Question

假设我在SQL中定义了一个日期时间范围：

declare @STARTTIME datetime = '2014-09-10 13:00'
declare @ENDTIME datetime = '2014-09-12 13:00'

我有时间片：

declare @sliceStart time = '12:00'
declare @sliceEnd time = '16:00'

（见图） enter image description here

我想计算这个切片在时间范围的持续时间内每天重叠多长时间。在第一天，重叠是3小时（从13到16），第二天是4小时，最后一天只有1小时。

范围实际上是在一个表中，并且有数百万个，我想计算考虑这些时间片的总持续时间。在我的例子中只有一个日期时间范围，但我目前最好的猜测是我需要创建一个我将应用于每一行的函数，因此该示例直接显示范围start和end作为变量。

Answer 1

我知道可以采用更有效的方式，但您可以尝试以下方法。这是 Fiddle 。

DECLARE @STARTTIME DATETIME
SET     @STARTTIME = '2014-09-10 13:00'
DECLARE @ENDTIME DATETIME
SET     @ENDTIME = '2014-09-12 13:00'

DECLARE @sliceStart TIME
SET     @sliceStart = '12:00'
DECLARE @sliceEnd TIME
SET     @sliceEnd = '16:00'

;WITH SlicesCTE AS
(
    SELECT @STARTTIME AS StartTime
    UNION ALL SELECT DATEADD(hh,1,StartTime) AS StartTime 
              FROM   SlicesCTE 
              WHERE  StartTime < @ENDTIME
)
SELECT CONVERT(DATE,StartTime),
       SUM
       (
           CASE 
           WHEN CONVERT(TIME,StartTime) >= @sliceStart 
                AND CONVERT(TIME,StartTime) < @sliceEnd 
           THEN 1 
           ELSE 0 
           END
       ) 
FROM   SlicesCTE 
GROUP BY CONVERT(DATE,StartTime)

Answer 2

这是一种方法。计算第一天和最后一天的重叠小时数，然后在中间添加完整的天数。

SQL Server使得它比必要的更难，因为很难添加时间。以下是使用小时作为差异的代码：

select t.*,
       (case when cast(starttime as date) = cast(endtime as date)
             then (case when sliceend > cast(starttime as time) and
                             slicestart < cast(endtime as time)
                        then datediff(hour,
                                      (case when slicestart < cast(starttime as time) then cast(starttime as time) else slicestart end),
                                      (case when sliceend > cast(endtime as time) then cast(endtime as time) else sliceend end)
                                     )
                        else 0
                   end)
             then datediff(hour,
                           (case when slicestart < cast(starttime as time) then cast(starttime as time) else slicestart end),
                           (case when sliceend > cast(endtime as time) then cast(endtime as time) else sliceend end)
                          )
             else (datediff(day, starttime, endtime) - 1) * datediff(hour, slicestart, sliceend) +
                  (case when sliceend > cast(starttime as time)
                        then datediff(hour, (case when slicestart < cast(starttime as time) then cast(starttime as time) else slicestart end),
                                      sliceend)
                        else 0
                   end) +
                  (case when slicestart < cast(endtime as time)
                        then datediff(hour, slicestart,
                                      (case when sliceend > cast(endtime as time) then cast(endtime as time) else sliceend end)
                                     )
                        else 0
                   end)
       end) as hours
from (select cast('2014-09-10 13:00' as datetime) as starttime,
             cast('2014-09-12 13:00' as datetime) as endtime,
             cast('12:00' as time) as slicestart,
             cast('16:00' as time) as sliceend
     ) t;

Here是查询的SQL小提琴。

Answer 3

    [sss_M_M_M_M_M_M_M_M_eee]
|- - - -|- - - -|- - - -|- - - -|

将每个“范围”视为以下集合：

一个开始时期（sss）
结束时期（eee）
中期持续时间为零或更多完整日期（_M_M_M _）

和

切片可能在之前，或重叠，或在开始期间内
切片可能在结束期间，或重叠，或在结束期之后
中间的每一天都会出现切片

开始时段以“范围”开始日期的时间已知，结束时段以“范围”结束日期的时间已知。如果“范围”开始日期和结束日期不同，则中间期间> 0并由(DATEDIFF(DAY, starttime, endtime) - 1)

