Using Hibernate UUIDGenerator via annotations

user_760923 0
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I'm using my uuid as following:

@Id
@GeneratedValue(generator = "uuid")
@GenericGenerator(name = "uuid", strategy = "uuid")
@Column(name = "uuid", unique = true)
private String uuid;

but I'm getting a smart Hibernate warning:

Using org.hibernate.id.UUIDHexGenerator which does not generate IETF RFC 4122 compliant UUID values; consider using org.hibernate.id.UUIDGenerator instead

So I want to switch to org.hibernate.id.UUIDGenerator, now my question is how should I tell it to Hibernate's generator. I saw some guy used it as a "hibernate-uuid" - so this is what I've tried, but with negative result:

@Id
@GeneratedValue(generator = "hibernate-uuid")
@GenericGenerator(name = "hibernate-uuid", strategy = "hibernate-uuid")
@Column(name = "uuid", unique = true)
private String uuid;
annotations uuid hibernate java
8 Answers

As @natan pointed out in a comment, if you are using Hibernate 5 the below code is sufficient:

@Id 
@GeneratedValue
private java.util.UUID id;

Define the id column with the type of BINARY(16) in MySQL or it's equivalent in other SQL implementations.

user_74515660

This will use UUID v4 and the auto generated uuid will be stored in the column as usual varchar(36):

@Id
@GeneratedValue(generator = "uuid2")
@GenericGenerator(name = "uuid2", strategy = "uuid2")
@Column(length = 36)
private String uuid;

This should have some performance impact:

  • consumed size is more than BINARY(16)
  • after hydration the java.lang.String instance consumes more memory than java.util.UUID: 112 bytes for UUID as string versus 32 bytes (i.e. two longs + obj header) for UUID.

But it's much more easier to work with string'ed UUID - easier to write queries and you can see the contents of the table.

Tested on Hibernate 5.3

user_10495420

With current 5.4.2 Hibernate version,

if you want a Human-Readable varchar(36) field in the database table,
but also a Serializable UUID data type in your Java Class,
you can use @Type(type = "uuid-char") at the same time you declare your field member with java.util.UUID type.

Note that @Column(length = 36) is important to reduce from 255 to 36 the field length in MySQL.

Note that with PostgreSQL you should use @Type(type = "pg-uuid") instead.

import org.hibernate.annotations.Type
import java.util.UUID
import javax.persistence.Column
import javax.persistence.GeneratedValue
import javax.persistence.Id

@Id @GeneratedValue
@Type(type = "uuid-char") @Column(length = 36)
private UUID id;
user_71202900
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