Question

我有下表：

+----+---------------------+-------------+-----------------+
| id | stat_time           |       reads |          writes |
+----+---------------------+-------------+-----------------+
|  1 | 2013-07-18 20:00:00 |    42614543 |         1342129 |
|  2 | 2013-07-18 21:00:00 |    23085319 |          326139 |
|  3 | 2013-07-25 12:00:00 |           0 |           39639 |
|  4 | 2013-07-25 13:00:00 |      754166 |           39639 |
|  5 | 2013-07-25 14:00:00 |      693382 |          295323 |
|  6 | 2013-07-25 15:00:00 |     1334462 |               0 |
|  7 | 2013-07-25 16:00:00 |    10748367 |          261489 |
|  9 | 2013-07-25 17:00:00 |     4337294 |               0 |
| 10 | 2013-07-25 18:00:00 |     3002796 |               0 |
| 11 | 2013-07-25 20:00:00 |     3002832 |               0 |
| 12 | 2013-07-25 23:00:00 |           0 |          333468 |
| 13 | 2013-07-26 17:00:00 |    10009585 |               0 |
| 15 | 2013-07-26 18:00:00 |     6005752 |               0 |
| 17 | 2013-07-26 21:00:00 |      333663 |               0 |
+----+---------------------+-------------+-----------------+

我想做这样的事情：

SELECT stat_time, SUM(reads), SUM(writes) from this_table GROUP BY stat_time HAVING 'the same day'..

所以一个输出三行的命令（一个用于2013-07-18，第二个用于2013-07-25，第三个用于2013-07-26）并且在列的每个这样的行中读/写这一天的读/写总和。

谢谢你，大卫

Answer 1

您希望将stat_time转换为一天来执行此操作。该方法取决于数据库。这是一种方式：

SELECT cast(stat_time as date) as stat_day, SUM(reads), SUM(writes)
from this_table
GROUP BY cast(stat_time as date)
order by stat_day;

以下是其他一些方法（适用于MySQL和Oracle）：

SELECT date(stat_time) as stat_day, SUM(reads), SUM(writes)
from this_table
GROUP BY date(stat_time)
order by stat_day;

SELECT trunc(stat_time) as stat_day, SUM(reads), SUM(writes)
from this_table
GROUP BY trunc(stat_time)
order by stat_day;

Answer 2

我们假设您有一个名为D-DATE_START的列

$query = 'SELECT cast(D_DATE_START as date) as stat_day ,'
        .'sum(I_REV_MODEL) as totalDayRevenu '
        .'FROM t_show WHERE FK_MODEL=136 '
        ."and t_show.D_DATE_START between '".$after."' and '".$before."'"
        .' GROUP BY cast(D_DATE_START as date) '
        .' ORDER BY stat_day ';

按日/月汇总的SQL命令

2 个答案: