在存储库中保存实体不起作用SPRING

Question

I'm trying to save entity in repository but it does not work at all. Repository is Autowired and in runtime I use saveAndFlush to save entity. I'm using PostgreSQL. Above test methods I added comments with explanation what is going on. I expected that method saveAndFlush should work but it did not. I can not find why.

@Transactional
public class TestClass{

    @Autowired private MyRepository repository;
    @Autowired private EntityManager entityManager;

    // Working version
    public void writingToRepositoryWorking() {
        entityManager.getTransaction().begin();
        entityManager.persist(new MyData(99));
        entityManager.getTransaction().commit();

    }

    // not working and throws exception : 
    // TransactionRequiredException: no transaction is in progress
    public void writingToRepositoryNotWorking() {
        repository.saveAndFlush(new MyData(99));
    }

    // not working, no exception, no data in repository, 
    // but auto generated ID is incremented
    public void writingToRepositoryNotWorkingToo() {
        repository.save(new MyData(99));
    }
}

repository interface file

@Repository
@Transactional
public interface MyRepository extends JpaRepository<MyData, Long> {}

MyData file

@Entity(name = "myData")
public class MyData {
    @Id @GeneratedValue(strategy = GenerationType.AUTO) long id;

    private int testValue;

    public MyData() { }

    public BugData(int testValue) {
        this.testValue = testValue;
    }

    public long getId() {
        return id;
    }

    public int getTestValue() {
        return testValue;
    }
}

ApplicationConfiguration file

@Configuration
@EnableJpaRepositories("com.mypackage.app")
@EnableTransactionManagement
@PropertySource("classpath:application.properties")
@EnableWebMvc
class ApplicationConfiguration extends WebMvcConfigurationSupport {

    @Value("${jdbc.url}") private String KEY_JDBC_URL;

    @Value("${jdbc.username}") private String KEY_JDBC_USERNAME;

    @Value("${jdbc.password}") private String KEY_JDBC_PASSWORD;

    @Bean
    public static PropertySourcesPlaceholderConfigurer propertySourcesPlaceholderConfigurer() {
        return new PropertySourcesPlaceholderConfigurer();
    }

    @Bean
    @Autowired
    public LocalSessionFactoryBean sessionFactory(DataSource dataSource) {
        LocalSessionFactoryBean factory = new LocalSessionFactoryBean();
        factory.setDataSource(dataSource);
        factory.setPackagesToScan("com.mypackage.app");
        factory.setHibernateProperties(hibernateProperties());
        return factory;
    }

    public Properties hibernateProperties() {
        Properties properties = new Properties();
        properties.setProperty("hibernate.dialect", "org.hibernate.dialect.PostgreSQLDialect");
        properties.setProperty("hibernate.show_sql", "true");
        properties.setProperty("hibernate.hbm2ddl.auto", "update");
        return properties;
    }

    @Bean
    @Autowired
    public HibernateTransactionManager transactionManager(SessionFactory sessionFactory) {
        return new HibernateTransactionManager(sessionFactory);
    }

    @Bean
    public DataSource dataSource() {
        BasicDataSource dataSource = new BasicDataSource();
        dataSource.setDriverClassName("org.postgresql.Driver");
        dataSource.setUrl(KEY_JDBC_URL);
        dataSource.setUsername(KEY_JDBC_USERNAME);
        dataSource.setPassword(KEY_JDBC_PASSWORD);
        return dataSource;
    }

    @Bean
    public EntityManagerFactory  entityManagerFactory() {
        LocalContainerEntityManagerFactoryBean em = new LocalContainerEntityManagerFactoryBean();

        em.setDataSource(dataSource());
        em.setPackagesToScan("com.mypackage.app");
        em.setJpaVendorAdapter(new HibernateJpaVendorAdapter());
        em.setJpaProperties(hibernateProperties());
        em.afterPropertiesSet();

        return em.getObject();
    }

    @Bean
    public EntityManager entityManager(EntityManagerFactory entityManagerFactory) {
        return entityManagerFactory.createEntityManager();
    }

    ...
}

Answer 1

For starter, you're actually working on 2 different EntityManager in your non-working test case:

1. EntityManager autowired into your test by Spring (this one is singleton and should be avoided anyway) ,other is
2. EntityManager created by the EntityManagerFactory configured in your ApplicationConfiguration.

At the same time, you also have another Session running along side the aforementioned 2 EntityManagers due to your configuration of Hibernate SessionFactory . Additionally, because of the configured HibernateTransactionManager , all transactions created by @Transactional are bound to the Hibernate's Session created by SessionFactory and the EntityManager used by your Repository certainly has no way to know about it. This is why TransactionRequiredException was thrown when your Repository tried to persist data.

To fix it, you may consider removing the Hibernate's SessionFactory and switch the transaction manager to a JpaTransactionManager . Then, @Transactional on your Repository will have the effect of creating a new transaction and binding it to the existing EntityManager that is known to Spring.

One side note is that the @Transactional on your TestClass doesn't help at all as the instance of this class is not instantiated and managed by Spring. To make this work, a proper configuration of transactional test class needs to be provided as described here: http://docs.spring.io/spring/docs/current/spring-framework-reference/html/testing.html .

Hope this helps.