JPA 简单的多对多与 eclipselink

Question

我有这个 ER 图：

这被翻译成这些类：

User.java

@Entity
@Table(name = "user")
@NamedQueries({
    @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)
    @Basic(optional = false)
    @Column(name = "id")
    private Integer id;
    @Column(name = "name")
    private String name;
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "user")
    private Collection<UserHasNotify> userHasNotifyCollection;

UserHasNotify.java

@Entity
@Table(name = "user_has_notify")
@NamedQueries({
    @NamedQuery(name = "UserHasNotify.findAll", query = "SELECT u FROM UserHasNotify u")})
public class UserHasNotify implements Serializable {
    private static final long serialVersionUID = 1L;
    @EmbeddedId
    protected UserHasNotifyPK userHasNotifyPK;
    @Column(name = "has_read")
    private String hasRead;
    @JoinColumn(name = "notify_id", referencedColumnName = "id", insertable = false, updatable = false)
    @ManyToOne(optional = false)
    private Notify notify;
    @JoinColumn(name = "user_id", referencedColumnName = "id", insertable = false, updatable = false)
    @ManyToOne(optional = false)
    private User user;

UserHasNotifyPK.java

@Embeddable
public class UserHasNotifyPK implements Serializable {
    @Basic(optional = false)
    @Column(name = "user_id")
    private int userId;
    @Basic(optional = false)
    @Column(name = "notify_id")
    private int notifyId;

Notify.java

@Entity
@Table(name = "notify")
@NamedQueries({
    @NamedQuery(name = "Notify.findAll", query = "SELECT n FROM Notify n")})
public class Notify implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    @Column(name = "id")
    private Integer id;
    @Column(name = "message")
    private String message;
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "notify")
    private Collection<UserHasNotify> userHasNotifyCollection;

现在，我要添加一个实体 User 和 Notify 并在它们之间创建一个关系。 所以我写了这个sn-p：

        User user = new User();
        user.setName("John");

        Notify notify = new Notify();
        notify.setMessage("Hello World");

        userFacade.create(user);
        notifyFacade.create(notify);

        UserHasNotify uhn = new UserHasNotify();
        uhn.setNotify(notify);
        uhn.setUser(user); 
        uhn.setHasRead("ok");
        uhnFacade.create(uhn);

但我收到此错误：

Internal Exception: com.mysql.jdbc.exceptions.jdbc4.MySQLIntegrityConstraintViolationException: Column 'user_id' cannot be null
Error Code: 1048
Call: INSERT INTO user_has_notify (has_read, user_id, notify_id) VALUES (?, ?, ?)
    bind => [3 parameters bound]
Query: InsertObjectQuery(com.test.entity.UserHasNotify[ userHasNotifyPK=null ])

为什么????????????

Answer 1

错误的原因可能在于远程通信。您使用外观的事实意味着您正在与后端进行远程通信。

这意味着您在uhn 实例上设置的user 和notify 实例被发送到远程系统上持久化，而本地实例永远不会收到生成的ID。

要验证和解决这个问题，您可以扩展您的示例：

保存user 和notify 后，从您的后端获取它们。这应该返回具有现有 ID 的持久实例。然后你可以使用它们来存储你的uhn 关系。

编辑：我错过了UserHasNotify 是带有嵌入 ID 的有状态关系这一事实。在代码中，您从未设置此 Id，因此提供者会错过它。

对于这种情况，我建议使用 IdClass 而不是嵌入的 Id - 映射更具可读性，因此您可能不会两次映射 User 和 Notify 关系 - 一次在嵌入PK并再次在实体中；）

这是它的样子：

public class UserHasNotifyPK implements Serializable {
    private Notify notify;
    private User user;
    ...
    }

.

@Entity
@Table(name = "user_has_notify")
@IdClass(UserHasNotifyPK.class)
@NamedQueries({
    @NamedQuery(name = "UserHasNotify.findAll", query = "SELECT u FROM UserHasNotify u")})
public class UserHasNotify implements Serializable {
    
    private static final long serialVersionUID = 1L;
        
    @Column(name = "has_read")
    private String hasRead;
    
    @Id
    @JoinColumn(name = "notify_id", referencedColumnName = "id", insertable = false, updatable = false)
    @ManyToOne(optional = false)
    private Notify notify;
    
    @Id
    @JoinColumn(name = "user_id", referencedColumnName = "id", insertable = false, updatable = false)
    @ManyToOne(optional = false)
    private User user;

然后您可以再次尝试相同的测试。