通过注释使用 Hibernate UUIDGenerator

我使用的 uuid 如下:

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

但我得到了一个聪明的冬眠警告:

吸毒 Org.hibernate.id.UUIDHexGenerator 它不生成 IETF RFC4122 符合 UUID 值; 考虑使用 而是 org.hibernate.id.UUIDGenerator

所以我想切换到 org.hibernate.id.UUIDGenerator，现在我的问题是，我应该如何告诉它 Hibernate 的生成器。我看到有人把它用作“休眠-uuid”——所以这就是我试过的方法，但结果是负面的:

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

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.

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最佳答案

It should be uuid2:

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

See 5.1.2.2.1. Various additional generators.

小开

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".

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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;
}
小开 
@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.

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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.

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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

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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;

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