java - 如何在现有 Spring Hibernate 应用程序中包含 Spring JPA

Question

我正在对我的应用程序进行一项增强。这个应用程序是 Spring Hibernate 实现。但是我现在正在工作的增强是 Spring-JPA。在不影响现有实现的情况下，我需要包含 JPA下面是我的 DBconfiguration 类。

@Configuration
@EnableJpaRepositories(basePackages = {"packagename","package name" })
@EnableTransactionManagement
public class DataBaseConfiguration {

    @Value("${postgres.driver}")
    private String postgresDriver;
    @Value("${postgres.url}")
    private String postgresUrl;
    @Value("${postgres.username}")
    private String postgresUsername;
    @Value("${postgres.password}")
    private String postgresPassword;

    @Value("${hibernate.dialect}")
    private String hibernateDialect;
    @Value("${hibernate.dialect.property}")
    private String dialectProperty;
    @Value("${hibernate.show-sql}")
    private String hibernateShowSql;
    @Value("${boolean.true}")
    private String booleanTrue;

    **\\ existing hibernate flow**
    @Autowired
    @Bean(name = "name")
    public SessionFactory getSessionFactory(DataSource dataSource) {

        LocalSessionFactoryBuilder sessionBuilder = new LocalSessionFactoryBuilder(
                dataSource);

        sessionBuilder.addAnnotatedClasses(entity.class);
        sessionBuilder.setProperty(hibernateDialect, dialectProperty);
        sessionBuilder.setProperty(postgresDriver, postgresUrl);
        sessionBuilder.setProperty(hibernateShowSql, booleanTrue);

        return sessionBuilder.buildSessionFactory();
    }

        **\\ existing hibernate flow**
    @Bean(name = "name")
    public DataSource getDataSource() {
        BasicDataSource dataSource = new BasicDataSource();
        dataSource.setDriverClassName(postgresDriver);
        dataSource.setUrl(postgresUrl);
        dataSource.setUsername(postgresUsername);
        dataSource.setPassword(postgresPassword);
        return dataSource;
    }

    **\\ existing hibernate flow**
    @Autowired
    @Bean(name = "manager")
    public HibernateTransactionManager getTransactionManager(
            SessionFactory sessionFactory) {
        HibernateTransactionManager transactionManager = null;
        transactionManager = new HibernateTransactionManager(sessionFactory);
        return transactionManager;
    }
    **\\included newly for JPA**
    @Bean
    public org.springframework.orm.jpa.JpaTransactionManager transactionManager() {
        JpaTransactionManager jpaTransactionManager = new JpaTransactionManager();
        jpaTransactionManager.setDataSource(getDataSource());
        jpaTransactionManager
                .setJpaDialect(new org.springframework.orm.jpa.vendor.HibernateJpaDialect());
        return jpaTransactionManager;
    }
        **\\included newly for JPA**
    @Bean
    public EntityManager entityManger() {
        return entityManagerFactory().getObject().createEntityManager();
    }
        \\included newly for JPA
    @Bean
    public LocalContainerEntityManagerFactoryBean entityManagerFactory() {
        LocalContainerEntityManagerFactoryBean entityManagerFactory = new LocalContainerEntityManagerFactoryBean();
        entityManagerFactory.setDataSource(getDataSource());

        entityManagerFactory.setPackagesToScan(
                "some package",
                "some package");
        Map<String, Object> jpaProperty = entityManagerFactory
                .getJpaPropertyMap();
        jpaProperty.put("hibernate.dialect", "org.hibernate.dialect.PostgreSQLDialect");

        HibernateJpaVendorAdapter va = new HibernateJpaVendorAdapter();
        va.setShowSql(Boolean.valueOf("true"));
        entityManagerFactory.setJpaVendorAdapter(va);
        return entityManagerFactory;
    }
}

当我部署此应用程序时，我遇到了异常

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class ------
Caused by: java.lang.NullPointerException
    at org.hibernate.hql.internal.antlr.HqlBaseParser.identPrimary(HqlBaseParser.java:4285)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.primaryExpression(HqlBaseParser.java:980)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.atom(HqlBaseParser.java:3609)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.unaryExpression(HqlBaseParser.java:3387)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.multiplyExpression(HqlBaseParser.java:3259)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.additiveExpression(HqlBaseParser.java:2964)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.concatenation(HqlBaseParser.java:597)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.relationalExpression(HqlBaseParser.java:2730)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.equalityExpression(HqlBaseParser.java:2591)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.negatedExpression(HqlBaseParser.java:2555)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.logicalAndExpression(HqlBaseParser.java:2471)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.logicalOrExpression(HqlBaseParser.java:2436)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.expression(HqlBaseParser.java:2146)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.exprList(HqlBaseParser.java:3994)
    at org.hibernate.hql.internal.antlr.HqlBaseParser.identPrimary(HqlBaseParser.java:4278)

Answer 1

我建议另一种方法，它可能更简单。让 JPA 领先并使用 LocalContainerEntityManagerFactoryBean 来配置所有内容(包括您现在用于 hibernate 的内容)。接下来使用 HibernateJpaSessionFactoryBean 公开底层的 SessionFactory。

这将节省您的重复配置，并允许您为所有事务使用单个 JpaTransactionManager，但是您仍然可以使用 SessionFactory 或 >HibernateTemplate 如果你愿意的话。

@Configuration
@EnableJpaRepositories(basePackages = {"packagename","package name" })
@EnableTransactionManagement
public class DataBaseConfiguration {

    @Autowired
    private Environment env;

    @Autowired
    @Bean(name = "name")
    public FactoryBean<SessionFactory> getSessionFactory(EntityManagerFactory emf) {
        HibernateJpaSessionFactoryBean factory = new HibernateJpaSessionFactoryBean();
        factory.setEntityManagerFactory(emf);
        return factory;
    }

        **\\ existing hibernate flow**
    @Bean(name = "name")
    public DataSource getDataSource() {
        BasicDataSource dataSource = new BasicDataSource();
        dataSource.setDriverClassName(env.getRequiredProperty("postgres.driver"));
        dataSource.setUrl(env.getRequiredProperty("postgres.url);
        dataSource.setUsername(env.getRequiredProperty("postgres.username");
        dataSource.setPassword(env.getRequiredProperty("postgres.password");
        return dataSource;
    }

    @Bean
    public JpaTransactionManager transactionManager() {
        JpaTransactionManager jpaTransactionManager = new JpaTransactionManager();
        jpaTransactionManager.setDataSource(getDataSource());
        jpaTransactionManager.setEntityManagerFactory(entityManagerFactory());
        return jpaTransactionManager;
    }

    @Bean
    public LocalContainerEntityManagerFactoryBean entityManagerFactory() {
        LocalContainerEntityManagerFactoryBean entityManagerFactory = new LocalContainerEntityManagerFactoryBean();
        entityManagerFactory.setDataSource(getDataSource());
        Map<String, String> properties = new HashMap<String, String>();
        properties.put("hibernate.current_session_context_class", SpringSessionContext.class.getName()); 

        entitytManagerFactory.setJpaPropertyMap(properties);
        entityManagerFactory.setPackagesToScan(
                "some package",
                "some package");

        HibernateJpaVendorAdapter va = new HibernateJpaVendorAdapter();
        va.setShowSql(env.getProperty("hibernate.show-sql", Boolean.class, true);
        va.setDatabasePlatform(env.getProperty("hibernate.dialect", "org.hibernate.dialect.PostgreSQLDialect");
        entityManagerFactory.setJpaVendorAdapter(va);
        return entityManagerFactory;
    }
}