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;
It should be uuid2:
...
#GenericGenerator(name = "uuid", strategy = "uuid2")
...
See 5.1.2.2.1. Various additional generators.
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.
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.
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".
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
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;
}
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;
#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.
Related
I know how to generate UUID in a relational database.
#Id
#GeneratedValue(generator = "UUID")
#GenericGenerator(name = "UUID", strategy = "org.hibernate.id.UUIDGenerator")
#Column(name = "id", updatable = false, nullable = false)
private UUID fileId;
However, GeneratedValue and GenericGenerator cannot be imported.
How can I do it in Mongo?
Mongo has ObjectId, It is a powerful data type that is incredibly useful as a unique identifier in MongoDB Documents.
For more information please check https://www.mongodb.com/developer/products/mongodb/bson-data-types-objectid/
Inspired by the suggestion given here - JPA Entity class giving error with 2 #GeneratedValue fields, my question is the OPPOSITE of this - How to generate uuids without dashes
I am using H2 DB and have this in my model:
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Long id;
#Column(name = "useruuid")
#GeneratedValue(generator = "uuid")
#GenericGenerator(name = "uuid", strategy = "uuid4")
private UUID userUUID;
And in my controller it goes like this:
#PostConstruct
private void postConstruct() {
AppUser appUser = new AppUser(1l, UUID.randomUUID());
appUserJPARepository.save(appUser);
}
Now when my Spring Boot app starts my H2 DB-Console shows this:
ID USERUUID
1 25b9b7f391d94825b349866fe9a9077c
Question: How do I get hyphens in the DB? So that my uuid in the DB will be - 25b9b7f3-91d9-4825-b349-866fe9a9077c
I fiddled with #GenericGenerator(name = "uuid4", strategy = "uuid4") uuid(1) to uuid5 but got the same result. What is going on here and what I am doing wrong or what should I do to get hyphens in the DB? Any help or related info/links relevant to this will be greatly appreciated.
You can just remove
#GeneratedValue(generator = "uuid")
#GenericGenerator(name = "uuid", strategy = "uuid4")
and the code will looks like
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Long id;
#Column(name = "useruuid")
#GeneratedValue
private UUID userUUID;
Actually, i want to change the entity id from Type Long to the UUID.
All my entites have a MappedSuperclass. To save time, I wanted to add the new solution into the MappedSuperclass object and remove the old solution from the specific entities.
Is this a good practice or is it associated with risks?
old solution
#Entity
public class CustomObject extends DomainObject {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "ID")
private Long id;
}
new solution
#MappedSuperclass
public class DomainObject {
#Id
#GenericGenerator(name = "uuid", strategy = "uuid2")
#GeneratedValue(generator = "uuid")
#Type(type="pg-uuid")
#Column(name = "ID")
private UUID id;
}
Can the following id generation be implemented without depending on Hibernate #GenericGenerator, with pure JPA only?
import java.io.Serializable;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import org.hibernate.annotations.GenericGenerator;
#Entity
public class MyObject implements Serializable {
#Id
#GeneratedValue(generator = "uuid")
#GenericGenerator(name = "uuid", strategy = "my.IdGenerator")
#Column(name = "ID", length = 36, nullable = false)
private String id;
// ...
}
You can use a #PrePersist callback
#Id
#Column(name = "ID", length = 36, nullable = false)
private String id;
// ...
#PrePersist
public void prePersist() {
id = UUID.randomUUID();
}
More on Life Cycle callbacks here: Configuring a Life Cycle Callback Method on a JPA Entity
There are a few JPA methods for doing this (meaning it could be implemented by EclipseLink, Hibernate, or somet other JPA implemntation)
Here's how you'd do it if you had a sequence object:
#Id
#javax.persistence.SequenceGenerator( name = "mySequence", sequenceName = "MY_TABLE_SEQ", allocationSize = 1, initialValue = 1 )
#javax.persistence.GeneratedValue( strategy = GenerationType.SEQUENCE, generator = "mySequence" )
#Column( name = "MY_TABLE_ID" )
private Integer myTableId;
And here's how if it were an identity column:
#Id
#javax.persistence.GeneratedValue( strategy = GenerationType.IDENTITY )
#Column( name = "MY_TABLE_SEQ" )
private Long myTableId;
Again, the thing to note here is that this is pure JPA, not Hibernate-specific.
I have a question about string uuid in database in binary form through hibernate persistence (JPA2). I'm using now this code:
private UUID id;
#Id
#Type(type="uuid-char")
#GeneratedValue(generator = "system-uuid")
#GenericGenerator(name = "system-uuid", strategy = "uuid")
#Column(length = 32, unique = true, nullable = false)
public final UUID getId() {
return id;
}
This work fine, but I must store it in binary form. Don't ask me why, but I must.
The type for binary UUID is uuid-binary. You must have Hibernate 3.6 for this to work.
For many more details and pitfalls, see the answers to this question.