在下面的查询中，通过使用cross apply，我们可以提供一些所需的计算列别名，可以在查询的其他位置重用。因此，通过使用在第一个交叉应用中建立的别名来简化第二次交叉应用。

第二个十字架中的案例表达适用于条件1,2和＆amp; 3.前面提到过：

if range start & end are the same day             (middle = 0)
    then calculate any overlap of slice and range            
else                                              (middle > 0)
    1 overlap for each day in the middle
  + any overlap in the start period (1 or less)
  + any overlap in the end period   (1 or less)

注意：为了满足切片和/或范围中的部分小时数，将未与完整小时分钟对齐的开始时间/结束时间用作度量单位;因此，切片同样被视为分钟，但是如果需要，我们可以恢复到几个小时，并且还可以计算“重叠”的数量（实际上是“重叠等价物”）。

示例结果：

| DATE_START |   DATE_END |TIME_START| TIME_END | SLICE_MIN | OVERLAP_MIN | OVERLAPS | OVERLAP_HOURS |
|------------|------------|----------|----------|-----------|-------------|----------|---------------|
| 2014-09-10 | 2014-09-12 | 13:00:00 | 13:00:00 |       240 |         480 |        2 |             8 |
| 2014-09-10 | 2014-09-12 | 12:00:00 | 16:00:00 |       240 |         720 |        3 |            12 |
| 2014-09-10 | 2014-09-12 | 13:15:00 | 17:00:00 |       240 |         645 |   2.6875 |         10.75 |
| 2014-08-10 | 2014-09-12 | 13:15:00 | 17:00:00 |       240 |        8085 |  33.6875 |        134.75 |

查询：（假设与切片进行比较的表中存在“范围”）

DECLARE @sliceStart TIME
SET     @sliceStart = '12:00'
DECLARE @sliceEnd TIME
SET     @sliceEnd = '16:00'

SELECT
      ca1.*
    , ca2.*
    , ca2.overlap_min / (ca1.slice_min * 1.0)  AS overlaps
    , ca2.overlap_min / 60.0                   AS overlap_hours
FROM Ranges
cross apply (
             select
                    CAST(starttime AS date)                  AS date_start
                  , CAST(endtime AS date)                    AS date_end
                  , CAST(starttime AS time)                  AS time_start
                  , CAST(endtime AS time)                    AS time_end
                  , DATEDIFF(MINUTE, @slicestart, @sliceend) AS slice_min
            ) ca1
cross apply (
             select
                  (CASE
                        WHEN ca1.date_start = ca1.date_end THEN 
                        (CASE
                              WHEN @sliceend > ca1.time_start AND @slicestart < ca1.time_end THEN DATEDIFF(MINUTE,
                                  (CASE
                                        WHEN @slicestart < ca1.time_start THEN ca1.time_start ELSE @slicestart END),
                                  (CASE
                                        WHEN @sliceend > ca1.time_end THEN ca1.time_end ELSE @sliceend END)
                                  )
                              ELSE 0 END)
                        ELSE (DATEDIFF(DAY, starttime, endtime) - 1) * ca1.slice_min
                        + (CASE
                              WHEN @sliceend > ca1.time_start THEN DATEDIFF(MINUTE, (CASE
                                            WHEN @slicestart < ca1.time_start THEN ca1.time_start ELSE @slicestart END),
                                      @sliceend)
                              ELSE 0 END)
                        + (CASE
                              WHEN @slicestart < ca1.time_end THEN DATEDIFF(MINUTE, @slicestart,
                                      (CASE
                                            WHEN @sliceend > ca1.time_end THEN ca1.time_end ELSE @sliceend END)
                                      )
                              ELSE 0
                         END)
                    END) AS overlap_min
             ) as ca2
;

See this SQLFiddle demo

顺便说一句，这种方法是计算工作日的变体

SQL计算日期时间范围的每一天的时间片

3 个答案: