一个select查询问题

[wukankan52]

tab1
name , count
A,1
B,1

tab2
name,count
A,NULL
B,NULL
C,NULL
D,NULL

在不改变tab2内容的情况下得到(将tab1中name和tab2匹配的值给tab2，不匹配赋值0)
A,1
B,1
C,0
D,0

[bancxc]

select tab2.name,isnull(tab1.[count],0) as [count]
from tab2
left join tab1 on tab2.name=tab1.name

[--小F--]

select
 tab2.name,isnull(tab1.[count],0) as [count]
from
 tab2
left join tab1 on
 tab2.name=tab1.name

[bancxc]

select tab2.name,isnull(tab1.[count],0) as [count]
from tab2
left join tab1 on tab2.name=tab1.name

[东那个升]

if  OBJECT_ID('tab1') is not null  drop table tab1
create table tab1
(name  varchar(10), [count] int)
if  OBJECT_ID('tab2') is not null  drop table tab2
create table tab2
(name  varchar(10), [count] int)

insert tab1 select 'A',1
insert tab1 select 'B',1


insert tab2 select 'A',NULL
insert tab2 select 'B',NULL
insert tab2 select 'C',NULL
insert tab2 select 'D',NULL


select tab2.name,isnull(tab1.[count],0) as [count]
from tab2
left join tab1 on tab2.name=tab1.name
 
 name       count
---------- -----------
A          1
B          1
C          0
D          0

(4 行受影响)