Question

通常我会在代码本身中执行此操作，但我很好奇是否可以在TSQL中有效地完成此操作。

Table 1 
Date - Value

Table 2
Date - Discount

表1包含每天的条目。表2仅在折扣更改时包含条目。在输入新折扣之前，应用于某个值的折扣被视为有效。

示例数据：

Table 1  
1/26/2010 - 10  
1/25/2010 - 9  
1/24/2010 - 8  
1/24/2010 - 9   
1/23/2010 - 7    
1/22/2010 - 10  
1/21/2010 - 11

Table 2
1/26/2010 - 2  
1/23/2010 - 1  
1/20/2010 - 0

我需要返回的内容如下：T1 Date - T1 Value - T2 Discount

示例数据：

1/26/2010 - 10 - 2    
1/25/2010 - 9  - 1  
1/24/2010 - 8  - 1  
1/24/2010 - 9  - 1  
1/23/2010 - 7  - 1    
1/22/2010 - 10 - 0  
1/21/2010 - 11 - 0

可能还是我最好继续在代码中执行此操作？

Answer 1

我相信这个子查询会做（未经测试）。

select *, 
   (select top 1 Discount 
    from table2 
    where table2.Date <= t.Date 
    order by table2.Date desc) as Discount
from Table1 t

然而，也许不是最高效的。

修改

测试代码：

create table #table1 ([date] datetime, val int) create table #table2 ([date] datetime, discount int) insert into #table1 ([date], val) values ('1/26/2010', 10) insert into #table1 ([date], val) values ('1/25/2010', 9) insert into #table1 ([date], val) values ('1/24/2010', 8) insert into #table1 ([date], val) values ('1/24/2010', 9) insert into #table1 ([date], val) values ('1/23/2010', 7) insert into #table1 ([date], val) values ('1/22/2010', 10) insert into #table1 ([date], val) values ('1/21/2010', 11) insert into #table2 ([date], discount) values ('1/26/2010', 2) insert into #table2 ([date], discount) values ('1/23/2010', 1) insert into #table2 ([date], discount) values ('1/20/2010', 0) select *, (select top 1 discount from #table2 where #table2.[date] <= t.[date] order by #table2.[date] desc) as discount from #table1 t drop table #table1 drop table #table2

结果：

2010-01-26 00:00:00.000 10 2 2010-01-25 00:00:00.000 9 1 2010-01-24 00:00:00.000 8 1 2010-01-24 00:00:00.000 9 1 2010-01-23 00:00:00.000 7 1 2010-01-22 00:00:00.000 10 0 2010-01-21 00:00:00.000 11 0

Answer 2

没有“最近”的查询会像“等于”查询一样高效，但这是可靠的ROW_NUMBER的另一项工作：

;WITH Discounts_CTE AS
(
    SELECT
        t1.[Date], t1.[Value], t2.Discount,
        ROW_NUMBER() OVER
        (
            PARTITION BY t1.[Date]
            ORDER BY t2.[Date] DESC
        ) AS RowNum
    FROM Table1 t1
    INNER JOIN Table2 t2
        ON t2.[Date] <= t1.[Date]
)
SELECT *
FROM Discounts_CTE
WHERE RowNum = 1

Answer 3

这是asof join的典型情况。在DolphinDB中，人们可以直接使用asof join有效地解决此问题。

测试代码：

table1 = table(2010.01.26 2010.01.25 2010.01.24 2010.01.24 2010.01.23 2010.01.22 2010.01.21 as date,  10 9 8 9 7 10 11 as val)
table2 = table(2010.01.26 2010.01.23 2010.01.20 as date, 2 1 0 as discount)
select date, val, discount from aj(table1, (select * from table2 order by date), `date)

Answer 4

添加到Joels答案...如果两个表中都存在ID，以下内容将提高性能：

select *, 
   (select top 1 Discount 
    from Table2 t2
    where t2.Date <= t1.Date 
    and t2.ID = t1.ID 
    order by t2.Date desc) as Discount
from Table1 t1

Answer 5

这适用于oracle XE。由于sql server确实具有分析功能，因此不应该很难移植它。

create table one (
    day date,
    value integer
);


create table two (
    day date,
    discount integer
);


insert into one values (trunc(sysdate), 10);
insert into one values (trunc(sysdate-1), 8);
insert into one values (trunc(sysdate-2), 1);
insert into one values (trunc(sysdate-3), 23);
insert into one values (trunc(sysdate-4), 3);
insert into one values (trunc(sysdate-5), 4);
insert into one values (trunc(sysdate-6), 8);
insert into one values (trunc(sysdate-7), 5);
insert into one values (trunc(sysdate-8),8);
insert into one values (trunc(sysdate-9), 8);
insert into one values (trunc(sysdate-10), 5);    


insert into two values (trunc(sysdate), 2);
insert into two values (trunc(sysdate-3), 1);
insert into two values (trunc(sysdate-5), 3);
insert into two values (trunc(sysdate-8), 1);


select day, value, discount, cnt,
    nvl(max(discount) over (partition by cnt) 
    ,0) as calc_discount
from (
    select day, value, discount,
        count(discount) over (order by day) as cnt
    from one
    left outer join two  
    using(day) 
)

SQL加入最近的日期

5 个答案: