How to Generate an auto UUID using Hibernate on spring boot

Question

What I am trying to achieve is generate a UUID which is automatically assigned during a DB Insert. Similar to the primary key column named "id" generating an id value.

The model values looks something like this:

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(nullable = false)
private Long id;


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

But when the DB insert is done. the "uuid" is empty.

Help is greatly appreciated. And if I am asking an obvious stupid question I am sorry.

`@GeneratedValue` and others are only to be used on `@Id` fields, for other fields they are ignored. — M. Deinum, Jul 13 '17 at 18:45
@M.Deinum So what is the alternative (solution)? Your answer is incomplete — TheRealChx101, May 26 '21 at 22:50

fg78nc · Answer 1 · 2018-07-28T17:28:55.673

27

Can you try?

    @Id
    @GeneratedValue(generator = "uuid2")
    @GenericGenerator(name = "uuid2", strategy = "org.hibernate.id.UUIDGenerator")
    @Column(name = "id", columnDefinition = "VARCHAR(255)")
    private UUID id;

edited Jul 28 '18 at 17:28

answered Jul 13 '17 at 17:13

fg78nc

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score 16 · Accepted Answer · answered Jul 13 '17 at 20:23

16

u could use some events like @PrePersist to populate UUID field https://docs.jboss.org/hibernate/orm/4.0/hem/en-US/html/listeners.html

but why just not assign uuid when object is created uuid = UUID.randomUUID() ?

answered Jul 13 '17 at 20:23

Bartun

549
2
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> but why just not assign uuid when object is created uuid = UUID.randomUUID() ? because sometimes you need to support other tools which doesn't use hibernate and do inserts directly – long Nov 01 '18 at 12:00

score 14 · Answer 3 · answered Mar 15 '20 at 00:24

There has been lot of changes in the framework and as tested in Spring Boot 2.2.5 with MySQL v5.7 (Should work with all 2.0 versions but need to check) UUID can be auto generated like below

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name="id", insertable = false, updatable = false, nullable = false)
private UUID id;

This will store it in binary format in compact manner (good for storage). If for some reason one needs to store UUID in Varchar field as human readable (dash separated values) it can be done as below

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Type(type="uuid-char")
@Column(name="id", columnDefinition = "VARCHAR(255)", insertable = false, updatable = false, nullable = false)
private String id;

By default Hibernate maps UUID with binary format, hence to change the format we need to provide hint using the Type annotation.

score 7 · Answer 4 · answered Dec 12 '19 at 14:21

Following worked for me:

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

score 0 · Answer 5 · answered Aug 22 '23 at 12:50

0

    @Id
    @GeneratedValue
    private UUID productUUID;

This will generate a random uuid when Entity record is saved

answered Aug 22 '23 at 12:50

slinger

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How to Generate an auto UUID using Hibernate on spring boot

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