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我一直在努力使用休眠创建一个复合键。我找到了这个链接： 如何制作复合主键（java持久性注释）

首先，我尝试了 Tim 接受的解决方案@NaturalId。不适合我。

其次，我尝试了 Arthur Ronald FD Garcia 的解决方案。我仍然有一些问题。这是我的代码，与 Arthur Ronald FD Garcia 的解决方案非常相似。

邮箱类：

@Entity
public class Mailbox {

    @Id
    @GeneratedValue
    private int mailboxId;
    @Column( unique=true, nullable=false )
    private String name;    
    @Column( unique=true, nullable=false )
    private String employeeId;  
    private String status;
    private Date createdOn;

    @OneToMany(cascade=CascadeType.ALL, mappedBy="joinedMailboxMessageId.mailboxId")
    private List<MailboxMessages> joinedMailboxMessageList = new ArrayList<MailboxMessages>();

    public Mailbox(){}
    public Mailbox(int id)  {this.mailboxId=id;}
    public void addSMail(SMail mail) {
        // Implementation
    }
    //Getters and setters
}

邮件类

@Entity
public class SMail {

    @Id
    @GeneratedValue
    private int messageId;

    private Date originDate;
    private String from;
    @ElementCollection
    private List<String> to=new ArrayList<String>();
    @Lob
    private String message;
    private String subject;

    @OneToMany(cascade=CascadeType.ALL, mappedBy="joinedMailboxMessageId.messageId")
    private List<MailboxMessages> joinedMailboxMessageList = new ArrayList<MailboxMessages>();

    public SMail() {}
    public SMail(int messageId) {
        this.messageId=messageId;
    }

    // addMailbox sets up bidirectional relationship
    public void addMailbox(Mailbox mailbox) { // Implementation}
//Getters and setters
}

最后是我的 MailboxMessages 类

@Entity
public class MailboxMessages {

    @ManyToOne
    @JoinColumn(name="MAILBOX_ID", insertable=false, updatable=false)
    private Mailbox mailboxId;

    @ManyToOne
    @JoinColumn(name="MESSAGE_ID", insertable=false, updatable=false)
    private SMail messageId;

    private boolean read;
    private boolean deleted;
    private boolean flagged;
    private String priority;
    private Date messageDate;

    @EmbeddedId
    // Implemented as static class - see bellow
    private MailboxMessagesId joinedMailboxMessageId;

    // INNER CLASS: required because this class contains composite id
    @Embeddable
    public static class MailboxMessagesId implements Serializable {

    @ManyToOne
        @JoinColumn(name="MAILBOX_ID")
        private Mailbox mailboxId;

        @ManyToOne
        @JoinColumn(name="MESSAGE_ID")
        private SMail messageId;

        // required no arg constructor
        public MailboxMessagesId() {}

        public MailboxMessagesId(Mailbox mailbox, SMail mail) {
            this.mailboxId = mailbox;
            this.messageId = mail;
        }

        public MailboxMessagesId(int mailboxId, int messageId) {
            this(new Mailbox(mailboxId), new SMail(messageId));
        }

        @Override
        public boolean equals(Object instance) {
            if (instance == null)
                return false;

            if (!(instance instanceof MailboxMessagesId))
                return false;

            final MailboxMessagesId other = (MailboxMessagesId) instance;

            if (!(mailboxId.getMailboxId()==(other.getMailboxId().getMailboxId())))
                return false;

            if (!(messageId.getMessageId()==(other.getMessageId().getMessageId())))
                return false;

            return true;
        }

        @Override
        public int hashCode() {
            int hash = 7;
            hash = 47 * hash + (this.mailboxId != null ? this.mailboxId.hashCode() : 0);
            hash = 47 * hash + (this.messageId != null ? this.messageId.hashCode() : 0);
            return hash;
        }

        //Getters and setters
    }
    //Constructors and getters and setters
}

现在真正的问题是休眠只是为邮箱类创建表。所以我想问题可能出在其他两个班级。我尝试了亚瑟罗纳德的确切解决方案；它工作得很好。你能帮我找出我在代码中可能犯的错误吗？并且如果在 MailboxMessage 类中创建复合键的这种技术有任何替代方法

编辑1：

发现的问题之一是我使用“from”作为我的成员字段，它是 sql 的保留关键字。谁会想到呢？但问题仍然存在。现在我确实有一个用于 smail 类的表，但仍然没有为 MailboxMessages 类创建表。请帮帮我。

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2 回答 2

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自己找到了解决方案：问题在于在 SMail 类中使用“from”作为数据成员，在 MailboxMessages 类中使用“read”作为数据成员，因为它们是 mysql 的保留关键字。现在我所有的表都是用复合键创建的，更不用说。（通过数据库确认）

于 2013-04-14T12:16:51.973 回答
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我认为这是例如字段名称只是示例的解决方案：

@Id
@Column(name = "EMP_NO_FK" , unique=true , nullable = false ,length=10)
private Long empno;

@Id
@Column(name = "GROUP_NO_FK" , unique=true , nullable = false ,length=5)
private Long  groupNo;

@Id
@Column(name = "DEPT_NO_FK" , unique=true , nullable = false ,length=10)
private Long deptno;

@Id
@Column(name = "YEAR_NAME_M_FK" , unique=true , nullable = false ,length=4)
private Integer  year;
于 2013-04-14T11:01:14.473 回答