How to implement a custom String sequence identifier generator with Hibernate

Question

I\'m using hibernate with spring, h2 and liquibase and I\'m trying to make a custom String id generator for my entities by taking example with this blog post but I\'m gettin

Answer 1

The problem is that SequenceStyleGenerator expects to return a numerical value, not a String.

I already tried a solution for this problem and it works like a charm. Therefore, you need to change your generator like this:

public class StringSequenceIdentifier implements IdentifierGenerator, Configurable {

    private String sequenceCallSyntax;

    @Override
    public void configure(Type type, Properties params, ServiceRegistry serviceRegistry) throws MappingException {
        final JdbcEnvironment jdbcEnvironment = serviceRegistry.getService(JdbcEnvironment.class);
        final Dialect dialect = jdbcEnvironment.getDialect();

        final String sequencePerEntitySuffix = ConfigurationHelper.getString(CONFIG_SEQUENCE_PER_ENTITY_SUFFIX, params, DEF_SEQUENCE_SUFFIX);

        final String defaultSequenceName = ConfigurationHelper.getBoolean(CONFIG_PREFER_SEQUENCE_PER_ENTITY, params, false)
                ? params.getProperty(JPA_ENTITY_NAME) + sequencePerEntitySuffix
                : DEF_SEQUENCE_NAME;

        sequenceCallSyntax = dialect.getSequenceNextValString(ConfigurationHelper.getString(SEQUENCE_PARAM, params, defaultSequenceName));
    }

    @Override
    public Serializable generate(SessionImplementor session, Object obj) {
        if (obj instanceof Identifiable) {
            Identifiable identifiable = (Identifiable) obj;
            Serializable id = identifiable.getId();
            if (id != null) {
                return id;
            }
        }
        long seqValue = ((Number) Session.class.cast(session)
            .createSQLQuery(sequenceCallSyntax)
            .uniqueResult()).longValue();

        return "CTC" + seqValue;
    }
}

Your mapping becomes:

@Entity(name = "Post")
@Table(name = "post")
public static class Post implements Identifiable<String> {

    @Id
    @GenericGenerator(
        name = "assigned-sequence",
        strategy = "com.vladmihalcea.book.hpjp.hibernate.identifier.StringSequenceIdentifier",
        parameters = @org.hibernate.annotations.Parameter(name = "sequence_name", value = "hibernate_sequence")
    )
    @GeneratedValue(generator = "assigned-sequence", strategy = GenerationType.SEQUENCE)
    private String id;

    @Version
    private Integer version;

    public Post() {
    }

    public Post(String id) {
        this.id = id;
    }

    @Override
    public String getId() {
        return id;
    }
}

Now, when you insert the following entities:

doInJPA(entityManager -> {
    entityManager.persist(new Post());
    entityManager.persist(new Post("ABC"));
    entityManager.persist(new Post());
    entityManager.persist(new Post("DEF"));
});

Hibernate generates the right identifier:

Query:["select nextval ('hibernate_sequence')"], Params:[()]
Query:["select nextval ('hibernate_sequence')"], Params:[()]
Query:["insert into post (version, id) values (?, ?)"], Params:[(0, CTC1)]
Query:["insert into post (version, id) values (?, ?)"], Params:[(0, ABC)]
Query:["insert into post (version, id) values (?, ?)"], Params:[(0, CTC2)]
Query:["insert into post (version, id) values (?, ?)"], Params:[(0, DEF)]

Code available on GitHub.

Answer 2

Yes, hibernate have prebuilt String generators. Just substitute your @GenericGenerator definition to another strategy.

@Entity
@Table(name = "contact")
public class Contact implements Serializable, Identifiable<String> {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(generator = "uuid")
    @GenericGenerator(name = "uuid", strategy = "uuid2")
    private String id;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }
}

For more information about different hibernate generators you can look at documentation.