获取MySQL中最不同的记录

Question

这可能是非常简单的事情，所以请原谅我的金发时刻:)

我有一张桌子“相册”

* albumId
* albumOwnerId (who created)
* albumCSD (create stamp date)

现在我想要做的是选择最近更新的前10张专辑。但是，我不希望同一个人的10张专辑回来-我只想每个独特的人一张专辑。即来自10个不同人的10张专辑。

所以，这就是我下面的，但它不能正常工作，我就是找不出原因。有什么想法吗？

谢谢

SELECT DISTINCT(albumOwnerId), albumId
FROM album
ORDER BY albumCSD DESC
LIMIT 0,10

下面是一些示例数据，后面是我想要获取的数据。希望这能让事情变得更清楚。

数据：

albumOwnerID, albumId, albumCSD
18, 194, '2010-10-23 11:02:30'
23, 193, '2010-10-22 11:39:59'
22, 192, '2010-10-12 21:48:16'
21, 181, '2010-10-12 20:34:11'
21, 178, '2010-10-12 20:20:16'
19, 168, '2010-10-12 18:31:55'
18, 167, '2010-10-11 21:06:55'
20, 166, '2010-10-11 21:01:47'
18, 165, '2010-10-11 21:00:32'
20, 164, '2010-10-11 20:50:06'
17, 145, '2010-10-10 18:54:24'
17, 144, '2010-10-10 18:49:28'
17, 143, '2010-10-10 18:48:08'
17, 142, '2010-10-10 18:46:54'
16, 130, '2010-10-10 16:17:57'
16, 129, '2010-10-10 16:17:26'
16, 128, '2010-10-10 16:07:21'
15, 119, '2010-10-10 15:24:28'
15, 118, '2010-10-10 15:24:11'
14, 100, '2010-10-09 18:22:49'
14, 99, '2010-10-09 18:18:46'
11, 98, '2010-10-09 15:50:13'
11, 97, '2010-10-09 15:44:09'
11, 96, '2010-10-09 15:42:28'
11, 95, '2010-10-09 15:37:25'

所需数据：

18, 194, '2010-10-23 11:02:30'
23, 193, '2010-10-22 11:39:59'
22, 192, '2010-10-12 21:48:16'
21, 181, '2010-10-12 20:34:11'
19, 168, '2010-10-12 18:31:55'
17, 145, '2010-10-10 18:54:24'
16, 130, '2010-10-10 16:17:57'
15, 119, '2010-10-10 15:24:28'
14, 100, '2010-10-09 18:22:49'
11, 98, '2010-10-09 15:50:13'

Answer 1

通过这个查询，我得到了您想要的结果

SELECT albumOwnerID, albumId, albumCSD
FROM album
WHERE albumCSD in
(SELECT  Max(album.albumCSD) AS MaxvonalbumCSD
FROM album
GROUP BY album.albumOwnerID);

但是在MS Access中

Answer 2

select albumOwnerID, albumID
from album
Group by albumOwnerID, albumID
Order by albumcsd desc
LIMIT 0,10

编辑：

select albumOwnerID, albumID 
from album
where albumOwnerID in (select distinct albumOwnerID from album order by albumCSD )
LIMIT 0,10