问题描述
到现在为止,我试图将单独的类中的所有数据库代码分开,例如
TableTeam或TableMember,它们的方法类似于 .getAllTeams()
或 .updateTeam(Team t)
。这些方法充当他们执行的sql查询的包装。
现在使用hibernate时,我试图尽可能地保留界面,只是将SQL查询更改为主要工作的休眠功能。
有一个例外:更新元素。
team = new Team(-1,Team Name,Collections。 <会员> emptySet());
@Test
public void testUpdateTeam()抛出异常{
table.addTeam(team);
team.setName(New Name);
table.updateTeam(team);
Team f = table.getAllTeams()。get(0);
Assert.assertEquals(team,f);
}
会导致以下异常:
org.hibernate.LazyInitializationException:无法延迟初始化角色集合:ch.tiim.sco.database.model.Team.members,无法初始化代理 - 无会话
at org.hibernate.collection.internal.AbstractPersistentCollection.throwLazyInitializationException(AbstractPersistentCollection.java:576)
at org.hibernate.collection.internal.AbstractPersistentCollection.withTemporarySessionIfNeeded(AbstractPersistentCollection.java:215)
at org.hibernate.collection.internal.AbstractPersistentCollection.readSize(AbstractPersistentCollection.java:156)
at org.hibernate.collection.internal.PersistentSet.size(PersistentSet.java:160)
at java.util .AbstractSet.equals(AbstractSet.java:92)
at org.hibernate.collection.internal.PersistentSet.equals(PersistentSet.java:441)
at ch.tiim.sco.database.model.Team .equals(Team.java:81)
在org.junit.Assert.isEqu als(Assert.java:131)
at org.junit.Assert.equalsRegardingNull(Assert.java:127)
at org.junit.Assert.assertEquals(Assert.java:111)
在org.junit.Assert.assertEquals(Assert.java:144)
at ch.tiim.sco.database.TableMemberTest.testEditTeam(TableMemberTest.java:45)
.updateTeam(Team t)
的代码如下:
public void updateTeam(Team t){
Session s = sessionFactory.getCurrentSession();
s.beginTransaction();
s.update(t);
s.getTransaction()。commit();
$ / code>
对于 .addTeam(Team t)
但是 s.save(t)
而不是 s.update(t)
团队看起来像这样:
@Entity
@ Table(name =team)
public class Team implements Model {
@Id
@GeneratedValue
@Column(name =team_id)
private int id ;
@Column(name =name)
私人字符串名称;
@ManyToOne
@JoinColumn(name =club_id)
私人俱乐部俱乐部;
@ManyToMany
@JoinTable(name =team_members,
joinColumns = {@JoinColumn(name =team_id)},
inverseJoinColumns = {@JoinColumn(name = member_id)}
)
private java.util.Set< Member>成员;
//我剥离了构造函数和setters / getters
}
现在我该如何摆脱这个错误,而不保证每次调用hibernate都来自同一个线程?
编辑这里是hibernate配置
<?xml version ='1.0'encoding ='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
- // Hibernate / Hibernate配置DTD 3.0 // EN
http://www.hibernate.org/dtd/hibernate-configuration -3.0.dtd>
< hibernate-configuration>
< session-factory>
< property name =connection.driver_class> org.h2.Driver< / property>
< property name =connection.username> user< / property>
< property name =connection.password> pass< / property>
< property name =connection.pool_size> 1< / property>
<! - SQL方言 - >
< property name =dialect> org.hibernate.dialect.H2Dialect< / property>
< property name =current_session_context_class>线程< / property>
< property name =cache.provider_class> org.hibernate.cache.internal.NoCacheProvider< / property>
< property name =hbm2ddl.auto>建立< / property>
< / session-factory>
< / hibernate-configuration>
对于Junit测试用例,您需要将打开的Session绑定到当前线程。
SessionFactory sf = null;
Session s = null;
SessionHolder持有者= null;
@Override
保护无效onSetUp()抛出异常{
System.out.println(On SetUP ----);
sf =(SessionFactory)beanFactory.getBean(sessionFactory);
s = sf.openSession();
TransactionSynchronizationManager.bindResource(sf,新的SessionHolder(s));
$ b $ @覆盖
保护无效onTearDown()抛出异常{
System.out.println(on onTearDown ----);
//取消绑定并关闭会话。
持有人=(SessionHolder)TransactionSynchronizationManager.getResource(sf);
s = holder.getSession();
s.flush();
TransactionSynchronizationManager.unbindResource(sf);
SessionFactoryUtils.releaseSession(s,sf);
//在此处拆除代码
}
如果您不想再小心关于LazyInitialisationException阅读,它很好地解释了会发生什么,以及如何解决这个问题。
I am porting a java fat client from JDBC SQLite to Hibernate H2.
