoinColumn(name="MODEL_ID") 
    private CarModel model; 

    .......... 
     Getters and setters go here 
    .......... 


} 

而国际泳联LLY，User.java

@Entity 
@Table(name="MASUSER") 
public class User { 
    @Id 
    @GeneratedValue(strategy=GenerationType.AUTO) 
    @Column(name="USER_REC_ID") 
    private int recordId; 

    @Column(name="USER_ID") 
    private String id; 

    @Column(name="USER_NAME") 
    private String name; 

    @Column(name="USER_EMAIL_ID") 
    private String emailId; 

    @Column(name="USER_PHONE") 
    private String phone; 


    private Collection<Car> cars = new ArrayList<Car>(); 

} 

现在，这里是我的问题。

在现实生活中，用户可能拥有许多汽车，每个汽车都有特定的品牌，型号和型号。我想用JPA/Hibernate实现这样的功能。

我可以定义一个@ManyToMany与User和Car的关系。在那种情况下，如果我做user.getCars().getModels()或user.getCars().getModels().get(0).getVariants()我最终会得到数据库中的所有内容。

如何映射User和Car以获得以下输出？

-------------------------------------------------------- 
| USER_ID  | CAR_ID  | MODEL_ID | VARIANT_ID | 
-------------------------------------------------------- 
| 1   | 1   | 2   | 3   | 
-------------------------------------------------------- 
| 1   | 1   | 2   | 2   | 
-------------------------------------------------------- 
| 3   | 2   | 1   | 1   | 
-------------------------------------------------------- 

此外，如果你觉得我没有设计我的实体班的权利，请随时让我知道我可以改进。

感谢

Answer 1

如果您获取的主要对象是这样

Criteria criteria=session.createCriteria(User.class); 
List<Object> list=criteria.list(); 
User user=(User)list.get(0); 
Car car=user.getCar(); 
CarModel carModel=car.getCarModel(); 
CarVarient carVarient=carModel.getCarVarient(); 

这样你可以得到每个对象，试图获取，但请记住，你正在使用FethType.Lazy而cascadeType.All在everyPojo

Answer 2

目前尚不清楚为什么你需要这样一个所有表格都嵌套关系的键。

我的两分钱：

@MappedSuperclass 
public abstract class AbstractEntity 
{ 
    @Id 
    @GeneratedValue(strategy=GenerationType.AUTO) 
    @Column 
    private Long id; 

    ...... 
    Getters and Setters go here 
    ...... 
} 

@Entity 
@Table(name = "MANUFACTURER") 
public class Manufacturer extends AbstractEntity 
{ 
    @Column 
    private String name; 

    @OneToMany(mappedBy = "manufacturer") 
    private Set<CarModel> models = new LinkedHashSet<>(); 

    ...... 
    Getters and Setters go here 
    ...... 
} 

@Entity 
@Table(name = "CAR_MODEL") 
public class CarModel extends AbstractEntity 
{ 
    @NotNull 
    @ManyToOne(optional = false) 
    @JoinColumn(name = "MANUFACTURER_ID") 
    private Manufacturer manufacturer; 

    @Column 
    private String name; 

    @OneToMany(mappedBy = "model") 
    private Set<CarVariant> variants = new LinkedHashSet<>(); 

    ...... 
    Getters and Setters go here 
    ...... 
} 

@Entity 
@Table(name = "CAR_VARIANT") 
public class CarVariant extends AbstractEntity 
{ 
    @NotNull 
    @ManyToOne(optional = false) 
    @JoinColumn(name = "CAR_MODEL_ID") 
    private CarModel model; 

    @Column 
    private String name; 

    @Column(name="FUEL_TYPE") 
    private String fuelType; 

    @OneToMany(mappedBy="variant") 
    private Set<Car> cars = new LinkedHashSet<>(); 

    ...... 
    Getters and Setters go here 
    ...... 
} 

@Entity 
@Table(name = "CAR") 
public class Car extends AbstractEntity 
{ 
    @NotNull 
    @ManyToOne(optional = false) 
    @JoinColumn(name = "CAR_VARIANT_ID") 
    private CarVariant variant; 

    @Column(name = "REGISTRATION_NUMBER") 
    private String registrationNumber; 


    @ManyToMany(mappedBy = "cars") 
    private Set<User> users = new LinkedHashSet<>(); 

    ...... 
    Getters and Setters go here 
    ...... 
} 

@Entity 
@Table(name = "USER") 
public class User extends AbstractEntity 
{ 
    @Column 
    private String name; 

    @Column 
    private String email; 

    @Column 
    private String phone; 

    @ManyToMany 
    @JoinTable(name = "USER_CAR", joinColumns = @JoinColumn(name = "USER_ID"), inverseJoinColumns = @JoinColumn(name = "CAR_ID")) 
    private Set<Car> cars = new LinkedHashSet<>(); 

    ...... 
    Getters and Setters go here 
    ...... 
} 

1. 避免Collection秒：使用Set S其中可能的（LinkedHashSet），或List那时你需要一个@OrderColumn
2. 给予简单的名字：避免MODEL_NAME，只是使用NAME
3. 使用@MappedSuperclass，特别是如果您有所有实体的代理ID

您现在可以：

Car car = ...; 
String manufacturerName = car.getVariant().getModel().getManufacturer().getName(); 

String jpql = "select c from Car c where c.variant.model.name = 'Golf' and c.variant.fuelType = 'Diesel'";