How can I use an UUID as an @Id for my @Entity properly?

Question

I want to store a User entity from Spring Boot into a MySQL database and I want to use an UUID as an Id. But when I follow the online solutions I only get The userId doesn't have a default value. And I just can't figure out whats wrong. Here is the code:

User entity:

@NoArgsConstructor
@AllArgsConstructor
@Entity
@Table(name = "user")
@Data
public class User {

    @JsonProperty("userId")
    @Column(name = "userId", columnDefinition = "BINARY(16)")
    @GeneratedValue(generator = "uuid2")
    @GenericGenerator(name = "uuid2", strategy = "uuid2")
    @Id
    private UUID userId;
    
    @JsonProperty("email")
    @Column(name = "email", nullable = false)
    private String email;
    
    @JsonProperty("name")
    @Column(name = "name", nullable = false)
    String name;

    @JsonProperty("surname")
    @Column(name = "surname", nullable = false)
    String surname;

    @JsonProperty("password")
    @Column(name = "password", nullable = false)
    private String password;
}

MySQL table:

create table if not exists user (
   userId binary(16) not null primary key,
   name varchar(80) not null,
   surname varchar(80) not null,
   email varchar(120) not null,
   password varchar(120) not null
);

The error message:

SQL Error: 1364, SQLState: HY000

2020-07-23 15:31:29.234 ERROR 16336 --- [nio-8080-exec-1] o.h.engine.jdbc.spi.SqlExceptionHelper   : Field 'userId' doesn't have a default value
2020-07-23 15:31:29.251 ERROR 16336 --- [nio-8080-exec-1] o.a.c.c.C.[.[.[/].[dispatcherServlet]    : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.orm.jpa.JpaSystemException: could not execute statement; nested exception is org.hibernate.exception.GenericJDBCException: could not execute statement] with root cause

user_6277104 · Answer

First of all I have to notice that:

According to JPA only the following types should be used as identifier attribute types:

any Java primitive type

any primitive wrapper type

java.lang.String

java.util.Date (TemporalType#DATE)

java.sql.Date

java.math.BigDecimal

java.math.BigInteger

Any types used for identifier attributes beyond this list will not be portable.

But, Hibernate supports UUID identifier value generation. This is supported through its org.hibernate.id.UUIDGenerator id generator.

You can use the default strategy, that is a version 4 (random) strategy according to IETF RFC 4122.

@Id
@Column(name = "userId", columnDefinition = "BINARY(16)")
@GeneratedValue
private UUID userId;

Or an alternative strategy which is a RFC 4122 version 1 (time-based) strategy (using IP address rather than mac address).

@Id
@Column(name = "userId", columnDefinition = "BINARY(16)")
@GeneratedValue(generator = "custom-uuid")
@GenericGenerator(
    name = "custom-uuid",
    strategy = "org.hibernate.id.UUIDGenerator",
    parameters = {
        @Parameter(
            name = "uuid_gen_strategy_class",
            value = "org.hibernate.id.uuid.CustomVersionOneStrategy"
        )
    }
)
private UUID userId;

How can I use an UUID as an @Id for my @Entity properly?

1 Answers