Until now I tried to separate all my database code in separate classes, likeTableTeam or TableMember, which have methods like .getAllTeams()
or .updateTeam(Team t)
. These methods acted as a wrapper around their sql queries they executed.
Now with hibernate I tried to leave the interface as good as possible and just change the SQL queries to hibernate functions, which mostly works.With one exception: updating elements.
team = new Team(-1, "Team Name", Collections.<Member>emptySet());
@Test
public void testUpdateTeam() throws Exception {
table.addTeam(team);
team.setName("New Name");
table.updateTeam(team);
Team f = table.getAllTeams().get(0);
Assert.assertEquals(team, f);
}
results in the following exception:
org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: ch.tiim.sco.database.model.Team.members, could not initialize proxy - no Session
at org.hibernate.collection.internal.AbstractPersistentCollection.throwLazyInitializationException(AbstractPersistentCollection.java:576)
at org.hibernate.collection.internal.AbstractPersistentCollection.withTemporarySessionIfNeeded(AbstractPersistentCollection.java:215)
at org.hibernate.collection.internal.AbstractPersistentCollection.readSize(AbstractPersistentCollection.java:156)
at org.hibernate.collection.internal.PersistentSet.size(PersistentSet.java:160)
at java.util.AbstractSet.equals(AbstractSet.java:92)
at org.hibernate.collection.internal.PersistentSet.equals(PersistentSet.java:441)
at ch.tiim.sco.database.model.Team.equals(Team.java:81)
at org.junit.Assert.isEquals(Assert.java:131)
at org.junit.Assert.equalsRegardingNull(Assert.java:127)
at org.junit.Assert.assertEquals(Assert.java:111)
at org.junit.Assert.assertEquals(Assert.java:144)
at ch.tiim.sco.database.TableMemberTest.testEditTeam(TableMemberTest.java:45)
The code for .updateTeam(Team t)
is the following:
public void updateTeam(Team t) {
Session s = sessionFactory.getCurrentSession();
s.beginTransaction();
s.update(t);
s.getTransaction().commit();
}
And the same for .addTeam(Team t)
but s.save(t)
instead of s.update(t)
And Team looks like this:
@Entity
@Table(name = "team")
public class Team implements Model {
@Id
@GeneratedValue
@Column(name = "team_id")
private int id;
@Column(name = "name")
private String name;
@ManyToOne
@JoinColumn(name = "club_id")
private Club club;
@ManyToMany
@JoinTable(name = "team_members",
joinColumns = {@JoinColumn(name = "team_id")},
inverseJoinColumns = {@JoinColumn(name = "member_id")}
)
private java.util.Set<Member> members;
//I stripped the constructor and the setters/getters
}
Now how do I get rid of this error, without guaranteeing that every call to hibernate is from the same thread?
Edit Here is the hibernate configuration
<?xml version='1.0' encoding='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<!-- Database connection settings -->
<property name="connection.driver_class">org.h2.Driver</property>
<!--property name="connection.url">jdbc:h2:./test</property-->
<property name="connection.username">user</property>
<property name="connection.password">pass</property>
<!-- JDBC connection pool (use the built-in) -->
<property name="connection.pool_size">1</property>
<!-- SQL dialect -->
<property name="dialect">org.hibernate.dialect.H2Dialect</property>
<!-- Enable Hibernate's automatic session context management -->
<property name="current_session_context_class">thread</property>
<!-- Disable the second-level cache -->
<property name="cache.provider_class">org.hibernate.cache.internal.NoCacheProvider</property>
<!-- Echo all executed SQL to stdout -->
<!--property name="show_sql">true</property-->
<!-- Drop and re-create the database schema on startup -->
<property name="hbm2ddl.auto">create</property>
</session-factory>
</hibernate-configuration>
For Junit testcases you need to bind an opened Session to the current Thread.
SessionFactory sf = null;
Session s = null;
SessionHolder holder = null;
@Override
protected void onSetUp() throws Exception {
System.out.println("On SetUP----");
sf = (SessionFactory) beanFactory.getBean("sessionFactory");
s = sf.openSession();
TransactionSynchronizationManager.bindResource(sf, new SessionHolder(s));
}
@Override
protected void onTearDown() throws Exception {
System.out.println("On onTearDown----");
// unbind and close the session.
holder = (SessionHolder)TransactionSynchronizationManager.getResource(sf);
s = holder.getSession();
s.flush();
TransactionSynchronizationManager.unbindResource(sf);
SessionFactoryUtils.releaseSession(s, sf);
// teardown code here
}
If you want to no more care about LazyInitialisationException read this, its well explain what happens, and how to resolve the problem definitely
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