Question

我目前有一张类似于此的表 -

RecordTime                    Running     Fault
-----------------------------------------------
2016-09-15 11:32:01.00        0           202
2016-09-15 11:32:08.00        1           202
2016-09-15 11:39:31.00        0           21
2016-09-15 11:40:07.00        1           4
2016-09-15 11:42:11.00        0           21
2016-09-15 11:42:39.00        1           45

然后我想计算每条记录的RecordTimes之间的时差。为此，我使用以下 -

WITH    rows AS
        (
        SELECT  *, ROW_NUMBER() OVER (ORDER BY RecordTime) AS rn
        FROM    dbo.Table1
        )
SELECT  DATEDIFF(second, mc.RecordTime, mp.RecordTime)
FROM    rows mc
JOIN    rows mp
ON      mc.rn = mp.rn - 1

返回 -

(No column name)
----------------
7
443
36
124
28
303

但实际上我想要做的是在原始表中创建一个计算列，它给出了这些值。这可能吗？

我认为我可以将查询转换为UDF，然后在列中引用它，但我对这种工作不太熟悉。

--- ---编辑

最终结果应为 -

RecordTime                    Running     Fault     Diff
--------------------------------------------------------
2016-09-15 11:32:01.00        0           202       7
2016-09-15 11:32:08.00        1           202       443
2016-09-15 11:39:31.00        0           21        36
2016-09-15 11:40:07.00        1           4         124
2016-09-15 11:42:11.00        0           21        28
2016-09-15 11:42:39.00        1           45        303

Answer 1

我建议您为此目的使用视图：

CREATE VIEW Table1_vw 
AS
WITH cte AS (
    SELECT  *, 
            ROW_NUMBER() OVER (ORDER BY RecordTime) AS rn
    FROM    dbo.Table1
)
SELECT  mc.RecordTime,
        mc.Running,
        mc.Fault,
        DATEDIFF(second, mc.RecordTime, mp.RecordTime) Diff
FROM cte mc
LEFT JOIN cte mp
ON mc.rn = mp.rn - 1

由于您使用的是 SQL Server 2012 ，因此您可以使用LEAD功能：

CREATE VIEW Table1_vw 
AS
SELECT  RecordTime,
        Running,
        Fault,
        DATEDIFF(second,RecordTime,LEAD(RecordTime,1,NULL) OVER (ORDER BY RecordTime ASC) ) as Diff
FROM Table1
GO

Answer 2

怎么样：

CREATE FUNCTION GetTimeDiff (@time datetime)
RETURNS INT 
AS BEGIN
    declare @t1 datetime 

    DECLARE @Ret int

    SELECT @t1 = RecordTime 
    FROM (
        SELECT  RecordTime, 
                ROW_NUMBER() OVER (ORDER BY RecordTime DESC) AS rn
        FROM Table1
        WHERE RecordTime < @time
    ) i
    WHERE i.rn = 1

    SELECT @Ret = DATEDIFF(second, @t1, @time)
    RETURN @Ret
END
GO

ALTER TABLE Table1
ADD TimeDiff AS dbo.GetTimeDiff(RecordTime) 
GO

您需要为函数添加逻辑以处理NULL @ t1等。

更新以使其适用于下一行：

 ROW_NUMBER() OVER (ORDER BY RecordTime ASC) AS rn
        FROM Table1
        WHERE RecordTime > @time

这使得所有价值观都是积极的而不是消极的：

SELECT @Ret = ABS(DATEDIFF(second, @t1, @time))

SQL Server DATEDIFF行之间的计算列

2 个答案: