在hibernate中映射一个表中的多个集合

时间:2010-11-05 14:54:52

标签: java hibernate hibernate-mapping

我有UserLog课程(我无法更改):

class User {
    private long id;
    private String name;
    private Set<Log> accessLogs;
    private Set<Log> actionLogs;
}

class Log {
    private String what;
    private Date when;
}

可能的映射如下所示:

<class name="com.example.User" table="users">
    <id name="id" access="field">
        <generator class="native" />
    </id>

    <property name="name" length="256" />

    <set name="accessLogs" table="user_access_logs" cascade="all-delete-orphan" order-by="`when`">
        <key column="user_id" />
        <composite-element class="com.example.Log">
            <property name="what" length="512" />
            <property name="when" column="`when`" />
        </composite-element>
    </set>

    <set name="actionLogs" table="user_action_logs" cascade="all-delete-orphan" order-by="`when`">
        <key column="user_id" />
        <composite-element class="com.example.Log">
            <property name="what" length="512" />
            <property name="when" column="`when`" />
        </composite-element>
    </set>

</class>

这样可以正常工作并映射到以下数据库表:

users
+----+------+
| id | name |
+----+------+
|  1 | john |
|  2 | bill |
|  3 | nick |
+----+------+

user_access_logs
+---------+------------+---------------------+
| user_id | what       | when                |
+---------+------------+---------------------+
|       1 | logged in  | 2010-09-21 11:25:03 |
|       1 | logged out | 2010-09-21 11:38:24 |
|       1 | logged in  | 2010-09-22 10:19:39 |
|       2 | logged in  | 2010-09-22 11:03:18 |
|       1 | logged out | 2010-09-22 11:48:16 |
|       2 | logged in  | 2010-09-26 12:45:18 |
+---------+------------+---------------------+

user_action_logs
+---------+---------------+---------------------+
| user_id | what          | when                |
+---------+---------------+---------------------+
|       1 | edit profile  | 2010-09-21 11:28:13 |
|       1 | post comment  | 2010-09-21 11:30:40 |
|       1 | edit profile  | 2010-09-21 11:31:17 |
|       1 | submit link   | 2010-09-22 10:21:02 |
|       2 | submit review | 2010-09-22 11:10:22 |
|       2 | submit link   | 2010-09-22 11:11:39 |
+---------+---------------+---------------------+

我的问题是如何在hibernate中将这两组(accessLogsactionLogs)映射到同一个表中,为日志生成以下模式:

user_logs
+---------+---------------+---------------------+--------+
| user_id | what          | when                | type   |
+---------+---------------+---------------------+--------+
|       1 | logged in     | 2010-09-21 11:25:03 | access |
|       1 | edit profile  | 2010-09-21 11:28:13 | action |
|       1 | post comment  | 2010-09-21 11:30:40 | action |
|       1 | edit profile  | 2010-09-21 11:31:17 | action |
|       1 | logged out    | 2010-09-21 11:38:24 | access |
|       1 | logged in     | 2010-09-22 10:19:39 | access |
|       1 | submit link   | 2010-09-22 10:21:02 | action |
|       2 | logged in     | 2010-09-22 11:03:18 | access |
|       2 | submit review | 2010-09-22 11:10:22 | action |
|       2 | submit link   | 2010-09-22 11:11:39 | action |
|       1 | logged out    | 2010-09-22 11:48:16 | access |
|       2 | logged in     | 2010-09-26 12:45:18 | access |
+---------+---------------+---------------------+--------+

编辑:我想保留Java代码和Set语义。我正在寻找像鉴别器,公式或hibernate API扩展这样的东西来解决这个问题而不涉及Java代码。也许就行了(假想的hibernate配置如下):

<set name="accessLogs" table="user_logs" ... formula="type='access'">
    <key column="user_id" />
    <composite-element class="Log">
        ...
    </composite-element>
</set>

<set name="actionLogs" table="user_logs" ... formula="type='action'">
    <key column="user_id" />
    <composite-element class="Log">
        ...
    </composite-element>
</set>

Edit2 :我还没有完整的解决方案。我确信可以通过扩展一些hibernate API来完成它。

3 个答案:

答案 0 :(得分:11)

您是否尝试以这种方式映射集:

<set name="accessLogs" table="user_logs" where="type='access'">    

答案 1 :(得分:5)

正如Maurizio所写,您需要使用where属性来仅检索属于每个集合的行。

要使用附加列 插入 行,您需要在<set>中添加以下元素:

<sql-insert>
  insert into user_logs(user_id, what, [when], type)
  values(?, ?, ?, 'access')
</sql-insert>

这实质上告诉Hibernate使用特定的SQL而不是生成它。请注意,我必须手动转义when列。

奖励:要将额外的列添加到数据库,请使用以下内容(在关闭<class>之后)

<database-object>
  <create>alter table user_logs add type char(6)</create>
  <drop></drop>
</database-object>

有趣的一点:我使用 NHibernate 编写并测试了它,但它是直接移植的,所以它应该完全相同。

答案 2 :(得分:1)

为什么不将列表合并到一个列表中呢?您的两个列表都属于同一类。

编辑 - 如果你想保留2个列表,那么创建一个组合的第三个列表,并且只保留它。您将必须控制对两个列表的访问者的访问权限,因此您知道何时更新第三个列表,但不应该太难。