unu*_*tbu 57
SELECT FLOOR(UNIX_TIMESTAMP(timestamp)/(15 * 60)) AS timekey
FROM table
GROUP BY timekey;
Run Code Online (Sandbox Code Playgroud)
小智 12
试试这个,15分钟间隔的记录分组,你可以改变15*60到你需要的间隔
SELECT sec_to_time(time_to_sec(datefield)- time_to_sec(datefield)%(15*60)) as intervals from tablename
group by intervals
Run Code Online (Sandbox Code Playgroud)
以下方法 1) 的改编:
select Round(date_format(date, "%i") / (15*60)) AS interval
from table
group by interval
Run Code Online (Sandbox Code Playgroud)
以下方法 3) 的改编:
SELECT Round(Convert(substring(date_column, 14, 2), UNSIGNED) / (15*60)) AS interval /* e.g. 2009-01-04 12:20:00 */
FROM table
GROUP BY interval;
Run Code Online (Sandbox Code Playgroud)
我在这里找到了一些方法:
1)
select date_format(date, "%W") AS `Day of the week`, sum(cost)
from daily_cost
group by `Day of the week`
order by date_format(date, "%w")
Run Code Online (Sandbox Code Playgroud)
2)
select count(*) as 'count',
date_format(min(added_on), '%Y-%M-%d') as 'week commencing',
date_format(added_on, '%Y%u') as 'week'
from system
where added_on >= '2007-05-16'
group by week
order by 3 desc;
Run Code Online (Sandbox Code Playgroud)
3)
SELECT substring(postdate, 1,10) AS dd, COUNT(id) FROM MyTable GROUP BY dd;
Run Code Online (Sandbox Code Playgroud)
(也在这里:http ://www.bradino.com/mysql/dayparting-on-datetime-field-using-substring/ )
编辑:所有解决方案在具有大量记录的表上都会表现不佳。
归档时间: |
|
查看次数: |
58601 次 |
最近记录: |