如何使用 ORACLE 数据库插入 JPA OneToOne 关系？

Question

我遇到了 JPA OneToOne 关系问题，我花了 4 天时间尝试解决但没有成功。
首先，看到oracle不支持主键Generate.Auto，我们必须在插入前使用序列创建触发器
这是我的 2 个表：

CREATE TABLE address (
 ID INTEGER PRIMARY KEY,
 CITY varchar(50) DEFAULT NULL,
 COUNTRY varchar(50) DEFAULT NULL,
 STREET varchar(50) DEFAULT NULL,
 SUBURB varchar(50) DEFAULT NULL
)

CREATE TABLE utilisateur (
 ID INTEGER PRIMARY KEY,
 EMAIL varchar(50) DEFAULT NULL,
 FIRSTNAME varchar(50) DEFAULT NULL,
 LASTNAME varchar(50) DEFAULT NULL,
 PASSWORD varchar(64) DEFAULT NULL,
 USERNAME varchar(50) NOT NULL,
 ADDRESS_ID INTEGER DEFAULT NULL,
 CONSTRAINT FK_USER_ADDRESS_ID FOREIGN KEY (ADDRESS_ID) REFERENCES address (ID)
)

以及插入的触发器：

create SEQUENCE address_seq;

CREATE OR REPLACE TRIGGER address_trig
  BEFORE INSERT ON address
  FOR EACH ROW
BEGIN
 :new.id := address_seq.nextval;
END

create SEQUENCE user_seq
/




CREATE OR REPLACE TRIGGER user_trig
  BEFORE INSERT ON utilisateur
  FOR EACH ROW
BEGIN
 :new.id := user_seq.nextval;
END;
/

实体：

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

    @Id
    @SequenceGenerator(name = "USER_SEQ", sequenceName = "USER_SEQ", allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "USER_SEQ")
    @Column(name = "id", nullable = false)
    private Integer id;

    @Column(nullable = false, length = 50)
    private String username;

    @JoinColumn(name = "ADDRESS_ID")
    @OneToOne(cascade = {CascadeType.ALL})
    private Address address;

    @ManyToMany
    @JoinTable(name = "user_roles", joinColumns = {
        @JoinColumn(name = "User_userid")}, inverseJoinColumns = {
        @JoinColumn(name = "Role_roleid")})
    private List<Role> roles;

    public User() {
        roles = new ArrayList<Role>();
        address = new Address();
    }

@Entity
public class Address implements Serializable {

    @Id
    @SequenceGenerator(name = "USER_SEQ", sequenceName = "USER_SEQ", allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "USER_SEQ")
    @Column(name = "id", nullable = false)
    private Integer id;
.....

创建方法：

public void create(T t) {
        this.em.persist(t);
        this.em.merge(t);
    }

为了创建新用户，调用此方法：

 public void doCreateUser() {
        try {
            addressService.create(address);
            newUser.setAddress(address);
            userService.create(newUser);
            FacesMessage message = new FacesMessage("utilisateur ajouté avec succès");
            FacesContext.getCurrentInstance().addMessage(null, message);
        } catch (Exception e) {
           ....}

使用 MySQL 插入成功，但是当我将池配置为使用 ORACLE 时，我收到一个错误，这意味着违反了外键，我知道在调试时我看到了创建和插入的对象地址进入用户，但在方法结束时我得到这个错误：

Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.3.2.v20111125-r10461): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLIntegrityConstraintViolationException: ORA-02291: integrity constraint (FK_USER_ADDRESS_ID) violated - parent key not found
Error Code: 2291
Call: INSERT INTO UTILISATEUR (id, EMAIL, FIRSTNAME, LASTNAME, PASSWORD, USERNAME, ADDRESS_ID) VALUES (?, ?, ?, ?, ?, ?, ?)
    bind => [7 parameters bound]

感谢您对此的任何帮助。
最好的问候。

Answer 1

从您的表中删除所有触发器 - JPA 将为您完成所有工作。仔细阅读有关通过序列生成主键值的信息。

如何使用 ORACLE 数据库插入 JPA OneToOne 关系？

How to do insertion with JPA OneToOne relationship using ORACLE database?

oracle

jpa

one-to-one