SQL加入最近的日期

Normally I would just do this in the code itself, but I am curious if this can be accomplished efficiently in TSQL.

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

Table 1 contains entries for each day. Table 2 contains entries only when the discount changes. A discount applied to a value is considered valid until a new discount is entered.

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。

Example data:

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

What I need returned is the following: T1 Date - T1 Value - T2 Discount

我需要返回的是以下内容：T1日期 - T1值 - T2折扣

Example data:

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

Possible or am I better off just continuing to do this in the code?

可能还是我最好继续在代码中执行此操作？

3 个解决方案

#1

I believe this subquery will do it (not tested).

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

Perhaps not the most performant however.

然而，也许不是最高效的。

Edit:

编辑：

Test code:

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

Results:

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

#2

No "nearest" query is going to be as efficient as an "equals" query, but this is another job for the trusty ROW_NUMBER:

没有“最近”的查询会像“等于”查询一样有效，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

#3

This works on oracle XE. Since sql server does have analytic functions, it shouldn't be to difficult to port it.

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) 
)

#1