SQL Server DATEDIFF计算行之间的列

Question

我目前有一个与此类似的表格-

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，我想计算每个记录的RecordTimes之间的时间差。为此，我正在使用以下内容-

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

哪个返回-

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

不过，我实际上要做的是在原始表中创建一个计算列，该列为我提供了这些值。这可能吗？

我以为我可以将查询转换为UDF，然后在列中进行引用，但是我对这种工作不是很有经验。

- -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 ，因此可以使用LEAD函数：

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