3

我有 3 个 bean：组织、角色、用户

角色 - 组织关系 - @ManyToOne

角色 - 用户关系 - @ManyToMany

组织 ：

    @Entity
    @Table(name = "entity_organization")
    public class Organization implements Serializable {

        private static final long serialVersionUID = -646783073824774092L;

        @Id
        @GeneratedValue(strategy = GenerationType.TABLE)
        Long id;

        String name;

        @OneToMany(targetEntity = Role.class, mappedBy = "organization")
        List<Role> roleList;

...

角色 ：

    @Entity
    @Table(name = "entity_role")
    public class Role implements Serializable {

        private static final long serialVersionUID = -8468851370626652688L;

        @Id
        @GeneratedValue(strategy = GenerationType.TABLE)
        Long id;

        String name;

        String description;

        @ManyToOne
        Organization organization;

...

用户：

    @Entity
    @Table(name = "entity_user")
    public class User implements Serializable {

        private static final long serialVersionUID = -4353850485035153638L;

        @Id
        @GeneratedValue(strategy = GenerationType.TABLE)
        Long id;
        @ManyToMany
        @JoinTable(name = "entity_user_role",
                joinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"),
                inverseJoinColumns = @JoinColumn(name = "role_id", referencedColumnName =                     "id"))
        List<Role> roleList;

...

所以我需要获取指定用户的所有组织（首先我需要选择所有用户角色，然后选择所有具有此角色的组织）

我有一个实现此逻辑的 sql 语句（例如，我选择 id = 1 的用户）：

SELECT * FROM entity_organization AS o 
INNER JOIN entity_role r ON r.organization_id = o.id 
INNER JOIN entity_user_role ur ON ur.role_id=r.id 
WHERE ur.user_id = 1

如何使用休眠命名查询机制来实现这一点？谢谢！

4

2 回答 2

19

@NamedQuery

我在@NamedQuery实体Organization类上创建了以下内容。

@NamedQuery(name = "query", query = "SELECT DISTINCT o " +
    "FROM Organization o, User u " +
    "JOIN o.roles oRole " +
    "JOIN u.roles uRole " +
    "WHERE oRole.id = uRole.id AND u.id = :uId")
public class Organization { ...

（我使用了标准的 JPA 注释，但我的提供者是 Hibernate。）

测试

这是我运行的测试。

EntityManager em = ...
TypedQuery<Organization> q = em.createNamedQuery("query", Organization.class);
q.setParameter("uId", 1); // try it with 1L if Hibernate barks about it
for (Organization o : q.getResultList())
  System.out.println(o.name);

使用下面的表格和示例数据，输出

A
B

请看看它是否适合你。

表

CREATE TABLE `organization` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) COLLATE utf8_unicode_ci DEFAULT NULL,
  PRIMARY KEY (`id`)
);

CREATE TABLE `role` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(255) COLLATE utf8_unicode_ci DEFAULT NULL,
  `description` varchar(255) COLLATE utf8_unicode_ci DEFAULT NULL,
  `organization_id` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`)
);

CREATE TABLE `user` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  PRIMARY KEY (`id`)
);

CREATE TABLE `user_has_role` (
  `user_id` int(11) NOT NULL DEFAULT '0',
  `role_id` int(11) NOT NULL DEFAULT '0',
  PRIMARY KEY (`user_id`,`role_id`)
);

ALTER TABLE `role` ADD CONSTRAINT `cst_organization_id` 
  FOREIGN KEY `fk_organiztaion_id` (`organization_id`)
    REFERENCES `organization` (`id`);

（我用的和你的有点不同，但应该没关系。）

样本数据

`organization`
+----+------+
| id | name |
+----+------+
|  1 | A    |
|  2 | B    |
+----+------+

`role`
+----+------+-------------+-----------------+
| id | name | description | organization_id |
+----+------+-------------+-----------------+
|  1 | A    | a           |               1 |
|  2 | B    | b           |               1 |
|  3 | C    | c           |               2 |
+----+------+-------------+-----------------+

`user`
+----+
| id |
+----+
|  1 |
|  2 |
|  3 |
+----+

`user_has_role`
+---------+---------+
| user_id | role_id |
+---------+---------+
|       1 |       1 |
|       1 |       2 |
|       1 |       3 |
|       2 |       1 |
|       3 |       1 |
|       3 |       3 |
+---------+---------+
于 2011-12-07T12:25:52.443 回答
0

尝试如下 HQL：

select ur.roleList.organization from User ur where ur.id = 1

它会给你List<Organization>.

于 2011-12-06T13:20:02.093 回答