Sql 今天和昨天的最新记录

我有一张有代码和日期的桌子

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

我需要得到今天的最新记录，以及昨天的最新记录A1，A2，A3，A4，如下所示

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以获取数据

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

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

---

这似乎给出了您的预期输出，包括昨天的两个“虚线”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来构造查询。第一个，

Dated

，只是从错误命名的

Date

列中删除时间部分

编号

然后根据日期和代码为每个结果分配行号

Codes

获取我们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   -

试试这个：

SELECT f.Name, c.code, MAX(y.[Date]) AS [Date]
FROM (SELECT -1 ID, 'yesterday' Name
      UNION ALL
      SELECT 0, 'today') f
     CROSS JOIN
     (SELECT code
      FROM yourTable
      GROUP BY code) c
      LEFT OUTER JOIN
      yourTable y ON c.code = y.code AND DATEDIFF(DAY, GETDATE(), y.[Date]) = f.ID
WHERE
    ISNULL(DATEDIFF(DAY, GETDATE(), y.[Date]), 0) > -2
GROUP BY
    f.Name, c.code, ISNULL(DATEDIFF(DAY, GETDATE(), y.[Date]), 0)

---

选择标志、代码、最大值（日期）。。。按旗帜分组，代码帮助我们帮助您：到目前为止您尝试了什么？对于“昨天”两张唱片，我只需要获取所有唱片。如果日期为将显示的日期，则其他wise将需要显示为空。