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/
Related
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;
I've tried to use custom id-generator as in Bypass GeneratedValue in Hibernate (merge data not in db?) and it works fine while working with Postgres DB. My code is equals to the code in example.
But while running test with H2 in-memory database I faced the problem, that id is not generated automaticaly.
Without custom generator
#Column(name = "id", nullable = false)
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
Generated create table script
create table db.schema.entity (id bigserial not null...
With custom generator
#Column(name = "id", nullable = false)
#Id
#GeneratedValue(generator = "idGenerator", strategy = GenerationType.IDENTITY)
#GenericGenerator(name="idGenerator", strategy = "...UseIdOrGenerate")
private Long id;
Generated create table script
create table db.schema.entity (id int8 not null...
As a result tests don't work.
Solved by changing #Column
#Column(name = "id", nullable = false, columnDefinition = "bigserial")
I've inherited a SQL Server database that I'm trying to map via JPA. Many of the tables have a uniqueidentifier column. I'm trying to map them like so:
#Id
#GenericGenerator(name = "generator", strategy = "guid", parameters = {})
#GeneratedValue(generator = "generator")
#Column(name = "APPLICATION_ID")
private String id;
Hibernate complains with:
Found: uniqueidentifier, expected: varchar(255)
The data type of the primary key property in the POJO determines the data type of its mapped DB column, which is specified by the Dialect class. According to the SQLServerDialect provided by hibernate, it does not have any data type that maps to uniqueidentifier, and String by default maps to varchar(255)
I think guid strategy on a String primary key only means that hibernate will generate a GUID value for POJO's primary key property and this generated GUID value will be inserted to the varchar(255) column to simulate the effect of uniqueidentifier
You can try to override the mapping specified by the Dialect class by using the columnDefinition attribute of #Column
#Id
#GenericGenerator(name = "generator", strategy = "guid", parameters = {})
#GeneratedValue(generator = "generator")
#Column(name = "APPLICATION_ID" , columnDefinition="uniqueidentifier")
private String id;
You need to use uuid2
#Id
#GenericGenerator(name = "generator", strategy = "uuid2", parameters = {})
#GeneratedValue(generator = "generator")
#Column(name = "APPLICATION_ID" , columnDefinition="uniqueidentifier")
private String id;
This works for me.
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.
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.