jpa postgresql 使用uuid作为主键

UUID

    @Id
    @Type(type = "org.hibernate.type.PostgresUUIDType")
    private UUID id = UUID.randomUUID();

dao

public interface DemoDao extends JpaRepository {
    public Demo findById(UUID uuid);
}

使用

Demo demo =demoDao.findById(UUID.fromString("127757ac-571c-9052-cdd9-a31f91d15972"));