Question

我正在编写一个查询，该查询会返回最新和上个月数据的数据。

这里是数据库中的日期，每个月的日期可能不同（并不总是在1日发布）。所以，我需要拉最新的（最大）（2017-07-03和之前的日期2017-06-01）

filedate
2017-05-01
2017-06-01
2017-07-03

以下是我尝试的内容：

declare @filedate date = '2017-07-03'

select distinct max(filedate) filedate
from Table
where filedate = @filedate
order by filedate asc


select filedate
from Table
where filedate < (select min(@filedate) from Table)
group by filedate

和结果：

filedate (from the first query)
2017-07-03

filedate (from the second query)
2017-05-01
2017-06-01

Answer 1

这是一个神奇的问题：

SELECT DISTINCT TOP 2 filedate
FROM TABLE
ORDER BY filedate DESC

Answer 2

如何选择前2名？像这样：

select top 2 filedate
from Table
order by filedate desc

Answer 3

感谢@CuriousKid. ..这就是我为完成这项工作所做的工作。

declare @filedate date = (select max(filedate) from Table)  --'2017-07-03'

select distinct top 1 filedate as PriorDate
from Table
where filedate <> @filedate
order by filedate desc

Answer 4

试试这个：

select max(filedate)from
(select '2017-05-01' as filedate union all
select '2017-06-01' union all
select '2017-04-10' union all
select '2017-07-03') as a where filedate<

(select max(filedate)from
(select '2017-05-01' as filedate union all
select '2017-06-01' union all
select '2017-04-10' union all
select '2017-07-03') as a limit 1)

union ALL

select max(filedate)from
(select '2017-05-01' as filedate union all
select '2017-06-01' union all
select '2017-04-10' union all
select '2017-07-03') as a

结果：

2017-06-01
2017-07-03

您只需将示例数据替换为您的表名，如下所示：

select max(filedate)from
(select filedate from yourtable) as a where filedate<

(select max(filedate)from
(select filedate from yourtable) as a limit 1)

union ALL

select max(filedate)from
(select filedate from yourtable) as a

SQL上一个日期记录

4 个答案: