按日/月汇总的SQL命令

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 |
+----+---------------------+-------------+-----------------+

我想执行这样的事情:

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

因此,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)并且在列中的每个这样的行中读取/写入的总和在这一天读/写.

谢谢你,大卫

Answer 1

您想要转换stat_time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;

以下是一些其他方法(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;