@OneToOne(mappedBy) 导致不创建列

Question

我正在尝试学习 JPA 和 Hibernate。当尝试将 mappedBy 注释用于 map 访问卡时，该列不会创建。 我是堆栈溢出的新手，所以如果我做错了请原谅我。

这是我的员工 class

@Entity
@Table(name = "EMPLOYEE_DATA")
public class Employee {

    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private int id;
    @Column(name = "EMPLOYEE_NAME")
    private String name;
    @Column(unique = true, length = 9, nullable = false, updatable = false)
    private String ssn;
    @Column(nullable = true)
    private int age;
    @Temporal(TemporalType.DATE)
    private Date dob;
    @Enumerated(EnumType.STRING)
    private EmployeeType type;
    @OneToOne
    private AccessCard card;
    @Transient
    public String debugString;

    //getters and setters, toString

这是我的门禁卡 class

@Entity
@Table(name = "ACCESS_CARD")
public class AccessCard {
    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private int id;
    @Temporal(TemporalType.DATE)
    private Date issueDate;
    private boolean isActive;
    private String firmwareVersion;
    @OneToOne(mappedBy = "card")
    private Employee owner;

   //getters and setters, toString

这是我的 JpaStarterMain class 用于创建对象并持久化它们

public class JpaStarterMain {
    public static void main(String[] args) {

        Employee employee2 = new Employee();
        employee2.setType(EmployeeType.FULL_TIME);
        employee2.setAge(24);
        employee2.setDob(new GregorianCalendar(1998, Calendar.OCTOBER, 20).getTime());
        employee2.setName("Louie Jenkins");
        employee2.setSsn("123456789");

        Employee employee1 = new Employee();
        employee1.setType(EmployeeType.PART_TIME);
        employee1.setAge(21);
        employee1.setDob(new GregorianCalendar(2001, Calendar.OCTOBER, 5).getTime());
        employee1.setName("Gigi Vad");
        employee1.setSsn("987654321");

        AccessCard card2 = new AccessCard();
        card2.setFirmwareVersion("2.7.0");
        card2.setIssueDate(new Date());
        card2.setActive(true);
        card2.setOwner(employee2);
        employee2.setCard(card2);

        AccessCard card1 = new AccessCard();
        card1.setFirmwareVersion("2.7.0");
        card1.setIssueDate(new Date());
        card1.setActive(false);
        card1.setOwner(employee1);
        employee1.setCard(card1);

        EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory("test");
        EntityManager entityManager = entityManagerFactory.createEntityManager();
        EntityTransaction transaction = entityManager.getTransaction();

        transaction.begin();

        entityManager.persist(employee1);
        entityManager.persist(employee2);
        entityManager.persist(card1);
        entityManager.persist(card2);

        transaction.commit();
        entityManager.close();
        entityManagerFactory.close();

我在控制台中得到这个

Hibernate: insert into EMPLOYEE_DATA (age, card_id, dob, EMPLOYEE_NAME, ssn, type, id) values (?, ?, ?, ?, ?, ?, ?)
Hibernate: insert into EMPLOYEE_DATA (age, card_id, dob, EMPLOYEE_NAME, ssn, type, id) values (?, ?, ?, ?, ?, ?, ?)
Hibernate: insert into ACCESS_CARD (firmwareVersion, isActive, issueDate, id) values (?, ?, ?, ?)
Hibernate: insert into ACCESS_CARD (firmwareVersion, isActive, issueDate, id) values (?, ?, ?, ?)

出于某种原因，当去掉@OneToOne 注释中的mappedBy 时，owner_id 被添加到ACCESS_CARD 表下的数据库中，但是当使用mappedBy 时，它不再起作用了！ 见下文

@Entity
@Table(name = "ACCESS_CARD")
public class AccessCard {
    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private int id;
    @Temporal(TemporalType.DATE)
    private Date issueDate;
    private boolean isActive;
    private String firmwareVersion;
    @OneToOne
    private Employee owner;

   //getters and setters, toString

然后我在控制台中得到这个

Hibernate: insert into EMPLOYEE_DATA (age, card_id, dob, EMPLOYEE_NAME, ssn, type, id) values (?, ?, ?, ?, ?, ?, ?)
Hibernate: insert into EMPLOYEE_DATA (age, card_id, dob, EMPLOYEE_NAME, ssn, type, id) values (?, ?, ?, ?, ?, ?, ?)
Hibernate: insert into ACCESS_CARD (firmwareVersion, isActive, issueDate, owner_id, id) values (?, ?, ?, ?, ?)
Hibernate: insert into ACCESS_CARD (firmwareVersion, isActive, issueDate, owner_id, id) values (?, ?, ?, ?, ?)

任何帮助都非常感谢！

Answer 1

当您使用mappedBy时，这意味着关系在关系的另一端持久存在。 这在数据持久化方面已经足够了。 您可以在OneToMany关系中找到相同的行为，其中mappedBy属性给出另一个 class 上相应成员的名称，并用ManytoOne注释。 在这第二种情况下，更明显的是持久化在manyToOne这边。