Question

所以我在这个表中列出了行：

IDCODE DATETIME DIFFERENCE (MINS)

1      02/03/2011 08:00-----10

2        02/03/2011 08:10-----13

3      02/03/2011 08:23 ----- 2

4       02/03/2011 08:25 -----  60

5       02/03/2011 09:25  -----      44

6       02/03/2011 10:20    -----    14

7       02/03/2011 10:34

我想计算（下一行 - 当前行的时间差）。因此，在第1行中，第2行的日期减去第1行的日期是10分钟的差异，因此它将在第1行中显示为10分钟的差异。

任何帮助将不胜感激

Answer 1

将表连接到自身，并在连接的On部分，您可以指定类似“ON MyTable.ID = MyTable.ID -1”的内容。

这取决于ID是连续的，没有间隙。

Answer 2

我使用旧版本的自联接来解决您的问题：

declare @t table (IDCODE int , dt datetime)

insert into @t
values
(1      ,'02/03/2011 08:00')-----10    
,(2        ,'02/03/2011 08:10')-----13    
,(3      ,'02/03/2011 08:23') ----- 2    
,(4       ,'02/03/2011 08:25') -----  60    
,(5       ,'02/03/2011 09:25')  -----      44    
,(6       ,'02/03/2011 10:20')    -----    14    
,(7       ,'02/03/2011 10:34')

;with cte as
    (
    select *,rn=ROW_NUMBER() over (order by IDCODE)
    from @t
    )

select base.IDCODE
      , base.dt
      , case when future.IDCODE is null 
           then null 
           else datediff(minute,base.dt,future.dt) 
        end as DIFFERENCE
from cte as base
left join cte as future on base.rn+1 = future.rn

结果：

IDCODE  dt  DIFFERENCE
1   2011-02-03 08:00:00.000 10
2   2011-02-03 08:10:00.000 13
3   2011-02-03 08:23:00.000 2
4   2011-02-03 08:25:00.000 60
5   2011-02-03 09:25:00.000 55
6   2011-02-03 10:20:00.000 14
7   2011-02-03 10:34:00.000 NULL

Answer 3

使用row_number 在Lead和CTE上进行演示

此解决方案不依赖于IDCODE作为标识符，而是依赖于序列，如果IDCODE与Lead and Row_number order by mydatetime
无关

declare @mytable table (idcode int, mydatetime datetime)

insert into @mytable
values
(1      ,'02/03/2011 08:00'),-----10
(2        ,'02/03/2011 08:10'),-----13
(3      ,'02/03/2011 08:23'), ----- 2
(4       ,'02/03/2011 08:25'), -----  60
(5       ,'02/03/2011 09:25'),  -----      44
(6       ,'02/03/2011 10:20'),    -----    14
(7       ,'02/03/2011 10:34')



-- using Lead
select
mydatetime,
lead(mydatetime,1,null) over (order by idcode) [mynextdatetime],
datediff(minute,mydatetime,lead(mydatetime,1,null) over (order by idcode)) [DatediffMinute]
from @mytable


-- using CTE

;with myDateList as (
  select row_number() over (order by idcode) rn,
  idcode,
  mydatetime
  from @mytable
 ) ,
myNextDate as
(
select  
  row_number() over (order by idcode) nextrn,
  idcode,
  mydatetime [nextdate]
  from
  myDateList
  where rn > 1
)
 select d.idcode,d.mydatetime, n.nextdate,
 datediff(minute,d.mydatetime,n.nextdate) datediffbyminutes
  from myDateList D
 left outer join myNextDate N
   on d.rn = n.nextrn 
 ;

Answer 4

使用CTE

的另一种解决方案

WITH DateCte AS
(
 SELECT 
  idcode,
  mydatetime,
  RANK() OVER ( ORDER BY idcode) AS CurrentRank,
  (RANK() OVER ( ORDER BY idcode) + 1 )AS NextRank  
  FROM Table1
 ) 

 SELECT
   ce.idcode,
   DATEDIFF (minute, ce.mydatetime, ne.mydatetime) AS DateDifference
 FROM DateCte AS ce LEFT OUTER JOIN  DateCte ne ON ce.NextRank = ne.CurrentRank

Sample Code

SQL中行之间的日期差异

4 个答案: