我有一个非常基本的LEFT OUTER JOIN来返回来自左表的所有结果和来自一个大得多的表的一些附加信息。左表包含4935条记录,但当我left OUTER JOIN到另一个表时,记录计数明显更大。

据我所知,这是绝对的福音,一个LEFT OUTER JOIN将返回所有记录从左表与匹配的记录从右表和空值的任何行不能匹配,因此,这是我的理解,它应该不可能返回更多的行比存在于左表,但它的发生都一样!

SQL查询如下:

SELECT     SUSP.Susp_Visits.SuspReason, SUSP.Susp_Visits.SiteID
FROM         SUSP.Susp_Visits LEFT OUTER JOIN
                      DATA.Dim_Member ON SUSP.Susp_Visits.MemID = DATA.Dim_Member.MembershipNum

也许我在语法上犯了一个错误,或者我对LEFT OUTER JOIN的理解是不完整的,希望有人能解释这是如何发生的?


当前回答

由于左边的表包含4935条记录,我怀疑您希望结果返回4935条记录。试试这个:

create table table1
(siteID int, 
SuspReason int)

create table table2
(siteID int, 
SuspReason int)

insert into table1(siteID, SuspReason) values 
(1, 678), 
(1, 186), 
(1, 723)
    
insert into table2(siteID, SuspReason) values 
(1, 678),
(1, 965)
   
select distinct t1.siteID, t1.SuspReason
from table1 t1 left join table2 t2 on t1.siteID = t2.siteID and t1.SuspReason = t2.SuspReason

union 

select distinct t2.siteID, t2.SuspReason 
from table1 t1 right join table2 t2 on t1.siteID = t2.siteID and t1.SuspReason = t2.SuspReason

其他回答

由于左边的表包含4935条记录,我怀疑您希望结果返回4935条记录。试试这个:

create table table1
(siteID int, 
SuspReason int)

create table table2
(siteID int, 
SuspReason int)

insert into table1(siteID, SuspReason) values 
(1, 678), 
(1, 186), 
(1, 723)
    
insert into table2(siteID, SuspReason) values 
(1, 678),
(1, 965)
   
select distinct t1.siteID, t1.SuspReason
from table1 t1 left join table2 t2 on t1.siteID = t2.siteID and t1.SuspReason = t2.SuspReason

union 

select distinct t2.siteID, t2.SuspReason 
from table1 t1 right join table2 t2 on t1.siteID = t2.siteID and t1.SuspReason = t2.SuspReason

您的查询将返回比左侧表(SUSP)更多行的唯一方法。在你的例子中是sus_visitors),是条件(sus_sus_visitors。MemID = DATA.Dim_Member. membershipnum)匹配右边表DATA.Dim_Member中的多行。因此,DATA中有多行。其中DATA.Dim_Member.MembershipNum的值相同。你可以通过执行下面的查询来验证:

选择DATA.Dim_Member。MembershipNum, count(DATA. dim_member .MembershipNum) from DATA。根据DATA.Dim_Member.MembershipNum指定成员组

Table1                Table2
_______               _________
1                      2
2                      2
3                      5
4                      6

SELECT Table1.Id, 
       Table2.Id 
FROM Table1 
LEFT OUTER JOIN Table2 ON Table1.Id=Table2.Id

结果:

1,null
2,2
2,2
3,null
4,null

这不是不可能的。左表中的记录数是它将返回的最小记录数。如果右表有两条记录与左表中的一条记录相匹配,则将返回两条记录。

DATA中似乎有多行。每个SUSP的Dim_Member表。Susp_Visits行。