9

I'm using spring boot and it perfectly makes me entity manager. And I decided to test getting session factory from the entity manager and to use it for an example. But I get the next problem:javax.persistence.TransactionRequiredException: no transaction is in progress

properties 

spring.datasource.url= jdbc:postgresql://localhost:5432/ring
spring.datasource.username=postgres
spring.datasource.password=root

spring.jpa.show-sql = false
spring.jpa.properties.hibernate.format_sql=false

#Note: The last two properties on the code snippet above were added to suppress an annoying exception
# that occurs when JPA (Hibernate) tries to verify PostgreSQL CLOB feature.
spring.jpa.database-platform=org.hibernate.dialect.PostgreSQL9Dialect
spring.jpa.properties.hibernate.temp.use_jdbc_metadata_defaults = false

spring.jpa.properties.hibernate.current_session_context_class = org.springframework.orm.hibernate5.SpringSessionContext

service class

package kz.training.springrest.service;

import kz.training.springrest.entity.User;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.springframework.stereotype.Service;
import org.springframework.transaction.annotation.Transactional;

import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.PersistenceContext;
import javax.persistence.PersistenceUnit;

@Service
public class UserService {

    @PersistenceContext
    private EntityManager entityManager;

    @Transactional
    public void insertUser(User user) {
        SessionFactory sessionFactory = entityManager.unwrap(Session.class).getSessionFactory();
        Session session = sessionFactory.getCurrentSession();
        session.save(user);
    }

}

runner

package kz.training.springrest.run;

import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.autoconfigure.domain.EntityScan;
import org.springframework.context.annotation.ComponentScan;
import org.springframework.data.jpa.repository.config.EnableJpaAuditing;
import org.springframework.transaction.annotation.EnableTransactionManagement;

@SpringBootApplication
@EntityScan("kz.training.springrest.entity")
@EnableTransactionManagement
@ComponentScan(basePackages="kz.training.springrest")
public class SpringrestApplication {

    public static void main(String[] args) {
        SpringApplication.run(SpringrestApplication.class, args);
    }
}

Do you have any ideas how to solve it?

Daimon
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  • You should not write dao code in service class. You should create a @Repository interface or class and perform dao operations there. – Shubham Kadlag Jun 01 '18 at 21:07
  • Should @EnableTransactionManagement be on a class also annotated with @Configuration? – gagan singh Jun 01 '18 at 22:00
  • I know that this code must be in @repository, but it doesn't solve my problem – Daimon Jun 02 '18 at 03:18
  • @EnableTransactionManagement can also be on spring boot runner – Daimon Jun 02 '18 at 03:19
  • @ShubhamKadlag That's honestly highly subjective statement. A lot of people view the JPA abstraction as _the DAO_ and simply use it inside their services, particularly if the application never intends to work with anything but JPA. To me adding a `@Repository` here is just another layer of abstraction atop of an existing abstraction which effectively just adds extra function calls to the stack with little to no meaning or reward. – Naros Jun 02 '18 at 04:40

3 Answers3

19

I don't quite understand why you're making your service method so unnecessarily complex. You should simply be able to do it this way

@Transactional
public void insertUser(User user) {
  entityManager.persist( user );
}

If there are points where you need access to the native Hibernate Session you can simply unwrap and use the Session directly like this:

@Transactional
public void doSomethingFancyWithASession() {
  Session session = entityManager.unwrap( Session.class );
  // use session as needed
}

The notion here is that Spring provides you an already functional EntityManager instance by you using the @PersistenceContext annotation. That instance will safely be usable by the current thread your spring bean is being executed within.

Secondly, by using @Transactional, this causes Spring's transaction management to automatically make sure that the EntityManager is bound to a transaction, whether that is a RESOURCE_LOCAL or JTA transaction is based on your environment configuration.

You're running into your problem because of the call to #getCurrentSession().

What is happening is Spring creates the EntityManager, then inside your method when you make the call to #getCurrentSession(), you're asking Hibernate to create a second session that is not bound to the transaction started by your @Transactional annotation. In short its essentially akin to the following:

EntityManager entityManager = entityManagerFactory.createEntityManager();
entityManager.getTransaction().begin();
Session aNewSession = entityManager.unwrap( Session.class )
  .getFactory()
  .getCurrentSession();
// at this point entityManager is scoped to a transaction
// aNewSession is not scoped to any transaction
// this also likely uses 2 connections to the database which is a waste

So follow the paradigm I mention above and you should no longer run into the problem. You should never need to call #getCurrentSession() or #openSession() in a Spring environment if you're properly allowing Spring to inject your EntityManager instance for you.

Naros
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0

I have same your error, when I deploy my spring boot app to WebLogic Server. (Even It works fine if I run it directly via Eclipse (Or deploy to Tomcat) ).

I solved the problem by adding @EnableTransactionManagement to UserService.

searching9x
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0

salved my problem using in hibernate conf "hibernate.allow_update_outside_transaction","true"

Ashish Rajput
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