我正在尝试将数据分组7天。
例如，。
我有一个数据，你可以在下面找到它。
count startDate finish_date
1247 2017-03-09 08:43:18 2017-03-09 16:05:34
1681 2017-03-10 08:30:13 2017-03-10 16:31:55
1464 2017-03-11 08:36:50 2017-03-11 16:42:03
1343 2017-03-12 08:26:57 2017-03-12 16:39:58
1333 2017-03-13 08:35:34 2017-03-13 16:26:18
1215 2017-03-14 08:36:58 2017-03-14 16:13:20
1817 2017-03-16 08:24:49 2017-03-16 17:18:19
1675 2017-03-17 08:22:30 2017-03-17 16:36:58
1546 2017-03-18 08:33:52 2017-03-18 16:51:52
1443 2017-03-20 08:11:00 2017-03-20 16:26:38
1481 2017-03-21 08:26:04 2017-03-21 16:57:30
1574 2017-03-23 08:19:07 2017-03-23 16:12:46
1270 2017-03-24 08:25:25 2017-03-24 16:37:59
1765 2017-03-25 08:22:58 2017-03-25 16:44:24
1200 2017-03-26 08:37:47 2017-03-26 14:59:51
1479 2017-03-27 08:17:50 2017-03-27 15:18:32
我想把他们分组7天。 我试过这个。为了它。
select count(*), min(locationDate) as startDate, max(locationDate) as finish_date from location where tagCode = 24901 and xLocation >= 278 and xLocation <= 354 and yLocation >= 239 and yLocation <= 426 and locationDate
>= DATE_SUB('2017-03-01 00:00:01',INTERVAL 7 day) and locationDate <= '2017-03-27 23:59:59' group by DATEDIFF(locationDate, '2017-03-01 00:00:01') div 7
数据就像。
count startDate finish_date
8283 2017-03-09 08:43:18 2017-03-14 16:13:20
7962 2017-03-16 08:24:49 2017-03-21 16:57:30
7291 2017-03-23 08:19:07 2017-03-27 15:22:05
问题是第二周它必须从2017-03-15开始，第三周需要开始2017-03-22但由于没有数据在几天没有开始我怎么能解决它？
答案 0 :(得分：1)
正如我在评论中提到的那样，我认为你写的结果对你提供的输入很好，但不会有不同的输入（比如2017-03-15而不是{{1} }）。
解决方案可能是编写类似这样的查询
2017-03-16我在输入的简化版本上测试了这个的简化版本，可能存在拼写错误......
修改
要显示间隔开始日期和结束日期，请尝试使用此
select sum(count) as count, min(location_date), max(location_date)
from (
select t1.location_date,
t1.count,
date_sub(location_date, interval (datediff(t1.location_date, t2.min_date) % 7) day) week_start
from location t1
cross join
(select min(location_date) as min_date from location) t2
where t1.tagCode = 24901 and
t1.xLocation between 278 and 354 and
t1.yLocation between 239 and 426 and
t1.locationDate >= DATE_SUB('2017-03-01 00:00:01',INTERVAL 7 day) and
t1.locationDate <= '2017-03-27 23:59:59'
) t3
group by week_start
答案 1 :(得分：0)
我认为你可以这样做： 您需要更改查询结果：
1 1247 2017-03-09 08:43:18 2017-03-09 16:05:34
2 1681 2017-03-10 08:30:13 2017-03-10 16:31:55
3 1464 2017-03-11 08:36:50 2017-03-11 16:42:03
4 1343 2017-03-12 08:26:57 2017-03-12 16:39:58
5 1333 2017-03-13 08:35:34 2017-03-13 16:26:18
6 1215 2017-03-14 08:36:58 2017-03-14 16:13:20
7 1817 2017-03-16 08:24:49 2017-03-16 17:18:19
8 1675 2017-03-17 08:22:30 2017-03-17 16:36:58
9 1546 2017-03-18 08:33:52 2017-03-18 16:51:52
10 1443 2017-03-20 08:11:00 2017-03-20 16:26:38
11 1481 2017-03-21 08:26:04 2017-03-21 16:57:30
12 1574 2017-03-23 08:19:07 2017-03-23 16:12:46
13 1270 2017-03-24 08:25:25 2017-03-24 16:37:59
14 1765 2017-03-25 08:22:58 2017-03-25 16:44:24
15 1200 2017-03-26 08:37:47 2017-03-26 14:59:51
16 1479 2017-03-27 08:17:50 2017-03-27 15:18:32
to this使用计算第一行最大日期和最小日期之间天数的逻辑：
-- max date of the row min date of the first row
select FLOOR(datediff('2017-03-12 16:05:34', '2017-03-09 08:43:18')/7);
select FLOOR( datediff('2017-03-16 17:18:19', '2017-03-09 08:43:18')/7);
-- what is important that you always compute the max date - the min date of the first row the same row like in your example is : '2017-03-09 08:43:18'
select FLOOR( datediff(max_date, '2017-03-09 08:43:18')/7);
rec_sum min_date max_date day_diff
1 1247 2017-03-09 08:43:18 2017-03-09 16:05:34 0
2 1681 2017-03-10 08:30:13 2017-03-10 16:31:55 0
3 1464 2017-03-11 08:36:50 2017-03-11 16:42:03 0
4 1343 2017-03-12 08:26:57 2017-03-12 16:39:58 0
5 1333 2017-03-13 08:35:34 2017-03-13 16:26:18 0
6 1215 2017-03-14 08:36:58 2017-03-14 16:13:20 0
7 1817 2017-03-16 08:24:49 2017-03-16 17:18:19 1
8 1675 2017-03-17 08:22:30 2017-03-17 16:36:58 1
9 1546 2017-03-18 08:33:52 2017-03-18 16:51:52 1
10 1443 2017-03-20 08:11:00 2017-03-20 16:26:38 1
11 1481 2017-03-21 08:26:04 2017-03-21 16:57:30 1
12 1574 2017-03-23 08:19:07 2017-03-23 16:12:46 2
13 1270 2017-03-24 08:25:25 2017-03-24 16:37:59 2
14 1765 2017-03-25 08:22:58 2017-03-25 16:44:24 2
15 1200 2017-03-26 08:37:47 2017-03-26 14:59:51 2
16 1479 2017-03-27 08:17:50 2017-03-27 15:18:32 2
-- now you can group the new result by the new field the result of division.
select
sum(rec_sum) ,
min(min_date),
max(max_date)
from (query result in the previous list)
group by day_diff
i know it's a little bit hard but i think you can do it the hard way is the day_diff computing .
答案 2 :(得分：0)
我只使用GROUP BY DATE_FORMAT：
SELECT someTimeStamp,SUM(amount) AS Total FROM sometable WHERE 1 GROUP BY DATE_FORMAT(someTimeStamp,"%Y%v")