Postgres + Hibernate + Java UUID

问题:

I want to use PostgreSQL's native UUID type with a Java UUID. I am using Hibernate as my JPA provider and ORM. If I try to save it directly, it is just saved as a bytea in Postgres.

How can I do this?

回答1:

Try use the latest development version of the JDBC driver (Currently 8.4dev-700), or wait for the next release version. (Edited to add: 8.4-701 has been released)

The release notes mention this change:

Map the database uuid type to java.util.UUID. This only works for relatively new server (8.3) and JDK (1.5) versions.

回答2:

Here's a repost of my answer on Postgresql UUID supported by Hibernate? ... I know this question is old, but if someone stumbles across it, this will help them out.

This can be solved by adding the following annotation to the UUID:

import org.hibernate.annotations.Type;
...
@Type(type="pg-uuid")
private java.util.UUID itemUuid;

As to why Hibernate doesn't just make this the default setting, I couldn't tell you...

UPDATE: There still seem to be issues using the createNativeQuery method to open objects that have UUID fields. Fortunately, the createQuery method so far has worked fine for me.

回答3:

In non-JPA usage I'd create a UserType to format the UUID as a string that PostgreSQL accepts, then just give the name of the UserType implementation as the column type.

回答4:

you could try with:

 @Column(name="foo", columnDefinition="uuid") 

where columnDefinition is a fragment of native SQL

回答5:

Perhaps you can get UUID type to be auto-converted to a String (uuid.toString() should give the canonical 36-char representation)? DBs usually convert things reliably between Strings and native types.

回答6:

Try this

@Column(name = "UUID", nullable=false, insertable = false, columnDefinition="uuid DEFAULT uuid_generate_v4()")
private String uuid;