Sql 计算两个连续行之间的时间差
我有一张这样的桌子:Sql 计算两个连续行之间的时间差,sql,datediff,Sql,Datediff,我有一张这样的桌子: RecordID TransDate 1 05-Oct-16 9:33:32 AM 2 05-Oct-16 9:33:37 AM 3 05-Oct-16 9:33:41 AM 4 05-Oct-16 9:33:46 AM 5 05-Oct-16 9:33:46 AM 我需要获得连续的TransDate值之间的差值。我使用的是SQL Server 2014,
RecordID TransDate
1 05-Oct-16 9:33:32 AM
2 05-Oct-16 9:33:37 AM
3 05-Oct-16 9:33:41 AM
4 05-Oct-16 9:33:46 AM
5 05-Oct-16 9:33:46 AM
我需要获得连续的TransDate
值之间的差值。我使用的是SQL Server 2014,我知道有一种方法可以使用滞后函数来实现这一点,但我不知道如何做到这一点
我需要这个输出:
RecordID TransDate Diff
1 05-Oct-16 9:33:32 AM 0:00:00
2 05-Oct-16 9:33:37 AM 0:00:05
3 05-Oct-16 9:33:41 AM 0:00:04
4 05-Oct-16 9:33:46 AM 0:00:05
5 05-Oct-16 9:33:46 AM 0:00:00
有什么想法吗
提前谢谢 这个怎么样:
select recordid, transdate,
cast( (transdate - lag(transdate) over (order by transdate)) as time) as diff
from t;
换句话说,您可以减去两个datetime
值,并将结果转换为时间。然后,您可以按自己的喜好格式化结果。可能有点老套:
SELECT RecordId / 2 AS id, min(TransDate) AS TransDate, max(TransDate) - min(TransDate) AS Diff GROUP BY RecordId / 2
未经测试,因为我现在没有可用的SQL Server。无滞后/超前的方法
select T1.recordId, T1.TransDate, datediff(ss, T1.TransDate, T2.Transdate) as Diff
from Table1 T1
left join Table1 T2
on T1.Recordid = T2.RecordId +1
尝试创建一个游标以获取日期差
--- Query to get date difference between two rows
declare @table table (olddate datetime, newdate datetime)
create table #table (olddate datetime, newdate datetime)
DECLARE db_cursor CURSOR FOR
SELECT CONVERT(date,[utl_recycle_date] ) as RecycleDate
FROM XYZ
WHERE account_number = '6900' AND match_status = 'F'
AND [utl_recycle_date] IS NOT NULL
AND [utl_recycle_date] > '11/01/2018'
GROUP BY DATEDIFF(DAY, CONVERT(date,[utl_recycle_date] ), CONVERT(date, GETDATE())),CONVERT(date,[utl_recycle_date] )
ORDER BY 1
DECLARE @RecycleDate datetime
DECLARE @NewDate datetime
OPEN db_cursor
FETCH next FROM db_cursor INTO @RecycleDate
WHILE @@FETCH_STATUS = 0
BEGIN
FETCH next FROM db_cursor INTO @NewDate
insert INTO #table (olddate, newdate) values (cast(@RecycleDate as date), cast(@NewDate as date))
set @RecycleDate = @NewDate
END
CLOSE db_cursor
DEALLOCATE db_cursor
select
olddate, newdate,
CASE
WHEN DATEDIFF(DAY, olddate, newdate) = 0 THEN 1
WHEN DATEDIFF(DAY, olddate, newdate) > 0 THEN DATEDIFF(DAY, olddate, newdate)
END AS RecyclerFrequency
FROM #table
drop table #table
优雅、简单、有效。谢谢我必须学习这些滞后函数,现在看来我会更频繁地使用它们。