我有一个表,其中有多列,但我需要找到StartTime列的当前行和EndTime列的上一行之间的差异。
示例是以下输出。
Batch Number Start Time End Time Difference
100004 8:00:00 8:03:30
100005 8:05:00 8:07:00 00:01:30
100006 8:08:40 8:15:00 00:01:40
32141 8:18:00 8:22:45 00:03:00
84230 8:25:10 8:33:42 00:02:25
23444 8:40:00 8:43:00 00:06:18
100001 8:50:00 8:52:00 00:07:00
我是SQL的新手,正在使用SQL SERVER 2008 R2。
请帮我在Simple Select Query中获取输出。
答案 0 :(得分:4)
CREATE TABLE #Batches
(
BatchID INT,
StartTime Datetime,
EndTime Datetime,
)
INSERT INTO #Batches
VALUES
(100004,'2016-05-16 08:00:00','2016-05-16 08:03:30'),
(100005,'2016-05-16 08:05:00','2016-05-16 08:07:00'),
(100006,'2016-05-16 08:08:40','2016-05-16 08:15:00'),
(32141 ,'2016-05-16 08:18:00','2016-05-16 08:22:45'),
(84230 ,'2016-05-16 08:25:10','2016-05-16 08:33:42'),
(23444 ,'2016-05-16 08:40:00','2016-05-16 08:43:00'),
(100001,'2016-05-16 08:50:00','2016-05-16 08:52:00')
;WITH CTE AS
(
SELECT
BatchID,
StartTime,
EndTime,
ROW_NUMBER() OVER (ORDER BY StartTime) AS Seq
FROM #Batches
)
SELECT
b.BatchID,
b.StartTime,
b.EndTime,
CONVERT(VARCHAR(20), DATEADD(SECOND,DATEDIFF(SECOND, bl.EndTime,b.StartTime),0),108) AS Diff,
DATEADD(SECOND,DATEDIFF(SECOND, bl.EndTime,b.StartTime),0) AS DiffDT
FROM CTE AS b
LEFT OUTER JOIN CTE AS bl ON bl.Seq = b.Seq - 1 -- Last batch
ORDER BY b.StartTime
答案 1 :(得分:1)
我关注此链接http://blog.sqlauthority.com/2011/11/24/sql-server-solution-to-puzzle-simulate-lead-and-lag-without-using-sql-server-2012-analytic-function/ 这将为您提供几秒钟的差异。
;with cteMain as (
select *, ROW_NUMBER() over (order by Start_time) sn
from table)
select m.batch_number, sLag.End_date, m.Start_time, convert(varchar,DateAdd(Second,DATEDIFF(SECOND, sLag.End_date, m.Start_time),0),108) as time_diff
from cteMain as m LEFT OUTER JOIN cteMain AS sLag ON sLag.sn = m.sn-1
order by m.batch_number