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我正在研究一个小的Spring-Hibernate-Mysql测试项目，并且由于某种原因，我的事务没有被提交给数据库。JPA事务没有提交给DB
在我的应用上下文我：
<!-- JTA -->
<bean id="txManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<tx:annotation-driven transaction-manager="txManager" />
<!-- JPA -->
<jee:jndi-lookup id="entityManagerFactory" jndi-name="persistence/myPU" />
<!-- In order to enable EntityManager injection -->
<bean
class="org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor">
<property name="persistenceUnits">
<map>
<entry key="myPU" value="persistence/myPU" />
</map>
</property>
</bean>
我的persistence.xml：
<persistence-unit name="myPU" transaction-type="JTA">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<jta-data-source>jdbc/mysqlResource</jta-data-source>
<properties>
<property name="hibernate.connection.shutdown" value="true" />
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLInnoDBDialect"></property>
<property name="hibernate.show_sql" value="true" />
</properties>
</persistence-unit>
我创建了一个名为 '人' 在我的分贝简单的表：
CREATE TABLE persons(
id VARCHAR(255) PRIMARY KEY,
version int,
full_name VARCHAR(255),
person_id VARCHAR(255),
email VARCHAR(255));
创建实体一个相应的实体和道：
@Entity
@Table(name = "persons")
public class Person implements Serializable {
private static final long serialVersionUID = 4349832844316517922L;
/*--- Members ---*/
/**
* Hibernate genetared UUID
*/
@Id
@GeneratedValue(generator = "system-uuid")
@GenericGenerator(name = "system-uuid", strategy = "uuid")
private String id;
@Version
private int version;
@Column(name = "full_name")
private String fullName;
@Column(name = "person_id")
private String personId;
@Column(name = "email")
private String eMail;
/*--- Constructor ---*/
public Person() {
}
/*--- Overridden Methods ---*/
@Override
public boolean equals(Object obj) {
if ((obj == null) || !(obj instanceof Person)) {
return false;
}
// reference comparison
if (obj == this) {
return true;
}
final Person other = (Person) obj;
return new EqualsBuilder().append(getPersonId(), other.getPersonId())
.append(geteMail(), other.geteMail())
.append(getFullName(), other.getFullName()).isEquals();
}
/**
* The unique hash code based on the clients' id and citizenship
*
* {@inheritDoc}
*/
@Override
public int hashCode() {
return new HashCodeBuilder().append(geteMail()).append(this.geteMail())
.append(this.getFullName()).toHashCode();
}
/*--- Getters & Setters ---*/
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public int getVersion() {
return version;
}
public void setVersion(int version) {
this.version = version;
}
public String getFullName() {
return fullName;
}
public void setFullName(String fullName) {
this.fullName = fullName;
}
public String getPersonId() {
return personId;
}
public void setPersonId(String personId) {
this.personId = personId;
}
public String geteMail() {
return eMail;
}
public void seteMail(String eMail) {
this.eMail = eMail;
}
}
道：
@Repository
public class PersonJpaDao extends BasicJpaDao<Person> implements IPersonDao {
public PersonJpaDao() {
super(Person.class);
}
}
这里是BasicJpaDao：
public class BasicJpaDao<T> implements IBasicDao<T> {
/* --- Members --- */
/** The JPA utility to work with the persistence layer. */
@PersistenceContext
protected EntityManager entityManager;
/** The type of the entity to which this DAO offers access. */
protected Class<T> entityClass;
/* --- Constructors --- */
/**
* Default constructor.
*
* @param entityClass
* The type of the entity to which this DAO offers access.
*/
public BasicJpaDao(Class<T> entityClass) {
super();
this.entityClass = entityClass;
}
/* --- Public methods --- */
/**
* {@inheritDoc}
*/
@Override
public void create(T entity) {
getEntityManager().persist(entity);
}
/**
* {@inheritDoc}
*/
@Override
public T read(Object primaryKey) {
return getEntityManager().find(getEntityClass(), primaryKey);
}
/**
* {@inheritDoc}
*/
@Override
public T update(T entity) {
return getEntityManager().merge(entity);
}
/**
* {@inheritDoc}
*/
@Override
public void delete(T entity) {
getEntityManager().remove(entity);
}
/**
* {@inheritDoc}
*/
@Override
public void flush() {
getEntityManager().flush();
}
/* --- Getters/Setters --- */
/**
* @return The JPA utility to work with the persistence layer.
*/
public EntityManager getEntityManager() {
return this.entityManager;
}
/**
* @param entityManager
* The JPA utility to work with the persistence layer.
*/
public void setEntityManager(EntityManager entityManager) {
this.entityManager = entityManager;
}
/**
* @return The type of the entity to which this DAO offers access.
*/
public Class<T> getEntityClass() {
return entityClass;
}
/**
* @param entityClass
* The type of the entity to which this DAO offers access.
*/
public void setEntityClass(Class<T> entityClass) {
this.entityClass = entityClass;
}
洙。基本上这工作，但没有被提交，我的意思是，如果我运行
@Transactional(propagation = Propagation.REQUIRED)
private void crearePerson() {
Person p1 = myDao.read("12345");
p1.setFullName("kiko too");
myDao.update(p1);
}
我可以看到（在调试中）p1从数据库中取回，但更新从不发生。 我能找到的唯一的东西接近的是：
JPA - transactions not being committed
我尝试添加
我的persistence.xml这个线程以下，但它并没有帮助。 我还向我的连接池（在我的应用程序服务器gui中）添加了一个属性，名为connection.shutdown，值为true，但它也没有帮助。
更新： 由于我使用JTA我想我的事务管理器是错误配置的。当我使用org.springframework.orm.jpa.JpaTransactionManager时，我应该使用org.springframework.transaction.jta.JtaTransactionManager。 所以我改变了我的应用程序上下文，我现在有：
<bean id="txManager"
class="org.springframework.transaction.jta.JtaTransactionManager" />
<tx:annotation-driven transaction-manager="txManager" />
不幸的是，我仍然遇到了同样的问题:( 在我的控制台我可以看到Hibernate查询如下（）我已经改变一些我原来的实体字段，但它并不重要）：
INFO：Hibernate：select user0_.id as id0_0_，user0_.email as email0_0_，user0_.full_name as full3_0_0_，user0_.password as password0_0_，user0_ .update_by_email as update5_0_0_，user0_.user_name as user6_0_0_，user0_.version as version0_0_ from users user0_ where user0_.id =？
任何想法？
由于提前， 瑜珈
嗨，谢谢你。我在交易中运行，只是编辑了我的问题。仍然无法看到我的表在运行代码后得到更新，我确信同意 - @Transactional注释通常应该在业务逻辑层 – forhas