SQL 两表关联，关联时取一字段的一部分做链接，求方法
如下面：
需要找出，#c中以#b表中domain为结尾的记录，数据量大希望有比较好的代码:
即找出：
1@abc.com
12@a.com
3@@b.com

create table #c(email varchar(10))
insert into #c values ('1@abc.com')
insert into #c values ('12@a.com')
insert into #c values ('3@@b.com')
insert into #c values ('123@d.com')
insert into #c values ('1@sina.com')

create table #b(domain varchar(10))
insert into #b values('abc.com')
insert into #b values('a.com')
insert into #b values('b.com')

select *
from #c
where RIGHT(email,len(email)-CHARINDEX('@',email,1)) in (select domain from #b)

[最优解释]
经过两位大神的引导，出来了一版。用join链接表，提高效能。

create table #c(email varchar(10))
insert into #c values ('1@abc.com')
insert into #c values ('12@a.com')
insert into #c values ('3@@b.com')
insert into #c values ('123@d.com')
insert into #c values ('4@b.com')
insert into #c values ('123@@A.com')
insert into #c values ('1@sina.com')
  
create table #b(domain varchar(10))
insert into #b values('abc.com')
insert into #b values('a.com')
insert into #b values('b.com')
  

select c.email
from #c c left join #b b on c.email like '%@'+b.domain
where b.domain is not null

[其他解释]

create table #c(email varchar(10))
insert into #c values ('1@abc.com')
insert into #c values ('12@a.com')
insert into #c values ('3@@b.com')
insert into #c values ('123@d.com')
insert into #c values ('1@sina.com')
 
create table #b(domain varchar(10))
insert into #b values('abc.com')
insert into #b values('a.com')
insert into #b values('b.com')
 
select *
from #c
where exists(select 1 from #b where #c.email like '%@'+domain)
/*
email
1@abc.com
12@a.com
3@@b.com
*/

[其他解释]
对字符串做这样的操作，其实无可避免性能底下。

--create table #c(email varchar(10))
 --insert into #c values ('1@abc.com')
 --insert into #c values ('12@a.com')
 --insert into #c values ('3@@b.com')
 --insert into #c values ('123@d.com')
 --insert into #c values ('1@sina.com')
  
 --create table #b(domain varchar(10))
 --insert into #b values('abc.com')
 --insert into #b values('a.com')
 --insert into #b values('b.com')
  
 SELECT  *
 FROM    #c
 WHERE   EXISTS ( SELECT 1
                  FROM   #b 
 
                  WHERE  REVERSE(SUBSTRING(REVERSE(email), 1,
                                           PATINDEX('%@%', REVERSE(email)) - 1)) = #b.DOMAIN )
 
 
 /*
 email
 ----------
 1@abc.com
 12@a.com
 3@@b.com
 
 (3 行受影响)
 
 */

[其他解释]
大版又出来了...
[其他解释]
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