JPA Hibernate映射超过 2 个实体

Question

我希望以以下方式映射实体：

我有一个用户，他买了一辆车（特定的品牌、型号和变体）。

为此，我创建了以下类。

Car.java汽车

@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
    ......
}

汽车变体.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
    ..........


}

最后，用户.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，我的问题是。

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

我可以定义@ManyToMany与User和Car的关系。在这种情况下，如果我执行user.getCars（）。getModels（）或user.getCars（）.getModels（（）.get（0）。getVariants（）我最终得到了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           |
--------------------------------------------------------

另外，如果您觉得我没有正确设计我的实体类，请随时告诉我如何改进。

谢谢

Answer 1

不清楚为什么需要这样一个包含嵌套关系中所有键的表。

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. 避免集合s：尽可能使用Sets（LinkedHashSet），或Lists当您需要@OrderCol的
2. 给出简单的名称：避免MODEL_NAME，只需使用NAME
3. 使用@MepdSuperclass，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'";

Answer 2

如果像这样获取主对象

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();

通过这种方式，您可以获取您试图获取的每个对象，但请记住，您使用的是＜code＞FethType。懒惰和cascadeType。所有在每个Pojo