TSQL：及时拆分值答案

【问题标题】：TSQL: split value in timeTSQL：及时拆分值
【发布时间】：2017-05-23 16:30:54
【问题描述】：

我有一个基本表 dbo.Basic

start_date                end_date                  task     value
----------------------------------------------------------------------------
2016-01-12 13:02:00.000   2016-01-12 17:18:00.000   400001   12675.59370000
2016-01-13 03:09:00.000   2016-01-13 07:48:00.000   400002   13689.93600000
2016-01-13 07:48:00.000   2016-01-13 12:13:00.000   400003   12001.64456400
2016-01-13 12:13:00.000   2016-01-13 20:45:00.000   400004   23189.46598800

我想按时间（以 1 分钟为单位）分配最大值。 1小时间隔

start_date                end_date                  task     value
----------------------------------------------------------------------------
2016-01-12 13:02:00.000   2016-01-12 14:00:00.000   400001   <calculated part of value for 400001>
2016-01-12 14:00:00.000   2016-01-12 15:00:00.000   400001   <calculated part of value for 400001>
2016-01-12 15:00:00.000   2016-01-12 16:00:00.000   400001   ...
2016-01-12 16:00:00.000   2016-01-12 17:00:00.000   400001   ...
2016-01-12 17:00:00.000   2016-01-12 17:18:00.000   400001   ...
2016-01-13 03:09:00.000   2016-01-13 04:00:00.000   400002   <calculated part of value for 400002>
2016-01-13 04:00:00.000   2016-01-13 05:00:00.000   400002  ...
...
etc.

【问题讨论】：

标签： sql sql-server tsql

【解决方案1】：

您可以像这样使用recursive CTE

DECLARE @SampleData AS TABLE
(
   start_date datetime,
   end_date datetime,
   task int
)

INSERT INTO @SampleData
VALUES
('2016-01-12 13:02:00.000','2016-01-12 17:18:00.000', 400001)  
,('2016-01-13 03:09:00.000','2016-01-13 07:48:00.000', 400002)  
,('2016-01-13 07:48:00.000','2016-01-13 12:13:00.000', 400003)  
,('2016-01-13 12:13:00.000','2016-01-13 20:45:00.000', 400004)

;WITH temp AS 
(
   SELECT sd.start_date, dateadd(hour, datediff(hour,0,sd.start_date) + 1,0) AS end_date, sd.end_date as actual_enddate, sd.task
   FROM @SampleData sd
   UNION ALL
   SELECT  t.end_date, 
         CASE 
            WHEN dateadd(hour, 1,t.end_date) < t.actual_enddate THEN dateadd(hour, 1,t.end_date) 
            ELSE t.actual_enddate 
         END AS end_date,
         t.actual_enddate, 
         t.task
   FROM temp t
   WHERE t.end_date != t.actual_enddate
)
SELECT t.start_date,t.end_date,t.task
FROM temp t
ORDER BY t.task 
OPTION (MAXRECURSION 0)

演示链接：http://rextester.com/GQN93710

【讨论】：

谢谢；完成最终选择 SELECT t.start_date ,t.end_date ,datediff(MINUTE, t.start_date, t.end_date) AS duration_min ,part.value_s ,cast((datediff(MINUTE, t.start_date, t.end_date)) / cast( (part.value_s) AS DECIMAL(15, 8)) AS DECIMAL(15, 8)) * cast(t.value AS DECIMAL(15, 8)) AS FINAL_VALUE ,t.task ,t.value FROM TEMP t LEFT JOIN ( SELECT tt.task ,sum(datediff(MINUTE, tt.start_date, tt.end_date)) AS value_s FROM TEMP tt GROUP BY tt.task ) part ON t.task = part.task ORDER BY t.task OPTION (MAXRECURSION 0 )
我认为最重要的是解决您的问题的“方法”，而不是完全回答。无论如何，谢谢分享...