休眠问题:@OneToMany批注返回重复项 [英] Hibernate issue: @OneToMany annotation returns duplicates
问题描述
我在使用Hibernate(4.3.0)时遇到问题，因为单向@OneToMany返回重复项.
I am facing a problem with Hibernate (4.3.0) where as unidirectional @OneToMany returns duplicates.
我的数据库结构(带有InnoDB的MySQL)的"entry"表与"entry_address"表具有1:N的关系."entry"表是主表，"entry_address"是"entry"表的子表.
My database structure (MySQL with InnoDB) where as 'entry' table has a 1:N relationship with 'entry_address' table. The 'entry' table is the main table and 'entry_address' is a sub-table of 'entry' table.
CREATE TABLE IF NOT EXISTS `entry` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(500) NOT NULL,
`active` int(1) NOT NULL DEFAULT '0',
`modifiedTS` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00' ON UPDATE CURRENT_TIMESTAMP,
`createdTS` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=4 ;
INSERT INTO `entry` (`id`, `name`, `active`, `modifiedTS`, `createdTS`) VALUES
(1, 'Test1', 0, '2012-11-05 13:41:03', '2012-11-01 10:11:22'),
(2, 'Test2', 1, '2012-11-05 11:19:37', '2012-11-01 10:11:33'),
(3, 'Test3', 1, '2012-11-05 11:19:37', '2012-11-01 10:11:44');
CREATE TABLE IF NOT EXISTS `entry_address` (
`id` int(10) unsigned NOT NULL,
`precedence` int(1) NOT NULL DEFAULT '0',
`line` varchar(255) DEFAULT NULL,
`line2` varchar(255) DEFAULT NULL,
`street` varchar(255) DEFAULT NULL,
`street2` varchar(255) DEFAULT NULL,
`zip` int(5) DEFAULT NULL,
`city` varchar(255) DEFAULT NULL,
`country` varchar(255) DEFAULT NULL,
UNIQUE KEY `entry_address_uq` (`id`,`precedence`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `entry_address` (`id`, `precedence`, `line`, `line2`, `street`, `street2`, `zip`, `city`, `country`) VALUES
(1, 0, 'Line4.1', 'Line4.2', 'Street4.1', 'Street4.2', 9488, 'Schellenberg', 'Liechtenstein'),
(2, 10, 'Line1.1', 'Line1.2', 'Street1.1', 'Street1.2', 9492, 'Eschen', 'Liechtenstein'),
(2, 20, 'Line2.1', 'Line2.2', 'Street2.1', 'Street2.2', 9490, 'Vaduz', 'Liechtenstein'),
(2, 30, 'Line3.1', 'Line3.2', 'Street3.1', 'Street3.2', 9494, 'Schaan', 'Liechtenstein'),
(3, 10, 'Line5.1', 'Line5.2', 'Street5.1', 'Street5.2', 9492, 'Eschen', 'Liechtenstein'),
(3, 20, 'Line6.1', 'Line6.2', 'Street6.1', 'Street6.2', 9492, 'Eschen', 'Liechtenstein');
ALTER TABLE `entry_address`
ADD CONSTRAINT `entry_address_fk` FOREIGN KEY (`id`) REFERENCES `entry` (`id`);
这是"entry"实体的最小代码.
Here's the minimal code of "entry" entity.
import java.util.Collection;
import javax.persistence.*;
@Entity
@Table(name = "entry")
public class Entry {
@Id
@GeneratedValue
@Column(name = "id")
private Integer id;
@Column(name = "name")
private String name;
@OneToMany(fetch=FetchType.EAGER)
@JoinColumn(name = "id")
private Collection<EntryAddress> addresses;
@Override
public String toString() {
return String.format("Entry [id=%s, name=%s, addresses=%s]", id, name, addresses);
}
}
这是"entry_address"实体的最小代码:
Here's the minimal code of "entry_address" entity:
import javax.persistence.*;
@Entity
@Table(name = "entry_address")
public class EntryAddress {
@Id
@Column(name = "id")
private Integer id;
@Column(name = "line")
private String line;
@Override
public String toString() {
return String.format("EntryAddress [line=%s]", line);
}
}
这是由Hibernate完成的查询(看起来不错！):休眠状态:从this_.id左边的外部连接entry_addresss2_项中选择this_.id作为id0_1_，this_.name作为name0_1_，addresses2_.id作为id0_3_，addresses2_.id作为id1_3_，addresses2_.id作为id1_0_，addresss2_.line作为line1_0_.= addresses2_.id
This is the query done by Hibernate (looks good!): Hibernate: select this_.id as id0_1_, this_.name as name0_1_, addresses2_.id as id0_3_, addresses2_.id as id1_3_, addresses2_.id as id1_0_, addresses2_.line as line1_0_ from entry this_ left outer join entry_address addresses2_ on this_.id=addresses2_.id
但是如果我使用以下命令运行JUnit:
But if I run JUnit using:
import java.util.Collection;
import junit.framework.Assert;
import li.pitschmann.transaction.dao.EntryDao;
import li.pitschmann.transaction.entity.Entry;
import org.junit.Test;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.test.context.ContextConfiguration;
import org.springframework.test.context.junit4.AbstractTransactionalJUnit4SpringContextTests;
@ContextConfiguration(locations={"file:**/web-spring.xml"})
public class EntryDaoTest extends AbstractTransactionalJUnit4SpringContextTests {
@Autowired
private EntryDao entryDao;
@Test
public void findAllEntries() {
Collection<Entry> entries = entryDao.findEntries();
Assert.assertNotNull(entries);
for (Entry e : entries) {
