Question

我使用Hibernate，我希望使用Hibernate过滤器选项过滤实体。但过滤器将被忽略。 Workspace实体包括用户实体。并且我希望只有具有doesent的用户的所有工作区实体在数据库的电子邮件字段中都具有NULL值。我期待着您的回音。感谢

工作区

@Entity
        @Table(name = "WORKSPACEENTITY", schema = "pwdmanager")
        @PrimaryKeyJoinColumn(name = "workspace_id", referencedColumnName = "id")
        @Setter
        @Getter

        @FilterDef(name = "FILTER")
        public class DBWorkspaceEntity{

            @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, mappedBy = "wkse_space")
            private List<DBPasswordEntity> pwde_password;

            @OneToOne(fetch = FetchType.EAGER)

            @Filters({
                    @Filter(name="FILTER", condition="{u}.usre_email is not null", aliases={@SqlFragmentAlias(alias="u", entity=DBUserEntity.class)}),
            })
            private DBUserEntity usre_user;
    }

用户

 @Entity
    @Table(name = "USERENTITY", schema = "pwdmanager")
    @Setter
    @Getter
    @PrimaryKeyJoinColumn(name = "user_id", referencedColumnName = "id")
    public class DBUserEntity{

        @Basic
        @Column(name = "USRE_EMAIL", nullable = true, length = 1024)
        private String usre_email;
    }

休眠

public class MasterDao{
    private Session session;
    private Transaction transaction;
    private SessionFactory sessionFactory;

    public void prepare() {
        sessionFactory = HibernateUtil.getSessionFactory();
        this.session = sessionFactory.openSession();
    }

    public void filter(final DBUserEntity usr) {
        try {
            session.enableFilter("FILTER");
            Query query = session.createQuery("FROM " + DBWorkspaceEntity.class.getSimpleName());
            List<DBWorkspaceEntity> result = query.list();
            session.disableFilter("FILTER_USER");

        } catch (final HibernateException ex) {
            log.error(ex);
        }
    }
}

Answer 1

我认为，过滤条件下的usre_email应该是列名。

 USRE_EMAIL

Hibernate过滤器不起作用

1 个答案: