急求一sql语句,解决后立马给分!!!!
链表查询,
表A
字段:
webid code regnum
232 1143 23
242 1143 4
200 1143 21
212 1143 7
221 1143 17
111 1143 12
101 1143 6
231 1143 7
265 1143 9
表B
字段:
webid code hynum
232 1143 2
242 1143 8
200 1143 21
212 1143 7
要查询结果为:
webid code regnum hynum
232 1143 23 2
242 1143 4 8
200 1143 21 21
212 1143 7 7
221 1143 17 0
111 1143 12 0
101 1143 6 0
231 1143 7 0
265 1143 9 0
求sql语句
------解决方案--------------------select a.*,b.hynum from a left join b on a.webid=b.webid
------解决方案--------------------select a.*,isnull(b.hynum,0)hynum
from a left join b on a.webid=b.webid and a.code=b.code
------解决方案--------------------select a.*, b.hynum
from a left join b on a.webid=b.webid and a.code=b.code
------解决方案--------------------if object_id('A') is not null
drop table A
create table A
(
webid int primary key ,
code int ,
regnum int
)
if object_id('B') is not null
drop table B
create table B
(
webid int primary key ,
code int ,
hynum int
)
insert A
select 232,1143,23 union all
select 242,1143,4 union all
select 200,1143,21 union all
select 212,1143,7 union all