选择具有按两列分组的最大值的行

Question

我已经看到了很多关于这类问题的解决方案(特别是这一个SQL只选择列上具有最大值的行),但这些似乎都不合适:

我有以下表格布局,附件的版本控制,绑定到实体:

TABLE attachments
+------+--------------+----------+----------------+---------------+
| id   | entitiy_id   | group_id | version_number | filename      |
+------+--------------+----------+----------------+---------------+
| 1    | 1            | 1        | 1              | file1-1.pdf   |
| 2    | 1            | 1        | 2              | file1-2.pdf   |
| 3    | 1            | 2        | 1              | file2-1.pdf   |
| 4    | 2            | 1        | 1              | file1-1.pdf   |
| 5    | 2            | 1        | 2              | file1-2.pdf   |
| 6    | 2            | 3        | 1              | file3-1.pdf   |
+------+--------------+----------+----------------+---------------+

输出应该是最大版本号,按group_id和entity_id分组,如果有帮助,我只需要单个entity_ids的列表:

+------+--------------+----------+----------------+---------------+
| id   | entitiy_id   | group_id | version_number | filename      |
+------+--------------+----------+----------------+---------------+
| 2    | 1            | 1        | 2              | file1-2.pdf   |
| 3    | 1            | 2        | 1              | file2-1.pdf   |
| 5    | 2            | 1        | 2              | file1-2.pdf   |
| 6    | 2            | 3        | 1              | file3-1.pdf   |
+------+--------------+----------+----------------+---------------+

我想出的是这个自我加入:

SELECT *
FROM   `attachments` `attachments`
       LEFT OUTER JOIN attachments t2
         ON ( attachments.group_id = t2.group_id
              AND attachments.version_number < t2.version_number )
WHERE  ( t2.group_id IS NULL )
   AND ( `t2`.`id` = 1 )
GROUP  BY t2.group_id

但是这个只有在不同的实体不共享相同的组号时才有效.不幸的是,这是必要的.

我在创建视图时遇到了一个有效的解决方案,但在我当前的设置中不支持此功能.

任何想法都受到高度赞赏.谢谢!

Answer 1

尝试这个：

select t1.* from attachments t1
left join attachments t2
on t1.entity_id = t2.entity_id and t1.group_id = t2.group_id and
   t1.version_number < t2.version_number
where t2.version_number is null