如何在适逢周末或假期的日期加入最近的货币兑换日?

How to join most recent currency exchange date on a date which falls on a weekend or holiday?

我有一个 table 使用不同货币进行交易的交易,其中一些日期恰逢周末或假期。示例:

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

我的 Currency table 的示例:

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   

是否可以在我的连接语句中使用一个简单的子查询,如果它是周末或假期,它将把最近的货币日期连接到我的交易日期?我想我应该在这里使用分区,但没有太多经验。

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

您可以使用派生的 table、ROW_NUMBER 和分区来解决此问题。这样做的目的是使用 DATEPART 函数和 dw(星期几)参数来消除周末,以忽略任何周六和周日。对于假期,您必须有 table 个假期日期,因为假期完全是主观的。

ROW_NUMBER 允许您在给定自定义排序和分区的情况下获取行号的索引。我们按货币划分,因此每次我们击中新货币时都会重置索引,我们按 DATE DESC 排序,因此每种货币的最近日期是 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;

首先,您需要JOIN Currency table 的所有行具有相同的CurrencyCode,然后您可以使用RANK() PARTITION (... 函数来select 与交易日期相比最近的日期。

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