LazyInitializationException trying to get lazy initialized instance

问题

I see the following exception message in my IDE when I try to get lazy initialized entity (I can't find where it is stored in the proxy entity so I can't provide the whole stack trace for this exception):

Method threw 'org.hibernate.LazyInitializationException' exception. Cannot evaluate com.epam.spring.core.domain.UserAccount_$$_jvste6b_4.toString()

Here is a stack trace I get right after I try to access a field of the lazy initialized entity I want to use:

org.hibernate.LazyInitializationException: could not initialize proxy - no Session

    at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:165)

    at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:286)

    at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:185)

    at com.epam.spring.core.domain.UserAccount_$$_jvstfc9_4.getMoney(UserAccount_$$_jvstfc9_4.java)

    at com.epam.spring.core.web.rest.controller.BookingController.refill(BookingController.java:128) 

I'm using Spring Data, configured JpaTransactionManager, database is MySql, ORM provider is Hibernate 4. Annotation @EnableTransactionManagement is on, @Transactional was put everywhere I could imagine but nothing works.

Here is a relation:

@Entity
public class User extends DomainObject implements Serializable {

    ..

    @OneToOne(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    @JoinColumn(name = "user_fk")
    private UserAccount userAccount;

    ..

@Entity
public class UserAccount extends DomainObject {

    ..

    @OneToOne(mappedBy = "userAccount")
    private User user;

    ..

.. a piece of configuration:

    @Bean
    public DataSource dataSource() {
        DriverManagerDataSource dataSource = new DriverManagerDataSource();
        dataSource.setDriverClassName(env.getRequiredProperty(PROP_NAME_DATABASE_DRIVER));
        dataSource.setUrl(env.getRequiredProperty(PROP_NAME_DATABASE_URL));
        dataSource.setUsername(env.getRequiredProperty(PROP_NAME_DATABASE_USERNAME));
        dataSource.setPassword(env.getRequiredProperty(PROP_NAME_DATABASE_PASSWORD));
        return dataSource;
    }

    @Bean
    public LocalContainerEntityManagerFactoryBean entityManagerFactory() {
        LocalContainerEntityManagerFactoryBean entityManagerFactoryBean = new LocalContainerEntityManagerFactoryBean();
        entityManagerFactoryBean.setDataSource(dataSource());
        entityManagerFactoryBean.setPersistenceProviderClass(HibernatePersistenceProvider.class);
        entityManagerFactoryBean.setPackagesToScan(env.getRequiredProperty(PROP_ENTITYMANAGER_PACKAGES_TO_SCAN));
        entityManagerFactoryBean.setJpaProperties(getHibernateProperties());            
        return entityManagerFactoryBean;
     }

    @Bean
    public JpaTransactionManager transactionManager(@Autowired DataSource dataSource,
                                                    @Autowired EntityManagerFactory entityManagerFactory) {
        JpaTransactionManager jpaTransactionManager = new JpaTransactionManager();
        jpaTransactionManager.setEntityManagerFactory(entityManagerFactory);
        jpaTransactionManager.setDataSource(dataSource);

        return jpaTransactionManager;
    }

.. and this is how I want to retrieve UserAccount:

    @RequestMapping(...)
    @Transactional()
    public void refill(@RequestParam Long userId, @RequestParam Long amount) {
        User user = userService.getById(userId);
        UserAccount userAccount = user.getUserAccount();
        userAccount.setMoney(userAccount.getMoney() + amount);
    }

Hibernate version is 4.3.8.Final, Spring Data 1.3.4.RELEASE and MySql connector 5.1.29.

Please, ask me if something else is needed. Thank you in advance!

回答1:

Firstly, you should understand that the root of the problem is not a transaction. We have a transaction and a persistent context (session). With @Transactional annotation Spring creates a transaction and open persistent context. After method is invoked a persistent context becomes closed.

When you call a user.getUserAccount() you have a proxy class that wraps UserAccount (if you don't load UserAccount with User). So when a persistent context is closed, you have a LazyInitializationException during call of any method of UserAccount, for example toString().

@Transactional working only on the userService level, in your case. To get @Transactional work, it is not enough to put the @Transactional annotation on a method. You need to get an object of a class with the method from a Spring Context. So to update money you can use another service method, for example updateMoney(userId, amount).

If you want to use @Transactional on the controller method you need to get a controller from the Spring Context. And Spring should understand, that it should wrap every @Transactional method with a special method to open and close a persistent context. Other way is to use Session Per Request Anti pattern. You will need to add a special HTTP filter.

https://vladmihalcea.com/the-open-session-in-view-anti-pattern/

回答2:

As @v.ladynev briefly explained, your issue was that you wanted to initialize a lazy relation outside of the persistence context.

I wrote an article about this, you might find it helpful: http://blog.arnoldgalovics.com/2017/02/27/lazyinitializationexception-demystified/

回答3:

For quick solutions despite of performance issues use @transactional in your service Sample:

@Transactional
public TPage<ProjectDto> getAllPageable(Pageable pageable) {
    Page<Project> data = projectRepository.findAll(pageable);
    TPage<ProjectDto> response = new TPage<>();
    response.setStat(data, Arrays.asList(modelMapper.map(data.getContent(), ProjectDto[].class)));
    return response;
}

it will get user details for project manager in the second query. For more advanced solution, you should read the blog post in the @galovics answer.

来源：https://stackoverflow.com/questions/42632648/lazyinitializationexception-trying-to-get-lazy-initialized-instance