我有一个包含代码和日期的表
Code Date
----------------------------
A1 21 May 2015 15:47
A2 21 May 2015 10:30
A3 20 May 2015 10:30
A4 21 May 2015 10:30
A1 19 May 2015 15:20
A2 21 May 2015 12:30
A3 19 May 2015 05:30
A4 18 May 2015 15:38
A1 19 May 2015 05:30
A2 20 May 2015 05:30
A3 21 May 2015 05:30
A4 21 May 2015 05:30
A3 21 May 2015 06:30
A1 21 May 2015 05:30
我需要获得今天的最新记录,以及昨天A1,A2,A3,A4的最新记录显示如下
Flag Code Date
-----------------------------------------
Today A1 21 May 2015 15:47
Today A2 21 May 2015 10:30
Today A3 21 May 2015 06:30
Today A4 21 May 2015 10:30
Yesterday A1 --
Yesterday A2 20 May 2015 05:30
Yesterday A3 20 May 2015 10:30
Yesterday A4 --
帮助我如何编写查询来获取数据
答案 0 :(得分:2)
这似乎给出了您的预期输出,包括两个"虚线"昨天的结果:
declare @t table (Code char(2),[Date] datetime)
insert into @t(Code,Date) values
('A1','2015-05-21T15:47:00'),
('A2','2015-05-21T10:30:00'),
('A3','2015-05-20T10:30:00'),
('A4','2015-05-21T10:30:00'),
('A1','2015-05-19T15:20:00'),
('A2','2015-05-21T12:30:00'),
('A3','2015-05-19T05:30:00'),
('A4','2015-05-18T15:38:00'),
('A1','2015-05-19T05:30:00'),
('A2','2015-05-20T05:30:00'),
('A3','2015-05-21T05:30:00'),
('A4','2015-05-21T05:30:00'),
('A3','2015-05-21T06:30:00'),
('A1','2015-05-21T05:30:00')
;With Dated as (
select *,DATEADD(day,DATEDIFF(day,0,[Date]),0) as BetterDate
from @t
), Numbered as (
select *,ROW_NUMBER() OVER (
PARTITION BY Code,BetterDate
ORDER BY [Date] desc) as rn
from Dated
), Codes as (
select distinct Code from @t
)
select
'Today' as Occasion,
c.Code,
COALESCE(CONVERT(varchar(20),n1.Date),'-') as Date
from
Codes c
left join
Numbered n1
on
c.Code = n1.Code and
n1.rn = 1 and
n1.BetterDate = DATEADD(day,DATEDIFF(day,0,GETDATE()),0)
union all
select
'Yesterday',
c.Code,
COALESCE(CONVERT(varchar(20),n1.Date),'-') as Date
from
Codes c
left join
Numbered n1
on
c.Code = n1.Code and
n1.rn = 1 and
n1.BetterDate = DATEADD(day,DATEDIFF(day,0,GETDATE()),-1)
order by Occasion,Code
在我们设置了样本数据之后,我们开始通过几个CTE构建查询。第一个Dated
只是从错误命名的Date
列中删除了时间部分。
Numbered
然后根据日期和代码为每个结果分配行号。
Codes
获取我们拥有数据的所有代码集,以便我们可以生成结果,无论特定代码是否具有今天或昨天的条目。
最后,我们使用这些CTE通过UNION ALL
结果:
Occasion Code Date
--------- ---- --------------------
Today A1 May 21 2015 3:47PM
Today A2 May 21 2015 12:30PM
Today A3 May 21 2015 6:30AM
Today A4 May 21 2015 10:30AM
Yesterday A1 -
Yesterday A2 May 20 2015 5:30AM
Yesterday A3 May 20 2015 10:30AM
Yesterday A4 -
答案 1 :(得分:0)
select case
when cast([Date] as date) >= cast(getdate() as date) then 'Today'
else 'Yesterday'
end as Flag
, Code
, Date
from (
select row_number() over (
partition by Code, cast([Date] as date)
order by [Date] desc) rn
, *
from YourTable
where cast([Date] as date) > dateadd(day, -1, cast(getdate() as date))
) as SubQueryAlias
where rn = 1
答案 2 :(得分:0)
以下是一些代码:
DECLARE @t TABLE(Code CHAR(2), Date DATETIME)
INSERT INTO @t
VALUES ( 'A1', '21 May 2015 15:47' ),
( 'A2', '21 May 2015 10:30' ),
( 'A3', '20 May 2015 10:30' ),
( 'A4', '21 May 2015 10:30' ),
( 'A1', '19 May 2015 15:20' ),
( 'A2', '21 May 2015 12:30' ),
( 'A3', '19 May 2015 05:30' ),
( 'A4', '18 May 2015 15:38' ),
( 'A1', '19 May 2015 05:30' ),
( 'A2', '20 May 2015 05:30' ),
( 'A3', '21 May 2015 05:30' ),
( 'A4', '21 May 2015 05:30' ),
( 'A3', '21 May 2015 06:30' ),
( 'A1', '21 May 2015 05:30' )
;WITH codes AS(SELECT DISTINCT Code, d FROM @t
CROSS JOIN (VALUES(CAST(GETDATE() AS DATE)),
(CAST(DATEADD(dd, -1, GETDATE()) AS DATE)))d(d))
SELECT CASE WHEN DAY(GETDATE()) = DAY(Date) THEN 'Today' ELSE 'Yestarday' END Day ,
c.Code ,
MAX(Date) AS Date
FROM codes c
LEFT JOIN @t t ON t.Code = c.Code AND CAST(t.Date AS DATE) = c.d
WHERE Date IS NULL OR Date > CAST(DATEADD(dd, -1, GETDATE()) AS DATE)
GROUP BY c.Code , DAY(Date)
ORDER BY Day, Code
输出:
Day Code Date
Today A1 2015-05-21 15:47:00.000
Today A2 2015-05-21 12:30:00.000
Today A3 2015-05-21 06:30:00.000
Today A4 2015-05-21 10:30:00.000
Yestarday A1 NULL
Yestarday A2 2015-05-20 05:30:00.000
Yestarday A3 2015-05-20 10:30:00.000
Yestarday A4 NULL
答案 3 :(得分:0)
试试这个:
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if (window != nullptr){
char* buffer = new char[100];
SendMessage(window , EM_GETSELTEXT, 0, LPARAM(buffer));
return buffer;
}
else{
return "";
}
}
else{
return "";
}
}