SQL - 选择倒数第二个日期

Question

我有了这样一个[ArchiveDate]列的表：

ArchiveDate 
2014-10-06 
2014-10-06 
2014-10-06 
2014-10-01 
2014-10-01 
2014-10-01 
2014-10-01 
2014-05-22 
2014-05-22 

我要选择倒数第二个日期，但是当我使用：

select max([ArchiveDate]) -1 'previousweek' 
from [PipelineArchive] 

我获得2014-10-05（该列不存在），而不是2014-10-01。

我不知道如何编码，以选择“最后一个”;任何帮助都感激不尽！

谢谢。

Answer 1

您需要按照ArchiveDate的降序排序，跳过一条记录，然后进行下一条记录。例如，在SQL Server 2012中，你可以这样来做：

SELECT DISTINCT [ArchiveDate] 
FROM [PipelineArchive] 
ORDER BY [ArchiveDate] DESC 
OFFSET (1) ROWS FETCH NEXT (1) ROWS ONLY 

Demo.

Answer 2

SELECT distinct([ArchiveDate]) 
FROM [PipelineArchive] 
WHERE [ArchiveDate] = (SELECT MAX([ArchiveDate]) AS second 
         FROM [PipelineArchive] 
         WHERE [ArchiveDate] < (SELECT MAX([ArchiveDate]) AS first 
               FROM [PipelineArchive]) 
         ) 

的最近日期是：

SELECT MAX([ArchiveDate]) AS first FROM [PipelineArchive] 

最大日起不少于那就是：

(SELECT MAX([ArchiveDate]) AS second 
FROM [PipelineArchive] 
WHERE [ArchiveDate] < (SELECT MAX([ArchiveDate]) AS first 
         FROM [PipelineArchive])) 

PRO

• 它的（相当）标准的SQL

CON

• 它不能简单地推广到n个日期