No bean named 'transactionManager' is defined

Question

I have configured two persistent units with the entity managers set up as show below:

Answer 1

Actually, there is a way to use named TransactionManager with Spring Data JPA. This works for me:

<bean id="myTransactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
    <property name="entityManagerFactory" ref="myEntityManagerFactory" />
</bean>
<tx:annotation-driven transaction-manager="myTransactionManager"/>

<jpa:repositories base-package="com.xxx.yyy" entity-manager-factory-ref="myEntityManagerFactory" transaction-manager-ref="myTransactionManager">
</jpa:repositories>

Answer 2

I use java configuration and specifying the transactionManagerRef was solution for me:

@Configuration
@EnableTransactionManagement
@EnableJpaRepositories(
    entityManagerFactoryRef = "myCustomEntityManagerFactory",
    basePackages = {"ua.demitt.other.path.to.repos"},
    transactionManagerRef = "myCustomTransactionManager" )

Answer 3

The default value for the transaction-manager attribute is transaction-manager. In your case, you should specify which transaction manager you want to use per method or service like this:

@Service
@Transactional(value="LiveTransactionManager") 
class someClass...

or

@Transactional(value="ArchiveTransactionManager") 
public void someMethod

Answer 4

In your @Configuration file, if this is your dataSource:

@Bean(name = "dataSource")
public DataSource getDataSource() {
    return DataSourceBuilder
            .create()
            .username(username)
            .password(password)
            .url(url)
            .build();
}

this would be your transaction manager bean:

@Bean(name = "DataSourceTransactionManager")
public DataSourceTransactionManager getDataSourceTransactionManager() {
    return new DataSourceTransactionManager(getDataSource());
}

Then annotate the service method you would like to be transactional:

@Transactional(transactionManager = "DataSourceTransactionManager", timeout = 60, rollbackFor = { Exception.class})