Question

Row       TimeStamp
____|________________________
1   | 2015-01-01 12:00:01.000
2   | 2015-01-01 12:00:02.000
3   | 2015-01-01 12:00:03.000
4   | 2015-01-01 12:00:04.000
5   | 2015-01-01 12:00:05.000
6   | 2015-01-01 12:00:06.000
7   | 2015-01-01 12:00:07.000
8   | 2015-01-01 12:00:08.000
9   | 2015-01-01 12:00:09.000

选择上一行的TimeStamp是一项相当简单的任务

e.g。

SELECT  MAX([TimeStamp])
FROM [MyTable] 
WHERE [TimeStamp] < '2015-01-01 12:00:02.000'

按预期获得2015-01-01 12:00:01.000。

但是我在从多个前面的行中选择一个TimeStamps列表时遇到了一些麻烦。

例如，如果我想获得Row＆gt; = 3＆amp;＆amp;的前面行的时间戳。＆lt; = 6

(i.e. 
    SELECT [TimeStamp]
    FROM [MyTable] 
    WHERE [Row] >= 3 AND [Row] <= 6

    =>

    TimeStamps     
    2015-01-01 12:00:03.000
    2015-01-01 12:00:04.000
    2015-01-01 12:00:05.000
    2015-01-01 12:00:06.000
)

我如何为这些结果行的每个获取前面的TimeStamp？

(i.e. TimeStamps 2015-01-01 12:00:02.000 2015-01-01 12:00:03.000 2015-01-01 12:00:04.000 2015-01-01 12:00:05.000 )

我见过很多与lag / lead相关的解决方案，但我对SQLServer 2008的使用很难改变。

Answer 1

你可以试试这个：

SELECT 
    t.TimeStamp,
    (
     SELECT MAX(t1.TimeStamp)
        FROM MyTable t1
        WHERE t1.Row < t.Row
    ) AS PrevTimeStamp
FROM MyTable t
WHERE t.Row >= 3 AND t.Row <= 6

这将为您提供并排的列，一个是当前列，另一个是前一列。

Answer 2

select b.id, max(a.timestamp)
from mytable a join mytable b 
on a.id < b.id
where b.id between --desired values
group by b.id

以下是样本数据的小提琴：fiddle

这对你有用吗？这将在所有先前的行上join，然后选择max时间戳，该时间戳始终是前一行的值，假设时间戳列按升序排序。

Answer 3

您可以使用CTE对行进行编号，并使用left join查找上一行：

; with  numbered as
        (
        select  row_number() over (order by TimeStamp) as rn
        ,       *
        from    YourTable
        )
select  cur.TimeStamp
,       prev.TimeStamp
from    numbered cur
left join
        numbered prev
on      prev.rn + 1 = cur.rn
where   cur.row between 3 and 6

为表/列表中的每个条目选择上一行

3 个答案: