如何在周末或假日的日期加入最近的货币兑换日期?

时间:2017-02-20 09:31:29

标签: sql tsql join greatest-n-per-group

我有一张包含不同货币交易的表格,其中一些日期属于周末或假日。例如:

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

我的Currency表格的示例:

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   

我可以在我的加入声明中使用一个简单的子查询,它会将最近的货币日期加入到我的交易日期,如果是在周末或假日吗?我想我应该在这里使用分区,但对它们没有多少经验。

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

2 个答案:

答案 0 :(得分:0)

您可以使用派生表ROW_NUMBER和分区来解决此问题。这样做是为了消除周末使用DATEPART函数和dw(星期几)参数来忽略任何星期六和星期日。对于假期,你必须有一个假日日期表,因为假期是完全主观的。

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。我们按货币进行分区,以便每次点击新货币时索引都会重置,我们会按DATE DESC排序,因此每种货币的最新日期为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;

答案 1 :(得分:0)

首先,您需要JOIN所有行与Currency表格相同CurrencyCode,然后您可以使用RANK() PARTITION (...函数选择最多的行与交易日期相比的最近日期。

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