How to discover fully qualified table column from Hibernate MetadataSources

Question

I have an entity, for which I have a Class<MyEntity> reference:

@Entity
class MyEntity {
    @Id int id;
    @Column String col1;
    @Column(name = "abc") String col2;
}

I'm currently using Hibernate to export my entities into an in-memory database as such:

MetadataSources metadata = new MetadataSources(...);
metadata.addAnnotatedClass(MyEntity.class);
SchemaExport export = new SchemaExport();
export.create(EnumSet.of(TargetType.DATABASE), metadata.buildMetadata());

Details about the Hibernate-specific API here.

Is there any reliable way to get a mapping from MyEntity.col2 (the annotated Java field reference) to the fully qualified column name in the database (and vice versa) through Hibernate API? I'd like to avoid re-implementing all the delicate details of how Java identifiers (including getters and setters) are mapped to SQL identifiers in the absence of explicit qualification.

Answer 1

The org.hibernate.boot.Metadata is what we are interested in since it contains the PersistentClass entity bindings.

First, you need to create an Integrator which will give you access to Metadata:

public class MetadataExtractorIntegrator 
    implements org.hibernate.integrator.spi.Integrator {
 
    public static final MetadataExtractorIntegrator INSTANCE = 
        new MetadataExtractorIntegrator();
 
    private Database database;
 
    private Metadata metadata;
 
    public Database getDatabase() {
        return database;
    }
 
    public Metadata getMetadata() {
        return metadata;
    }
 
    @Override
    public void integrate(
            Metadata metadata,
            SessionFactoryImplementor sessionFactory,
            SessionFactoryServiceRegistry serviceRegistry) {
 
        this.database = metadata.getDatabase();
        this.metadata = metadata;
 
    }
 
    @Override
    public void disintegrate(
        SessionFactoryImplementor sessionFactory,
        SessionFactoryServiceRegistry serviceRegistry) {
 
    }
}

If you use JPA, you can register it as follows:

Map<String, Object> configuration = new HashMap<>();
 
Integrator integrator = integrator();
if (integrator != null) {
    configuration.put("hibernate.integrator_provider", 
        (IntegratorProvider) () -> Collections.singletonList(
            MetadataExtractorIntegrator.INSTANCE
        )
    );
}
 
EntityManagerFactory entityManagerFactory = new EntityManagerFactoryBuilderImpl(
    new PersistenceUnitInfoDescriptor(persistenceUnitInfo), 
    configuration
)
.build();

Now, when running the following test case:

Metadata metadata = MetadataExtractorIntegrator.INSTANCE.getMetadata();

for ( PersistentClass persistentClass : metadata.getEntityBindings()) {
 
    Table table = persistentClass.getTable();
     
    LOGGER.info( "Entity: {} is mapped to table: {}",
                 persistentClass.getClassName(),
                 table.getName()
    );
 
    for(Iterator propertyIterator = persistentClass.getPropertyIterator(); 
            propertyIterator.hasNext(); ) {
        Property property = (Property) propertyIterator.next();
         
        for(Iterator columnIterator = property.getColumnIterator(); 
                columnIterator.hasNext(); ) {
            Column column = (Column) columnIterator.next();
             
            LOGGER.info( "Property: {} is mapped on table column: {} of type: {}",
                         property.getName(),
                         column.getName(),
                         column.getSqlType()
            );
        }
    }
}

Against on the following entities:

We get the following output:

Entity: com.vladmihalcea.book.hpjp.util.providers.entity.BlogEntityProvider$Tag is mapped to table: tag
Property: name is mapped on table column: name of type: varchar(255)
Property: version is mapped on table column: version of type: integer
 
Entity: com.vladmihalcea.book.hpjp.util.providers.entity.BlogEntityProvider$PostComment is mapped to table: post_comment
Property: post is mapped on table column: post_id of type: bigint
Property: review is mapped on table column: review of type: varchar(255)
Property: version is mapped on table column: version of type: integer
 
Entity: com.vladmihalcea.book.hpjp.util.providers.entity.BlogEntityProvider$Post is mapped to table: post
Property: title is mapped on table column: title of type: varchar(255)
Property: version is mapped on table column: version of type: integer
 
Entity: com.vladmihalcea.book.hpjp.util.providers.entity.BlogEntityProvider$PostDetails is mapped to table: post_details
Property: createdBy is mapped on table column: created_by of type: varchar(255)
Property: createdOn is mapped on table column: created_on of type: datetime(6)
Property: version is mapped on table column: version of type: integer

Cool, right?

You can check out this example on GitHub as well.