How to generate Custom Id in JPA

Question

i want generate Custom Id in JPA it must be primary key of table. there are many examples to create Custom Id using hibernate like this i want same implementation but in JPA.The id must be alphanumeric like STAND0001

Thanks.

There is no such thing as spring JPA. There is JPA, the specification, and some implementations of JPA, like Hibernate or EclipseLink. Spring doesn't have, and is not a JPA implementation. So check the documentation of your **actual** JPA implementation. — JB Nizet, Nov 13 '17 at 07:34
JB Nizet thanks for your reply ..i will edit question.. can you please tell me how to generate custom id in JPA — Amol Raje, Nov 13 '17 at 07:40
i am using dependency spring-boot-starter-data-jpa version 1.4.7.RELEASE and implementation in hibernate — Amol Raje, Nov 13 '17 at 07:50
Which itself depends on hibernate. So you're using hibernate. So check the documentation of hibernate. — JB Nizet, Nov 13 '17 at 07:53
Can you post some code, i.e your entity/bo/model and how you want to generate your primary key? i.e number or string? — Sreenath Reddy, Nov 13 '17 at 07:58
I think the question was how to create generator not depending on specific JPA implementation. Looks like JPA doesn't have such functionality which is a pity, because I see no problem to define it in JPA. — Marx, Sep 21 '18 at 07:23

score 21 · Accepted Answer · answered Nov 13 '17 at 08:10

You can do it using GenericGenerator like this :

 @Entity
public class Client {

    @Id
    @GenericGenerator(name = "client_id", strategy = "com.eframe.model.generator.ClientIdGenerator")
    @GeneratedValue(generator = "client_id")  
    @Column(name="client_id")
    private String clientId;
}

and the custom generator class (will add prefix to the ID, you can make it do what you like):

public class ClientIdGenerator implements IdentifierGenerator {

@Override
public Serializable generate(SessionImplementor session, Object object)
        throws HibernateException {

    String prefix = "cli";
    Connection connection = session.connection();

    try {
        Statement statement=connection.createStatement();

        ResultSet rs=statement.executeQuery("select count(client_id) as Id from Client");

        if(rs.next())
        {
            int id=rs.getInt(1)+101;
            String generatedId = prefix + new Integer(id).toString();
            return generatedId;
        }
    } catch (SQLException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    return null;
}
}

GenericGenerator is from hibernate package, and the question was how to do that independently of JPA provicder — Marx, Sep 21 '18 at 07:24
@Marx I think you'll have to use the SessionCustomizer in EclipseLink. There is no mention of the said behavior in the JSR 338. — LalakaJ, Nov 05 '18 at 08:51
In Spring Boot 2.7 with Hibernate 5.6.9 you have to use `SharedSessionContractImplementor` instead of `SessionImplementor` — swissbuechi, Jun 22 '22 at 07:49

How to generate Custom Id in JPA

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