Issue
I have the following scenario. I am using JPA, Spring:
@Autowired
SampleService service;
@Transactional(propagation = Propagation.REQUIRED, rollbackFor = Exception.class)
public void PerformLogic(LogicData data) throws SIASFaultMessage
{
SampleObject so = createSampleObject();
try{
.//do some logic to persist things in data
.
.
persistData(data);
.
.
.
updateSampleObject(so);
}
catch(Exception){
updateSampleObject(so);
throw new SIASFaultMessage();
}
}
@Transactional(propagation = Propagation.REQUIRES_NEW)
public createSampleObject()
{
SampleObject so = new SampleObject();
.
.//initialize so
.
service.persist(so);
return so;
}
@Transactional(propagation = Propagation.REQUIRES_NEW)
public updateSampleObject(SampleObject so)
{
service.persist(so);
return so;
}
When everything works fine, data is persisted in database without problems. However, when an exception is thrown, I need that the method updateSampleObject(so) persist the information in the database. This is not what is happening. If an exception is thrown the method updateSampleObject gets rolled back also, which is not what I need. I need that these two methods (createSampleObject and updateSampleObject) get persisted all the time, no matter whether an exception got thrown or not. How can I achieve this?
Moreover, if I anotate the methods createSampleObject and updateSampleObject with:
@Transactional(propagation = Propagation.NEVER)
the idea is that an exception is thrown and I get no exception thrown. Where is the problem? Analizing the logs I see this line:
org.springframework.orm.jpa.JpaTransactionManager ==> Creating new transaction with name [com.test.PerformLogic]: PROPAGATION_REQUIRED,ISOLATION_DEFAULT....
which means this transaction is created, but I see no hint of the other transaction.
This is the part of my configuration file for Spring regarding transactions
<bean id="myDataSource"
class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="${jdbc.driverClassName}"/>
<property name="url" value="${jdbc.url}"/>
<property name="username" value="${jdbc.username}"/>
<property name="password" value="${jdbc.password}"/>
</bean>
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="myDataSource"/>
<property name="packagesToScan" value="cu.jpa"/>
<property name="persistenceProviderClass" value="org.hibernate.ejb.HibernatePersistence"/>
<property name="jpaProperties">
<props>
<prop key="hibernate.dialect">${hibernate.dialect}</prop>
<prop key="hibernate.show_sql">true</prop>
<prop key="hibernate.hbm2ddl.auto">${hdm2ddl.auto}</prop>
</props>
</property>
<property value="/META-INF/jpa-persistence.xml" name="persistenceXmlLocation"/>
<property name="persistenceUnitName" value="jpaPersistenceUnit"/>
</bean>
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory"/>
<property name="nestedTransactionAllowed" value="true" />
</bean>
<tx:annotation-driven transaction-manager="transactionManager"/>
Solution
Spring transactions are proxy-based. Here's thus how it works when bean A causes a transactional of bean B. A has in fact a reference to a proxy, which delegates to the bean B. This proxy is the one which starts and commits/rollbacks the transaction:
A ---> proxy ---> B
In your code, a transactional method of A calls another transactional method of A. So Spring can't intercept the call and start a new transaction. It's a regular method call without any proxy involved.
So, if you want a new transaction to start, the method createSampleObject()
should be in another bean, injected into your current bean.
This is explained with more details in the documentation.
Answered By - JB Nizet
Answer Checked By - Dawn Plyler (JavaFixing Volunteer)