System.out.println("++: " + e);
}
// Assert.assertEquals(3, entries.size());
}
}
import java.util.Collection;
import org.hibernate.SessionFactory;
import org.springframework.transaction.annotation.Transactional;
import li.pitschmann.transaction.dao.EntryDao;
import li.pitschmann.transaction.entity.Entry;
public class EntryDaoImpl implements EntryDao {
private SessionFactory sessionFactory;
/**
* {@inheritDoc}
*/
@SuppressWarnings("unchecked")
@Transactional
public Collection<Entry> findEntries() {
return sessionFactory.getCurrentSession().createCriteria(Entry.class).list();
}
public void setSessionFactory(SessionFactory sessionFactory) {
this.sessionFactory = sessionFactory;
}
}
Spring XML(最重要的部分，Spring 3.1.2.RELEASE):
Spring XML (most important part, Spring 3.1.2.RELEASE):
<tx:annotation-driven transaction-manager="transactionManager" />
<context:annotation-config />
<!-- MySQL DataSource -->
<bean id="dataSource" class="com.mchange.v2.c3p0.ComboPooledDataSource">
<property name="driverClass" value="com.mysql.jdbc.Driver" />
<property name="jdbcUrl" value="jdbc:mysql://localhost:3306/entry_db" />
<property name="user" value="root" />
<property name="password" value="" />
</bean>
<!-- Session Factory -->
<bean id="sessionFactory" class="org.springframework.orm.hibernate4.LocalSessionFactoryBean">
<property name="dataSource" ref="dataSource" />
<property name="packagesToScan">
<list>
<value>li.pitschmann.transaction.entity</value>
</list>
</property>
<property name="hibernateProperties">
<props>
<prop key="hibernate.dialect">org.hibernate.dialect.MySQL5InnoDBDialect</prop>
<prop key="hibernate.show_sql">true</prop>
</props>
</property>
</bean>
<!-- Transaction Manager -->
<bean id="transactionManager" class="org.springframework.orm.hibernate4.HibernateTransactionManager">
<property name="sessionFactory" ref="sessionFactory" />
</bean>
控制台日志为:
++: Entry [id=1, name=Test1, addresses=[EntryAddress [line=Line4.1]]]
++: Entry [id=2, name=Test2, addresses=[EntryAddress [line=Line1.1], EntryAddress [line=Line1.1], EntryAddress [line=Line1.1]]]
++: Entry [id=2, name=Test2, addresses=[EntryAddress [line=Line1.1], EntryAddress [line=Line1.1], EntryAddress [line=Line1.1]]]
++: Entry [id=2, name=Test2, addresses=[EntryAddress [line=Line1.1], EntryAddress [line=Line1.1], EntryAddress [line=Line1.1]]]
++: Entry [id=3, name=Test3, addresses=[EntryAddress [line=Line5.1], EntryAddress [line=Line5.1]]]
++: Entry [id=3, name=Test3, addresses=[EntryAddress [line=Line5.1], EntryAddress [line=Line5.1]]]
我还尝试使用@OneToMany(fetch = FetchType.LAZY)而不是FetchType.EAGER-重复地址的相同问题.
I also tried to use @OneToMany(fetch=FetchType.LAZY) instead of FetchType.EAGER - same issue with duplicate addresses.
++: Entry [id=1, name=Test1, addresses=[EntryAddress [line=Line4.1]]]
++: Entry [id=2, name=Test2, addresses=[EntryAddress [line=Line1.1], EntryAddress [line=Line1.1], EntryAddress [line=Line1.1]]]
++: Entry [id=3, name=Test3, addresses=[EntryAddress [line=Line5.1], EntryAddress [line=Line5.1]]]
期望
这是我的期望(3个具有不同地址的入口对象):
This is my expecation (3 entry objects with different addresses):
++: Entry [id=1, name=Test1, addresses=[EntryAddress [line=Line4.1]]]
++: Entry [id=2, name=Test2, addresses=[EntryAddress [line=Line1.1], EntryAddress [line=Line2.1], EntryAddress [line=Line3.1]]]
++: Entry [id=3, name=Test3, addresses=[EntryAddress [line=Line5.1], EntryAddress [line=Line6.1]]]
Hibernate中是否有bug，或者我做错了什么?希望有人可以帮助我找到根本原因吗?谢谢:-)
Is there a bug in Hibernate or am I doing something wrong? Hope someone can help me to find the root cause?! Thank you :-)
推荐答案
您可能需要像这样修改您的方法:
You might need to modify your method like so:
@SuppressWarnings("unchecked")
@Transactional
public Collection<Entry> findEntries() {
return sessionFactory.getCurrentSession()
.createCriteria(Entry.class)
.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY)
.list();
}
另外，将 addresses 更改为 Set :
@OneToMany(fetch=FetchType.EAGER)
@JoinColumn(name = "id")
private Set<EntryAddress> addresses;
哦...在 EntryAddress 中，您已将 id 定义为 @Id ，但这不是唯一的.您应该将 id 作为主键，并使其自动递增，就像在 Entry 中所做的那样.然后在 EntryAddress 中创建另一个字段，该字段是 Entry 的外键，名称类似于 entry_id .
Oh...In EntryAddress you have id defined as the @Id but it is not unique. You should make id the primary key and have it auto increment like you do in Entry. Then create another field in EntryAddress that is the foreign key to Entry called something like entry_id.
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