String ID Primary Key generator in JPA/Hibernate

Question

I know that we can generate a random UUID -

@Id
@GeneratedValue(generator="system-uuid")
@GenericGenerator(name="system-uuid", strategy = "uuid")
private String myId;

But UUID is a String if size 32. How can i generate a random alphanumeric string of size 6 and store as ID?

I want to store this in MongoDB

Possible duplicate of [How to generate Custom Id using hibernate while it must be primary key of table](https://stackoverflow.com/questions/31158509/how-to-generate-custom-id-using-hibernate-while-it-must-be-primary-key-of-table) — Ubercool, Apr 23 '18 at 12:18

score 3 · Accepted Answer · answered Apr 23 '18 at 12:39

3

You will have to create a custom ID generator by implementing hibernate's IdentifierGenerator.

public class SomeCustomGenerator implements IdentifierGenerator {

    @Override
    public Serializable generate() {...}
}

And then use it:

@Id
@GeneratedValue(generator = "cust-generator")
@GenericGenerator(name = "cust-generator", strategy = "com...generator.SomeCustomGenerator")
private String myId;

Take a look at the example

answered Apr 23 '18 at 12:39

veljkost

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can i asking you, is there alternative to get sequence in hibernate beside using query ? – masbrojo Feb 03 '20 at 02:26

String ID Primary Key generator in JPA/Hibernate

1 Answers1