Question

我有一个监控系统，每n秒收集一次数据（n大约为10但是有所不同）。我想以15分钟的间隔汇总收集的数据。有没有办法将时间戳值合并为15分钟的块，以便分组工作？

Answer 1

SELECT   FLOOR(UNIX_TIMESTAMP(timestamp)/(15 * 60)) AS timekey
FROM     table
GROUP BY timekey;

Answer 2

试试这个，将间隔15分钟的记录分组，你可以将15 * 60更改为你需要的时间间隔

SELECT sec_to_time(time_to_sec(datefield)- time_to_sec(datefield)%(15*60)) as intervals from tablename
group by intervals

Answer 3

以下方法1）的改编：

select Round(date_format(date, "%i") / (15*60)) AS interval  
from table
group by interval

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;

我找到的一些方法here：

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")

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;

3）

SELECT substring(postdate, 1,10) AS dd, COUNT(id) FROM MyTable GROUP BY dd;

（也在这里：http://www.bradino.com/mysql/dayparting-on-datetime-field-using-substring/）

编辑：所有解决方案都会在包含大量记录的表格上表现不佳。

Answer 4

我开始使用unutbu上面给出的答案，但没有得到我需要的东西，不得不添加一点。

Select Created, from_unixtime(FLOOR(UNIX_TIMESTAMP(Created)/(15*60))*(15*60)) GroupTime, COUNT(*) as Cnt FROM issue i GROUP BY GroupTime

此代码除以15分钟跨度内的900秒，然后将该值放大并将其乘以900，基本上向下舍入到最接近的15分钟增量。

Answer 5

以下查询对行进行分组，并以15分钟的间隔创建时间戳。

Select concat( date(created_dt) , ' ', sec_to_time(time_to_sec(created_dt)- time_to_sec(created_dt)%(15*60) + (15*60)))as created_dt_new from table_name group by created_dt_new

例如，时间戳 2016-11-09 13:16:29
2016-11-09 13:16:49
2016-11-09 13:17:06
2016-11-09 13:17:26
2016-11-09 13:18:24
2016-11-09 13:19:59
2016-11-09 13:21:17

分为2016-11-09 13:30:00

sec_to_time(time_to_sec(created_dt)- time_to_sec(created_dt)%(15*60) + (15*60)))
上限时间最接近15分钟。例如12:10 - ＆gt; 12:15
concat( date(created_dt) , ' ', sec_to_time(time_to_sec(created_dt)- time_to_sec(created_dt)%(15*60) + (15*60)))
生成从时间戳字段中获取日期的时间戳。

Answer 6

Unix时间戳：使用以下其中一种方法将它们设置为最接近15分钟：

timestamp div (15 * 60) * (15 * 60) -- div is integer division operator
timestamp - timestamp % (15 * 60)

日期时间：假设数据类型没有小数秒，请使用以下时间将其设置为最接近的15分钟：

date - INTERVAL EXTRACT(SECOND FROM date) SECOND - INTERVAL EXTRACT(MINUTE FROM date) % 15 MINUTE

DBFiddle

Answer 7

这对我有用

mysql> **SELECT FROM_UNIXTIME(UNIX_TIMESTAMP(NOW())- UNIX_TIMESTAMP(NOW())%(15*60));**
+---------------------------------------------------------------------+
| FROM_UNIXTIME(UNIX_TIMESTAMP(NOW())- UNIX_TIMESTAMP(NOW())%(15*60)) |
+---------------------------------------------------------------------+
| 2012-02-09 11:15:00                                                 |
+---------------------------------------------------------------------+
1 row in set (0.00 sec)

Answer 8

为我工作

SELECT CONCAT (
        YEAR(transactionDate)
        ,'-'
        ,MONTH(transactionDate)
        ,'-'
        ,DAYOFMONTH(transactionDate)
        ,' '
        ,HOUR(transactionDate)
        ,':'
        ,((floor((MINUTE(transactionDate) / 15)) + 1) * 15) - 1
        ,':59'
        ) AS tmp1
    ,count(*)
FROM tablename
GROUP BY tmp1 limit 20;

Answer 9

将“ 15”更改为所需的时间间隔。

select count(*),
CONCAT(HOUR(col_date),":",(MINUTE(create_date) div 15)*15) as date
from tablename
GROUP BY date
ORDER BY col_date ASC;

Answer 10

我对GROUP BY不满意。

SELECT   datetime
FROM     table
WHERE MOD(MINUTE(TIME(datetime)),15) = 0 AND SECOND(TIME(datetime)) = 0;

以15分钟为间隔对mysql进行分组查询

10 个答案: