Question

我在这里有下表，但在获取结果时遇到了一些问题。

@Entity
@Table(name = "USER_VW")
public class WorkspaceUserImpl
{
    @JoinColumn(name = "USER_ID", insertable=false, updatable=false)
    @OneToOne(targetEntity = UserImpl.class, fetch = FetchType.EAGER)
    private User user;

}

@Table(name = "IK_USER")
@Inheritance(strategy = InheritanceType.JOINED)
@AttributeOverride(name = "id", column = @Column(name = "USER_ID") )


    public class UserImpl extends BaseAuditable<UserIdentifier>implements User, UserAuthentication {

        private static Logger log = LoggerFactory.getLogger(UserImpl.class);

        @Id
        @Type(type = "com.commons.UserIdentifierTypeMapper")
        @Column(name = "USER_ID")
        private UserIdentifier id;
    }

和用户

Public Inteface User
{
      UserIdentifier getId();
}

现在我编写了一个HQL查询来从WorkspaceUserImpl类中获取所有数据，并为UserImpl类提供给定的用户ID，如下所示。

SELECT w from WorkspaceUserImpl w where w.user.id = : user_id;

并尝试了

SELECT w from WorkspaceUserImpl as w INNER JOIN w.user as u where u.id = : user_id;

甚至尝试使用JOIN FETCH

并将参数 user_id 设置为1234。

但是我得到了List作为部分ID的emply，但在DB中它有5个记录。

我在这里犯了任何查询错误吗？善意的建议..

Answer 1

您是否尝试过以下查询：

from WorkspaceUserImpl as w JOIN FETCH w.user as u where u.id = : user_id;

关联的HQL查询

1 个答案: