Question

我遇到了一些我尚未解决的问题。看看下表。

+---+------------+-------------------------------------+
|id |   likes    |   fb_likes    |     created_time    |
+---+------------+-------------------------------------+
|10 |    0       |      0        | 2013-04-07 13:53:36 |
| 9 |    3       |      2        | 2013-04-07 13:17:18 |
| 8 |    2       |      1        | 2013-04-06 21:49:30 |
| 7 |    2       |      0        | 2013-04-06 20:25:24 |
| 6 |    6       |      5        | 2013-04-06 18:52:27 |
| 5 |    3       |      0        | 2013-04-05 21:19:34 |
| 4 |    5       |      2        | 2013-04-05 19:09:13 |
| 3 |    9       |     10        | 2013-04-03 14:18:56 |
| 2 |    0       |      1        | 2013-04-02 14:28:36 |
| 1 |    0       |      0        | 2013-04-02 08:39:32 |
+---+------------+-------------------------------------+

我想选择每天最喜欢的ID。我已经看过很多想要做类似事情的帖子，但还没有找到解决方案。

我的尝试失败（这选择了正确的喜欢，但错误的ID）：

SELECT id, MAX(likes+fb_likes) as Tlikes, created_time FROM table GROUP BY DAY(created_time);

我也试过这样的事情： http://dev.mysql.com/doc/refman/5.0/en/example-maximum-column-group-row.html

任何帮助将不胜感激

更新我忘了添加fb_likes列。我多么愚蠢。

Answer 1

SELECT  a.*
FROM    TableName a
        INNER JOIN
        (
            SELECT  DATE(created_time) date, MAX(likes) max_likes
            FROM    TableName
            GROUP   BY DATE(created_time)
        ) b ON  DATE(a.Created_Time) = b.date AND
                a.likes = b.max_Likes

SQLFiddle Demo

如果您只想显示至少有1票的记录，则需要在外部选择中添加WHERE子句，例如

SQLFiddle Demo

Answer 2

SELECT id, likes, day(created_time)
FROM Likes
WHERE likes =
(SELECT MAX(maxLikes.likes) 
 FROM Likes maxLikes 
 WHERE day(maxLikes.created_time) = day(Likes.created_time)
);

http://sqlfiddle.com/#!2/3e35f/8

每天最喜欢的行

2 个答案: