Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
533 views
in Technique[技术] by (71.8m points)

sql server - SQL Join on Nearest less than date

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

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.

Example data:

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

Example data:

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?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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

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.

Edit:

Test code:

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:

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

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...