@OneToOne映射与Hibernate共享主键用户 - 帐户

Question

我已经尝试将this example为我的情况工作。 Primus是User，Secundus是Account。用户应该与帐户共享主键。一切正常，直到我尝试级联坚持：

User user = new User(); 
user.setName("Andy"); 
this.uDao.create(user); 

是好的和作品，但是......

User user = new User(); 
user.setName("Andy"); 
Account account = new Account(); 
account.setUsername("xyz"); 
user.setAccount(account); 
this.uDao.create(user); 

给出了错误：

01:14:35,844 INFO [stdout] (ServerService Thread Pool -- 80) Hibernate: insert into user (name) values (?) 

01:14:35,860 INFO [stdout] (ServerService Thread Pool -- 80) Hibernate: insert into account (username, user_id) values (?, ?) 

01:14:35,861 WARN [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (ServerService Thread Pool -- 80) SQL Error: 1452, SQLState: 23000 
01:14:35,861 ERROR [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (ServerService Thread Pool -- 80) Cannot add or update a child row: a foreign key constraint fails (`shitstorm`.`account`, CONSTRAINT `fk_account_user` FOREIGN KEY (`user_id`) REFERENCES `user` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION) 

我能做些什么？我的错误是什么？我在我的硕士论文，并在我的数据库中有一对一的关系，需要处理这个。让Cascading.persist工作对我来说很重要。我尝试了很多教程并阅读了很多解释，但我无法处理这种情况。我在链接中发布的教程是我的最后一次尝试。 Wildfly 8.2.1上运行着这个复制品，非常感谢！

这里是SQL：

CREATE TABLE IF NOT EXISTS `shitstorm`.`user` (
    `id` INT(11) NOT NULL AUTO_INCREMENT, 
    `name` VARCHAR(45) NULL DEFAULT NULL, 
    PRIMARY KEY (`id`)) 

-- ----------------------------------------------------- 
-- Table `shitstorm`.`account` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `shitstorm`.`account` (
    `user_id` INT(11) NOT NULL, 
    `username` VARCHAR(45) NULL DEFAULT NULL, 
    PRIMARY KEY (`user_id`), 
    INDEX `fk_account_user_idx` (`user_id` ASC), 
    CONSTRAINT `fk_account_user` 
    FOREIGN KEY (`user_id`) 
    REFERENCES `shitstorm`.`user` (`id`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 

这里是我的实体：

@Entity 
@Table(name="user") 
@NamedQuery(name="User.findAll", query="SELECT u FROM User u") 
public class User implements Serializable { 
    private static final long serialVersionUID = 1L; 

    @Id 
    @GeneratedValue(strategy=GenerationType.IDENTITY) 
    @Column(unique=true, nullable=false) 
    private int id; 

    @Column(length=45) 
    private String name; 

    //bi-directional one-to-one association to Account 
    @OneToOne(cascade=CascadeType.PERSIST, mappedBy="user") 
    private Account account; 

    public User() { 
    } 

    public int getId() { 
     return this.id; 
    } 

    public void setId(int id) { 
     this.id = id; 
     if(this.account != null){ 
      this.account.setUserId(id); 
     } 
    } 

    public String getName() { 
     return this.name; 
    } 

    public void setName(String name) { 
     this.name = name; 
    } 

    public Account getAccount() { 
     return this.account; 
    } 

    public void setAccount(Account account) { 
     this.account = account; 
     if(account != null){ 
      account.setUser(this); 
     } 
    } 

} 

和

@Entity 
    @Table(name="account") 
    @NamedQuery(name="Account.findAll", query="SELECT a FROM Account a") 
    public class Account implements Serializable { 
     private static final long serialVersionUID = 1L; 

     @Id 
     @Column(name="user_id", unique=true, nullable=false) 
     private int userId; 

     @Column(length=45) 
     private String username; 

     //bi-directional one-to-one association to User 
     @OneToOne 
     @PrimaryKeyJoinColumn(name="user_id") 
//@JoinColumn(name="user_id", nullable=false, insertable=false, updatable=false) -- was generated by JPA TOOLS 
     private User user; 



     public Account() { 
     } 

     public int getUserId() { 
      return this.userId; 
     } 

     public void setUserId(int userId) { 
      this.userId = userId; 
     } 

     public String getUsername() { 
      return this.username; 
     } 

     public void setUsername(String username) { 
      this.username = username; 
     } 

     public User getUser() { 
      return this.user; 
     } 

     public void setUser(User user) { 
      this.user = user; 
     } 

    } 

Answer 1

您可以在JPA 2+更简单地实现这一目标如下：

public class User{ 

    @Id 
    @GeneratedValue(strategy=GenerationType.IDENTITY) 
    @Column(unique=true, nullable=false) 
    private int id; 

    @OneToOne(cascade=CascadeType.PERSIST, mappedBy = "user") 
    private Account account; 
} 

public class Account{ 

    @Id 
    @OneToOne 
    @JoinColumn(name = "user_id") 
    private User user; 
} 

参见：

https://en.wikibooks.org/wiki/Java_Persistence/Identity_and_Sequencing#Primary_Keys_through_OneToOne_and_ManyToOne_Relationships