Defining database independent JPA object uid

Question

It turns out that the following example works when using mysql 5.x, however it doesn't when using an oracle 10g database.

Is there a way to define a unique identifier field that is independent of the database technology?

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name="id")
private long id;

I have tested this in hibernate and the following exception occurs only when using Oracle:

org.hibernate.MappingException: Dialect does not support identity key generation

Answer 1

Using a database table is a portable way to generate identifiers.

The simplest way to use a table to generate identifiers is to specify TABLE as the generation strategy:

@Id
@GeneratedValue(strategy=GenerationType.TABLE)
@Column(name="id")
private long id;

The provider will create the default table if you're using schema generation; if not, you must specify an existing table:

@TableGenerator(name="InvTab",
    table="ID_GEN",
    pkColumnName="ID_NAME",
    valueColumnName="ID_VAL",
    pkColumnValue="INV_GEN")
@Id 
@GeneratedValue(generator="InvTab")
@Column(name="id")
private long id;

http://www.oracle.com/technology/products/ias/toplink/jpa/howto/id-generation.html#table