CODEWITHSUNDEEP
javaspringhibernateinterceptor
Lucas Noetzold
Lucas Noetzold

Reputation: 1740

Programmatically adding hibernate interceptor in JpaProperties - Spring

I'm writing a library with spring boot, and I need to programmatically insert a hibernate interceptor through it (because I can't use a .properties in a lib).

I want to avoid providing my own sessionFactory bean, I think it would be good to leave that possibility to an implementing project, also saves me from manually scanning for entities.

My dumb idea was that I could "inject" my interceptor into the JpaProperties. That didn't work at all, it ran the @PostConstruct but nothing changed. I had a feeling this wouldn't work, but I would like to understand why, and how I may get this to work.

@Autowired private JpaProperties properties;
@Autowired private MyInterceptor myInterceptor; //yep a bean

@PostConstruct public void add() {
    ((Map) properties.getProperties())
            .put(
                    "hibernate.session_factory.interceptor",
                    myInterceptor
            );
}

Upvotes: 2

Views: 1463

Answers (1)

Ryan Garner
Ryan Garner

Reputation: 46

As this is using an @PostConstruct annotation, the addition to the JpaProperties will only occur after the EntityManagerFactoryBuilder has been created in JpaBaseConfiguration. This means that changes to the property map will not be present in the builder after this point.

To customize the JpaProperties, you should instantiate a bean which adds your configuration in, like:

    @Primary
    @Bean
    public JpaProperties jpaProperties() {
        JpaProperties properties = new JpaProperties();
        properties.getProperties().put("hibernate.session_factory.interceptor", myInterceptor);
        return properties;
    }

This will then be injected into HibernateJpaConfiguration and used when constructing the EntityManagerFactoryBuilder.

Upvotes: 1

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