Question

我在MVC SpringBoot应用程序中遇到了JPA关系的问题。

我正在尝试创建一个应用程序，其中我有许多用户，每个用户都可以拥有多个汽车。用户可以拥有多辆汽车。我在用户和汽车对象之间建立了@OneToOne关系，这就是我所做的：

这是User类：

@Entity
@Table(name = "user")
public class User implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @Column(name = "username", nullable = false)
    private String username;

    @Column(name = "password", length = 500, nullable = false)
    private String password;

    @OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
    private List<Car> cars;

}

然后这是Car类：

@Entity
@Table(name = "car")
public class Car implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @Column(length = 11)
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @Column(name = "make", nullable = false)
    private String make;

    @Column(name = "model", nullable = false)
    private String model;

    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name = "id")
    private User user;

}

然后，这是实际的服务实现

@Component
@Transactional(readOnly = true)
public class CarServiceImpl implements CarService {

@Inject
private CarRepository carRepository;

@Inject
private UserRepository userRepository;

@Override
@Transactional(readOnly = false)
public Car addCar(Long userId, Car car) {
    User user = userRepository.findOne(userId);      
    user.getGpsLocationModels().add(car);

    car.setUser(user);
    carRepository.save(car);

    return car;
}

然后我有端点，但完全有效。添加方法看起来像是应该工作，至少我得到了预期的输出，但是我不知道如何编写它的find方法，很好无法弄清楚如何根据用户检索汽车，我知道如何通过他们的ID来获取它们，但不是分别为每个用户获取它们。

这是我的尝试：

@Override
public Car findCar(Long userId, Long carId) {

    //get the current user (that comes as JSON Request Param)
    User user = userRepository.findOne(userId);

    //get the car based on its ID, here's the problem, I want the car based on its ID AND its user, I can't display cars which aren't for that particular user
    Car car = carRepository.findOne(carId);

    return car;
}

以下是为特定用户获取所有汽车的方法：

@Override
public List<Car> displayAllCars(Long userId) {
    return userRepository.findOne(userId).getCars();
}

我非常感谢您提供的任何帮助。

非常感谢

Answer 1

您的映射不正确。汽车＆gt;用户是@ManyToOne。如果你也使这个双向，你也可以通过用户检索汽车：

@Entity
@Table(name = "user")
public class User implements Serializable {

    @OneToMany(mappedBy ="user",cascade = CascadeType.ALL)
    private Set<Car> cars;

    public Set<Car> getCars(){
        return cars;
    }

    public void addCar(Car car){
        cars.add(car);
        car.setUser(this);
    }
}

@Entity
@Table(name = "car")
public class Car implements Serializable {

    @ManyToOne(fetch=FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumn(name="user_id")
    private User user;
}

@Override
public Set<Car> findCars(Long userId) {
    return userRepository.findOne(userId).getCars();
}

Answer 2

您可以拥有一个接受用户ID的方法，并返回汽车存储库中的列表，如：

List<Car> findCarByUser(Long userID);

然后你会有

@Override
public List<Car> displayAllCars(Long userId) {
return carRepository.findCarByUser(userId);
}

MVC SpringBoot和JPA关系

2 个答案: