左加入自我与标准

Question

设置

create table #history (
adddttm date, 
number int 
) 
insert into #history values ('2013-01-01 08:56:00.000',1); 
insert into #history values ('2013-01-01 08:56:00.000',2); 
insert into #history values ('2013-02-13 08:56:00.000',2); 
insert into #history values ('2013-02-13 08:56:00.000',3); 

查询

select * 
from #history new 
left join #history old 
    on new.number = old.number 
where new.adddttm = '2013-02-13 08:56:00.000' 
and old.adddttm = '2013-01-01 08:56:00.000' 

我希望下面的查询返回：

----------|-|----------|- 
2013-02-13|2|2013-01-01|2 
2013-02-13|3|null  |null 

，但我从来没有得到过第二行。为什么这会留下连接跳过缺失的行？

Answer 1

很简单，因为在where子句中引用了old.adddttm。我写了一篇关于为什么的文章Fun with Outer Joins。再次非常简单，这是因为ON子句仅用于两个表之间的连接，而WHERE子句仅用于对结果集的限制。虽然这两者很容易混淆。

此查询将起作用。

select * 
from #history new 
left join #history old 
    on new.number = old.number 
    and old.adddttm = '2013-01-01 08:56:00.000' 
WHERE new.adddttm = '2013-02-13 08:56:00.000' 

Answer 2

您的where子句排除行。

select old.* 
from history new 
left join history old 
    on new.number = old.number and old.adddttm = '2013-01-01 08:56:00.000' 
where new.adddttm = '2013-02-13 08:56:00.000' 

现在，当它没有得到连接行时，您会得到预期的空值。