Question

我有一个表格列：

ID   UserId Serial ModifiedDate
---- ------ ----- ----------------

我需要计算每行时间戳之间的差异。例如，如果我有这样的表：

ID   UserId Serial ModifiedDate
---- ------ ----- ----------------
001    1     1111  2015-07-20 10:56:53.0000000
002    1     1111  2015-07-21 18:49:24.0000000
003    1     1111  2015-07-22 08:49:23.0000000

我需要区分001和002的时间戳，然后是002和003，结果需要如下：

ID   UserId Serial ModifiedDate                Difference 
---- ------ ----- ----------------             --------
001    1     1111  2015-07-20 10:56:53.0000000 
002    1     1111  2015-07-21 18:49:24.0000000
003    1     1111  2015-07-22 08:49:23.0000000

我尝试使用游标来执行此操作，但我找不到在新别名列中获得差异结果的方法。

这是我的疑问：

DECLARE @id bigint, @lastmodif datetime2(7), @id2 bigint, @lastmodif2 datetime2(7),@total int;

DECLARE status_cursor CURSOR FOR 

SELECT [Id], [ModifiedDate], '0' AS Difference 
FROM [Registrations]
  where p.Serial = 1111  
  Order by ModifiedDate

OPEN status_cursor

DECLARE status_cursor2 CURSOR FOR 

SELECT [Id], [ModifiedDate],'0' AS Difference 
FROM [Registrations]
  where p.Serial = 1111  
  Order by ModifiedDate

OPEN status_cursor2

FETCH NEXT FROM status_cursor2

WHILE @@FETCH_STATUS = 0
BEGIN
FETCH NEXT FROM status_cursor INTO @regid, @lastmodif
FETCH NEXT FROM status_cursor2 INTO @regid2, @lastmodif2
SET @total = 
        (
           DATEDIFF(second,@lastmodif,@lastmodif2)
        )
UPDATE status_cursor SET Result = @total

Answer 1

根据您的评论，您的SQL Server版本是2012.所以您可以通过LEAD()

执行此操作

您需要将HH替换为所需的值

SELECT ID,
       UserId,
       Serial,
       ModifiedDate,
       DATEDIFF(HH,ModifiedDate,LEAD(ModifiedDate) over(ORDER BY ID)) AS [Difference]
FROM Times

如果LAG()由于配置问题而无法在您的数据库上运行，请尝试以下查询。

WITH CTE AS
(
    SELECT *,ROW_NUMBER() OVER(ORDER BY CAST(ID AS INT)) AS RN
    FROM times
)

SELECT C1.*,DATEDIFF(HH,C1.MODIFIEDDATE,C2.MODIFIEDDATE) AS [DIFFERENCE]
FROM CTE C1 LEFT JOIN CTE C2 ON C1.RN+1 = C2.RN

Answer 2

用于sql server 2008及更高版本

with cte as 
(
    select 1 as Id , 1 as UserID, 1111 as Serial, ' 2015-07-20 10:56:53.0000000' as ModifiedDate
    union all 
    select 
    002,    1,     1111 , '2015-07-21 18:49:24.0000000'
    union all
    select 
    003,    1,     1111,  '2015-07-22 08:49:23.0000000'
)
,
cte2 as (
select *, 
        (select min(ModifiedDate) from cte as nextRow where nextRow.Serial = cte.Serial and nextRow.Id > cte.Id)  as NextModifiedDate

    from 
    cte 
)
select id,USERid,Serial,ModifiedDate,DATEDIFF(SECOND,ModifiedDate,isnull(NextModifiedDate,ModifiedDate)) as Difference 
 from cte2

此外，您可以直接在表格中使用它

    select * 
            -- ,(select min(ModifiedDate) from Registrations as nextRow where nextRow.Serial = Registrations.Serial and nextRow.Id > Registrations.Id)  as NextModifiedDate
            , DATEDIFF(second,Registrations.ModifiedDate, 
                        isnull((select min(ModifiedDate) from Registrations as nextRow where nextRow.Serial = Registrations.Serial and nextRow.Id > Registrations.Id),Registrations.ModifiedDate)
                       ) as Difference 
        from 
        Registrations

Answer 3

这适用于2008年，并且在几秒钟内返回更大的差异。

 WITH cte AS (SELECT ID, ModifiedDate, 
   DENSE_RANK() OVER (ORDER BY ID ASC) AS Ranking from Registrations),

 SELECT cte1.ID, cte1.ModifiedDate, 
   DATEDIFF(ss, cte1.ModifiedDate, cte2.ModifiedDate) AS TimeDiff 
   FROM cte cte1, cte cte2 
   WHERE cte1.Ranking = cte2.Ranking - 1

计算同一表中SQL中两个时间戳的差异

3 个答案: