How can I translate JPA xml configuration into a java class?

Question

I want to a configure JPA in a java class (not using the usual persistence.xml approach)

Here is an example of what I'm doing:

public SessionFactory getSessionFactory(){
    Configuration configuration=new Configuration();

    Properties settings= new Properties();

    settings.put(Environment.DRIVER, "");
    settings.put(Environment.URL, "");
    settings.put(Environment.USER, "");
    settings.put(Environment.PASS, "");
    settings.put(Environment.DIALECT, "");
    settings.put(Environment.SHOW_SQL, "true");
    settings.put(Environment.DEFAULT_SCHEMA, "");
    configuration.setProperties(settings);
     configuration.addAnnotatedClass(Passenger.class);  
     ServiceRegistry serviceRegistry=new StandardServiceRegistryBuilder()
             .applySettings(configuration.getProperties()).build();
     sessionFactory=configuration.buildSessionFactory(serviceRegistry);

}

You can follow https://dzone.com/articles/hibernate-5-java-configuration-example this article. — kamlesh pandey, Feb 26 '19 at 09:45
Possible duplicate of [JPA without persistence.xml](https://stackoverflow.com/questions/30905908/jpa-without-persistence-xml) — Peter Šály, Feb 26 '19 at 09:46
To state the blindingly obvious, what you post IS NOT THE JPA API. Neither is the DZone article listed above. If you really mean HIBERNATE API then kindly state it and remove the JPA tag. — , Feb 26 '19 at 11:19

score 0 · Answer 1 · answered Feb 26 '19 at 09:30

If you cannot (or don't want) use the standard XML configuration, you can bootstrap JPA via a custom implementation. Here is an example. If you provide more info about your desired result (dialect, persistence unit name etc.) I can give you a more detailed example.

How can I translate JPA xml configuration into a java class?

1 Answers1