Unable to generate UUID id for my entities

Question

I'm working on a web project using:

Spring 4.x (Spring Boot 1.3.3)
Hibernate 4.x
PostgreSQL 9.x

I'm new to Postgres DB and for my tables I decided to use (for the first time) UUID identifiers, but I'm having some troubles...

For ID field I used Postgres uuid type and I set as default value uuid_generate_v4(). All works correctly when I generate a new row directly by a PSQL insert, but I cannot create a new record by my application.

As an example, this is how I declared an entity in my application:

@Entity
@Table(name = "users")
public class User {

    @Id
    @Type(type = "pg-uuid")
    private UUID id;

    // other fields and methods...

}

For this declaration I'm using the Type Hibernate annotation.

Find operations work well, but when I try to make an insert I get this exception:

org.springframework.orm.jpa.JpaSystemException: ids for this class must be manually assigned before calling save(): net.myapp.User;

Followiong this tutorial I tried to solve the issue. I changed my entity definition to:

@Entity
@Table(name = "users")
public class User {

    @GeneratedValue(generator = "uuid2")
    @GenericGenerator(name = "uuid2", strategy = "uuid2")
    @Column(columnDefinition = "BINARY(16)")
    @Id
    private UUID id;

    // other fields and methods...
}

But now I'm getting wrong ID values when I retrieve (existing) data from DB (still haven't tried insert...).

So what is the right way to define ID field in my case?

UPDATE

Using the second definition...

When I try to find by ID, I'm getting:

org.postgresql.util.PSQLException: ERROR: operator does not exist: uuid = bytea Hint: No operator matches the given name and argument type(s). You might need to add explicit type casts.

When I try to create new record:

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute statement; SQL [n/a];

Did you try to add just `@Id @GeneratedValue` without the `Column` and `GenericGenerator`? — Ali Dehghani, Apr 10 '16 at 11:11
yes, working with uuid in postgres through ORM always was a pain in a .. head. i've spent a few days with this problem two years ago. a few tips: 1. you need uuid enabled in postgresql (just ensure you can generate them from console) 2. uuid - bytea error means that your insert was incorrect and it saved id as bytes 3. to exclude generator problems first try to create them in java: id = UUID.randomUUID(); — AdamSkywalker, Apr 11 '16 at 13:27

score 9 · Answer 1 · answered Jun 01 '17 at 09:18

9

I used this configuration and this is working for me and I am using

Spring Boot 1.5.2
Hibernate 5
PostgreSQL 9.6

Entity.java

@Id  
@Column(updatable = false, nullable = false, columnDefinition = "uuid DEFAULT uuid_generate_v4()")
@GeneratedValue(generator = "UUID")
@GenericGenerator(name = "UUID", strategy = "org.hibernate.id.UUIDGenerator")
private UUID id;

before that I am suggesting you to load uuid-ossp it into your database to use it.

CREATE EXTENSION IF NOT EXISTS "uuid-ossp";

answered Jun 01 '17 at 09:18

Vinit Solanki

1,863
2
15
29

Could you explain a bit more about the "uuid-ossp"? – Prasannjeet Singh Aug 07 '22 at 14:26
1

uuid-ossp is postgress module which provides some reusable functions for UUID. – Vinit Solanki Aug 08 '22 at 04:59

score 5 · Accepted Answer · edited Apr 11 '16 at 14:46

5

I use the following configuration and it works. I am using Glassfish as application server.

@Id
@GenericGenerator(name = "uuid", strategy = "uuid2")
@GeneratedValue(generator = "uuid")
@Column(name = "key", unique = true, nullable = false)
@Type(type="pg-uuid")
private UUID key;

The table key has the Postgres uuid type:

CREATE TABLE billuser.billingresult (key uuid NOT NULL,   .... )...

edited Apr 11 '16 at 14:46

davioooh

23,742
39
159
250

answered Apr 10 '16 at 16:01

S.Stavreva

577
4
8

score 0 · Answer 3 · answered Apr 11 '16 at 14:43

Worked as a charm:

Create a custom configuration class

package mypackage;

public class PostgresUuidDialect extends PostgreSQL9Dialect {

@Override
public void contributeTypes(final TypeContributions typeContributions, final ServiceRegistry serviceRegistry) {
    super.contributeTypes(typeContributions, serviceRegistry);
    typeContributions.contributeType(new InternalPostgresUUIDType());
}

protected static class InternalPostgresUUIDType extends PostgresUUIDType {

        @Override
        protected boolean registerUnderJavaType() {
            return true;
        }
    }
}

In your application.properties add this

spring.jpa.database-platform=mypackage.PostgresUuidDialect

(change mypackage according to your needs)

Pros: you don't need to use any special annotation in your JPA entity

Cons: this obscure class in your project...

Unable to generate UUID id for my entities

3 Answers3