Question

您好
我正在使用JPA来保存一些java类，如下所示。只要数据库不包含具有与Y.x.id相同id的类X的元素，Y的持久性似乎工作正常。我为此感到困惑。似乎我正在使用的JPA实现（EclipseLink）似乎没有弄清楚虽然Y是新的，但Y.x并不是新的，即Y应该被持久化但是关系项X应该被更新。任何有助于弄清楚是什么导致这种情况的帮助将不胜感激。如果有什么不清楚的话，我已经引用了我的程序源代码的相关摘要，请不要犹豫，请我澄清。

由于提前。

public static interface YY {
    Object id();
}
@Entity
@Access(AccessType.FIELD)
@Table(name = "X")
public class X implements YY {

    @Id
    long id;

    protected X() {
        this.id = Test.orderId++;
    }


    @Override
    public Long id() {
        return id;
    }
    }

public class YPK {


    private final Date date;
    private final Long x;

    public YPK(X x, Date date) {
        this.date = date;
        this.x = x.id();
    }

    @Override
    public boolean equals(Object arg0) {
        if (arg0 == this) {
            return true;
        } else if (arg0 instanceof YPK) {
            return date.equals(((YPK) arg0).date)
                && x.equals(((YPK) arg0).x);
        }
        return false;
    }

    @Override
    public int hashCode() {
        return date.hashCode() ^ x.hashCode();
    }
}

@Entity
@Access(AccessType.FIELD)
@Table(name = "Y")
@IdClass(YPK.class)
public class Y implements YY {

    @Id
    @Temporal(TemporalType.TIMESTAMP)
    Date date;


    @Id
    @ManyToOne(cascade = CascadeType.ALL)
    private X x;

    public Y(X x) {
        this.x = x;
        date = new DateTime().toDate();
    }

    protected Y() {
    }

    @Override
    public YPK id() {
        return new YPK(x, date);
    }

}

public class Test {

public static long orderId = 0;
public static long customerId = 0;

public static void main(String[] args) {
    Map<String, String> properties = new HashMap<String, String>(){
        {
        put("javax.persistence.jdbc.url", "jdbc:derby://localhost:1527/myDB;create=false;");
        put("javax.persistence.jdbc.driver", "org.apache.derby.jdbc.ClientDriver");
        put("javax.persistence.jdbc.user", "myDbUser");
        put("javax.persistence.jdbc.password", "passwd");
//        put(""eclipselink.ddl-generation", "drop-and-create-tables");
        put(""eclipselink.ddl-generation", "none");
        }
    };

    EntityManagerFactory emf =
        Persistence.createEntityManagerFactory("myPersistenceUnit", properties);
    final EntityManager em = emf.createEntityManager();

    X x =  new X();;
    saveItem(em, X.class, x);

    Y y =  new Y(x);;


    Y y2 =  new Y(x);;

    saveItems(em, Y.class, Arrays.asList(y, y2));
    saveItems(em, Y.class, Arrays.asList(y, y2));

    Z z =  new Z(y2);;
    saveItems(em, Z.class, Arrays.asList(z));
}

private static <T extends YY> void saveItem(final EntityManager em,
        Class<T> clazz, T item) {
    synchronized(em) {
        EntityTransaction tx = em.getTransaction();
        saveItem(em, clazz, item, tx);
    }
}

private static <T extends YY> void saveItems(final EntityManager em,
        Class<T> clazz, Collection<T> items) {
    synchronized (em) {
        EntityTransaction tx = em.getTransaction();
        try {
            tx.begin();
            for (T item : items) {
                saveItem(em, clazz, item, tx);
            }
            tx.commit();
        } finally {
            if (tx.isActive()) {
                tx.rollback();
            }
        }
    }
}

private static <T extends YY> void saveItem(final EntityManager em,
            Class<T> clazz,T item, EntityTransaction tx) {
    Object id = item.id();
    T existing = em.find(clazz, id);

    if (existing != null) {
        em.merge(item);
    } else {
        em.persist(item);
    }
}

}

我运行程序一次，“drop-and-create-tables”取消注释，一切正常，因为数据库没有X条目。我注释掉drop-and-create-tables并再次运行它，但这次我得到以下错误。感谢

[EL Warning]: 2011-03-24 06:52:56.047--UnitOfWork(20391510)--Exception [EclipseLink-4002]
(Eclipse Persistence Services -
    2.2.0.v20110202-r8913): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLIntegrityConstraintViolationException:
The statement was aborted because it would have caused a duplicate key value in a unique or primary key constraint or unique index identified by 'SQL110324065226450' defined on 'X'.
 Error Code: -1 Call:
 INSERT INTO X (ID) VALUES (?)  bind => [1 parameter bound]
 Query: InsertObjectQuery(com.test.X@1b6101e)
 Exception in thread "main" javax.persistence.RollbackException: Exception 
[EclipseLink-4002] (Eclipse Persistence Services -
    2.2.0.v20110202-r8913):
 org.eclipse.persistence.exceptions.DatabaseException
 Internal Exception: java.sql.SQLIntegrityConstraintViolationException:
 The statement was aborted because it would have caused a duplicate key value in a unique or primary key constraint or unique index identified by 'SQL110324065226450' defined on 'X'.
 Error Code: -1 Call: INSERT INTO X (ID) VALUES (?)     bind => [1 parameter bound]
 Query: InsertObjectQuery(com.test.X@1b6101e)   at
org.eclipse.persistence.internal.jpa.transaction.EntityTransactionImpl.commitInternal(EntityTransactionImpl.java:102)
at org.eclipse.persistence.internal.jpa.transaction.EntityTransactionImpl.commit(EntityTransactionImpl.java:63)
at com.test.Test.saveItems(Test.java:80)
at com.test.Test.main(Test.java:56)

日志在这里： 1st run successful 2nd run failure

Answer 1

注意到你正在打电话，

saveItems（em，Y.class，Arrays.asList（y，y2））;

两次？第一次调用或第二次调用时会发生错误吗？

请在最佳状态下包含您的日志。

Answer 2

首先，我不明白为什么使用受保护的构造函数。第二 - 你使用PostgreSQL吗？也许在你的数据库中你必须使用序列？旁观 http://en.wikibooks.org/wiki/Java_Persistence/Identity_and_Sequencing#Sequencing

我曾经在PostgreSQL中遇到过类似的问题。我通过使用这些注释解决了它。

JPA错误：持久关系时出现重复键错误

2 个答案: