从一系列日期时间中获取十分钟的间隔

Question

我有一个包含日期时间值范围的日期时间列。我想用所有这些日期时间值创建另一列，但要向下舍入到十分钟。

所以，是这样的：

datetimesent              |     ten_minute_column

2012-06-11 18:27:58.000   |     2012-06-11 18:20:00.000
2012-06-15 15:19:08.000   |     2012-06-15 15:10:00.000
...                       |

我玩得最远的就是把它放在分钟的桶槽中。 我是这样做的：

SELECT DatetimeSent,
    DATEADD(Minute, DATEDIFF(Minute, 0, DatetimeSent), 0) AS Minute_bucket
FROM allrequests

但我需要十分钟的桶槽。

Answer 1

试试这个：

select dateadd(minute, datepart(minute, datetimesent) / 10 * 10, 
dateadd(hour, datediff(hour, 0,datetimesent), 0)) ten_minute_column
from 
(select cast('2012-06-11 18:27:58.000' as datetime) datetimesent
union all
 select cast('2012-06-15 15:19:08.000' as datetime)) a

Answer 2

您可以使用许多功能来做到这一点：

WITH D AS
(   SELECT  CURRENT_TIMESTAMP [DateField]
    UNION ALL
    SELECT  DATEADD(MINUTE, 5, CURRENT_TIMESTAMP)
)
SELECT  DATEADD(MINUTE, (10 * FLOOR(DATEPART(MINUTE, DateField) / 10.0)) - DATEPART(MINUTE, DateField), DATEADD(MINUTE, DATEDIFF(MINUTE, 0, DateField), 0)) AS RoundedDate
FROM    D

这里的要点是去掉超过 10 分钟间隔的分钟数，然后从实际分钟数中减去（除去秒数）。

这可以通过将一些函数移动到一个连接来稍微整理一下。但是，我认为这不会带来任何性能提升（根本没有进行任何测试）

;WITH T AS
(   SELECT  Number,
            (10 * FLOOR(Number / 10.0)) - Number [RoundedDifference]
    FROM    (   SELECT  ROW_NUMBER() OVER(ORDER BY Object_ID) - 1 [Number]
                FROM    sys.All_Objects
            ) n
    WHERE   Number < 60
), D AS
(   SELECT  CURRENT_TIMESTAMP [DateField]
    UNION ALL
    SELECT  DATEADD(MINUTE, 5, CURRENT_TIMESTAMP)
)
SELECT  DateField,
        DATEADD(MINUTE, RoundedDifference, DATEADD(MINUTE, DATEDIFF(MINUTE, 0, DateField), 0)) [RoundedDate]
FROM    D
        INNER JOIN T
            ON DATEPART(MINUTE, DateField) = Number

Answer 3

SELECT  DatetimeSent,
        Dateadd(ms, -Datepart(ms, Dateadd(minute, Datediff(minute, 0, 
                                                    Dateadd(minute, 
                                                    -Datepart(minute, 
                                                     datetimesent) 
                                                    %10, 
                                                    datetimesent)), 0 
                                                ) 
                ), Dateadd(minute, Datediff(minute, 0, Dateadd(minute, 
                                                       - 
                                   Datepart( 
                                   minute, 
                                                   datetimesent 
                                   )% 
                                   10, 
                                     datetimesent 
                                                   )), 0)) AS Minute_bucket
FROM   allrequests 

Answer 4

假设从不存在毫秒，您可以通过这种方式去除分钟和秒，然后对结果进行分组：

SELECT DATEADD(SECOND, -(CONVERT(INT, RIGHT(CONVERT(CHAR(2),
  DATEPART(MINUTE, GETDATE())),1))*60)-(DATEPART(SECOND,GETDATE())), GETDATE());

这是一个查询，它根据表（或表的子集）的最小和最大日期获取正确的时隙数：

DECLARE @x TABLE(datetimesent DATETIME);
INSERT @x SELECT '2012-06-11 18:27:58.000'
UNION ALL SELECT '2012-06-15 15:19:08.000';

DECLARE @start SMALLDATETIME, @end SMALLDATETIME, @i INT;

SELECT @start = CONVERT(DATE, MIN(datetimesent)), @end = CONVERT(DATE,
MAX(datetimesent))
FROM @x
-- WHERE ...;

SELECT @i = DATEDIFF(DAY, @start, @end) * 144;

;WITH slots(ten_minute_column) AS
(
   SELECT TOP (@i * 144) DATEADD(MINUTE, 10 * (ROW_NUMBER() OVER
   (ORDER BY s1.[object_id])-1), @start)
   FROM sys.all_columns AS s1
   -- you may need to cross join to another table if this doesn't
   -- provide enough rows. Depends on overall datediff...
)
SELECT x.datetimesent, slots.ten_minute_column
 FROM @x AS x
 INNER JOIN slots
 ON x.datetimesent >= slots.ten_minute_column
 AND x.datetimesent < DATEADD(MINUTE, 10, slots.ten_minute_column)
 -- WHERE ...;

结果：

datetimesent              ten_minute_column
-----------------------   -------------------
2012-06-11 18:27:58.000   2012-06-11 18:20:00
2012-06-15 15:19:08.000   2012-06-15 15:10:00