2017-02-20 49 views
1

我有一张桌子,上面有不同货币的交易,其中一些日期属于周末或假期。例如:如何参加周末或节假日的最新汇率日期?

Date  currency_code ammount 
20-02-2017 EUR    55 
18-02-2017 GBP    33 
17-02-2017 EUR    44.55 

什么我Currency表看起来像一个例子:

SELECT rate,date, currency_code FROM exchangeTable: 

rate  Date   currency_code 
53,35  13-02-2017 ADP 
53,35  14-02-2017 ADP  
182,4  16-02-2017 ADP  
192,45 17-02-2017 ADP  
191,31 20-02-2017 ADP 

有一个简单的子查询我可以在我加入语句中使用,这将加入最新的货币日期到我的交易日,如果它在周末或假期?我想我应该在这里使用分区,但是没有太多经验。

left join (?????????) a on a.date = b.date and a.currency_code= b.currency_code 
+0

您是否试过在最后包含'WHERE'子句?就像'WHERE a.Date = b.Date'? – David

+0

替换???????用GROUP BY查询。 – jarlh

回答

0

首先,你需要JOIN所有行形成Currency表相同CurrencyCode,那么你可以使用RANK() PARTITION (...功能,选择一个与最近的日期相比,交易的日期。

SELECT 
    * 
FROM 
    (SELECT 
     t.*, 
     c.*, 
     RANK() OVER (PARTITION BY t.ID ORDER BY ABS(DATEDIFF(d, t.[Date], c.[Date])) ASC, c.[Date] DESC) rn 
    FROM 
     Transactions t 
    INNER JOIN 
     Currency c ON t.CurrencyCode = c.CurrencyCode) t 
WHERE 
    t.rn = 1 
0

您可以使用派生表ROW_NUMBER和分区来解决此问题。这样做可以消除使用DATEPART函数和dw(星期几)参数的周末来忽略任何周六和周日。对于假期,你必须有假期日期表,因为假期完全是主观的。

ROW_NUMBER允许您获得给定自定义排序和分区的行号索引。我们按货币进行分割,因此每次我们点击一​​种新货币时索引都会重置,我们按DATE DESC订购,因此每种货币的最近日期为1

-- create a sample table with the date, currency, and exchange rate 
create table rates (
    id int identity(1,1) primary key, 
    date date not null, 
    currency char(3) not null, 
    rate decimal(10,2) not null 
) 
go 

-- create table of holidays we'll use for excluding rates records later 
create table holidays (
    id int identity(1, 1) primary key, 
    date date not null, 
    name varchar(100) not null 
) 

-- create some sample data 
-- Feb 18 and 19 are Saturday and Sunday 
insert into rates (date, currency, rate) values 
('2017-02-16', 'GBP', 1.23), 
('2017-02-17', 'GBP', 1.24), 
('2017-02-18', 'GBP', 1.25), 
('2017-02-19', 'GBP', 1.26), 
('2017-02-20', 'GBP', 1.27), 
('2017-02-16', 'SGD', 2.23), 
('2017-02-17', 'SGD', 2.24), 
('2017-02-18', 'SGD', 2.25), 
('2017-02-19', 'SGD', 2.26), 
('2017-02-20', 'SGD', 2.27); 

insert into holidays (date, name) values 
('2017-02-20', 'National Cherry Pie Day'); -- this is a real thing 


with t as (
    select id, 
      date, 
      currency, 
      rate, 
      row_number() over (partition by currency order by date desc) as age 
    from rates 
    where datepart(dw, date) not in (1, 7) -- sunday, saturday 
      and date not in (select date from holidays) -- exclude holiday rates 
) 
select * from t where age = 1; 
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