如何分别将时间列分为5分钟间隔和最大值/最小值SQL? [英] How to group time column into 5 minute intervals and max/min value respectively SQL?

问题描述

我有一个数据库，该数据库的 Datetime 列包含+/- 30秒的间隔，而 Value 列包含10到100之间的随机数.我的表如下所示:

I have a database with a Datetime column containing intervals of +/- 30 seconds and a Value column containing random numbers between 10 and 100. My table looks like this:

datetime               value
----------------------------
2016-05-04 20:47:20    12
2016-05-04 20:47:40    44
2016-05-04 20:48:30    56
2016-05-04 20:48:40    25
2016-05-04 20:49:30    92
2016-05-04 20:49:40    61
2016-05-04 20:50:00    79
2016-05-04 20:51:20    76
2016-05-04 20:51:30    10
2016-05-04 20:51:40    47
2016-05-04 20:52:40    23
2016-05-04 20:54:00    40
2016-05-04 20:54:10    18
2016-05-04 20:54:50    12
2016-05-04 20:56:00    55

我想要以下输出:

datetime               max_val    min_val
-----------------------------------------
2016-05-04 20:45:00    92         12
2016-05-04 20:50:00    79         10
2016-05-04 20:55:00    55         55

在我什至可以继续获取最大值和最小值之前，我首先必须将 datetime 列 GROUP 间隔5分钟.根据我的研究，我想到了这一点:

Before I can even continue getting the maximum value and the minimum value, I first have to GROUP the datetime column into 5 minute intervals. According to my research I came up with this:

SELECT
  time,
  value
FROM random_number_minute
GROUP BY
  UNIX_TIMESTAMP(time) DIV 300

实际上将 GROUPS 的datetime列间隔为5分钟，如下所示:

Which actually GROUPS the datetime column into 5 minute intervals like this:

datetime
-------------------
2016-05-04 20:47:20
2016-05-04 20:50:00
2016-05-04 20:56:00

这非常接近，因为它需要下一个最接近的日期时间，在这种情况下为20:45:00，20:50:00等.我想向下舍入 datetime (日期时间)到最近的5分钟，而与 seconds (秒)无关，例如，分钟数是:

This comes very close as it takes the next closest datetime to, in this case, 20:45:00, 20:50:00, etc. I would like to rounddown the datetime to the nearest 5 minutes regardless of the seconds, for instance if the minutes are:

minutes    rounddown
--------------------
10         10
11         10
12         10
13         10
14         10
15         15
16         15
17         15
18         15
19         15
20         20

时间可能是 14:59 ，我想向下舍入到 10:00 .经过数小时的研究，我也尝试使用此方法:

The time could be 14:59 and I would like to rounddown to 10:00. I also tried using this after hours of research:

SELECT
    time,
    time_rounded =
    dateadd(mi,(datepart(mi,dateadd(mi,1,time))/5)*5,dateadd(hh,datediff(hh,0,dateadd(mi,1,time)),0))

但是不幸的是，这没有用.我收到此错误:

But sadly this did not work. I get this error:

对本地函数'datediff'的调用中参数计数不正确

我也尝试过:

SELECT  
    time, CASE  
          WHEN  DATEDIFF(second, DATEADD(second, DATEDIFF(second, 0, time_out) / 300 * 300, 0), time) >= 240
            THEN    DATEADD(second, (DATEDIFF(second, 0, time) / 300 * 300) + 300, 0)
            ELSE    DATEADD(second, DATEDIFF(second, 0, time) / 300 * 300, 0)
          END

返回相同的错误.

我该怎么做?在对 datetime 进行分组之后，如何获得数据分组的最大值和最小值?

How can I do this? And after the datetime is grouped, how can I get the max and min value of the data grouping?

推荐答案

很抱歉，如果我重复另一个答案.我会删除的.

Sorry if I'm repeating another answer. I'll delete if I am..

SELECT FROM_UNIXTIME(FLOOR(UNIX_TIMESTAMP(datetime)/300)*300) x
     , MIN(value) min_value
     , MAX(value) max_value 
  FROM my_table 
 GROUP 
    BY x;

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