你好
我正在使用 JPA 来保留一些 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 的持久性似乎就可以正常工作。我对此感到困惑。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 不是新的,即应该保留 Y 但应该更新关系项 X。任何有助于找出造成这种情况的原因的帮助将不胜感激。
如果有任何不清楚的地方,我在下面引用了我的程序源代码的相关片段,请不要犹豫,请我澄清。
谢谢 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);
}
}
我在取消注释“drop-and-create-tables”的情况下运行程序一次,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,但这次我收到以下错误。 谢谢
[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)
最佳答案
注意到你在打电话,
saveItems(em, Y.class, Arrays.asList(y, y2));
两次?错误发生在第一次调用还是第二次?
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关于java - JPA 错误 : duplicate key error when persisting a relationship,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5411337/