通过注释使用Hibernate UUIDGenerator

Question

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;

Answer 1

It should be uuid2 :

...
@GenericGenerator(name = "uuid", strategy = "uuid2")
...

See 5.1.2.2.1. Various additional generators .

Answer 2

HibernateDoc says you can use following:

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

I hope you are using Hibernate 3.5.

Answer 3

Try...

@Id
@GeneratedValue(generator = "uuid2")
@GenericGenerator(name = "uuid2", strategy = "uuid2")
@Column(name = "uuid", columnDefinition = "BINARY(16)")
public UUID getId()
{
    return id;
}

public void setId(UUID i)
{
    id = i;
}

Note the "uuid2" as opposed to "uuid".

Answer 4

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.

Answer 5

Unknown Id.generator: hibernate-uuid

@Id
@GeneratedValue(generator = "uuid")
@GenericGenerator(name = "uuid", strategy = "org.hibernate.id.UUIDGenerator")
@Column(name = "id", unique = true)
public String getId() {
    return id;
}

public void setId(String id) {
    this.id = id;
}

Answer 6

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 (ie 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

Answer 7

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;

Answer 8

@Id
@GeneratedValue(generator = "uuid")
@GenericGenerator(name = "uuid", strategy = "uuid")
@Column(name = "UUID_ID")
public String getId(){
return id;
}

This is the proper way to use annotation for uuid generators in Hibernate 5.0.11.FINAL.

Note: IT is deprecated.