今日最新及昨日最新纪录

时间:2015-05-21 09:30:04

标签: sql sql-server sql-server-2008 sql-server-2005 sql-server-2012

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

4 个答案:

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

构建结果集

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

Example at SQL Fiddle.

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:

 char* shortcut::getSelectedText(){
  POINT mouse;
  if (GetCursorPos(&mouse)){
  HWND window = WindowFromPoint(mouse);//get the HWND from the active application
    if (window != nullptr){
        char* buffer = new char[100];
        SendMessage(window , EM_GETSELTEXT, 0, LPARAM(buffer));
        return buffer;  
    }
    else{
        return "";
    }
  }
  else{
    return "";
  }
}