select 
    [user],
    avg(rating) as [avg]
from 
(select     a.[user],b.rating from      u_near a join     u1_base b on a.near1=b.userid and b.itemid='?'
union all 
select     a.[user],b.rating from      u_near a join     u1_base b on a.near2=b.userid and b.itemid='?'
union all 
select     a.[user],b.rating from      u_near a join     u1_base b on a.near3=b.userid and b.itemid='?'
union all 
select     a.[user],b.rating from      u_near a join     u1_base b on a.near4=b.userid and b.itemid='?'
union all 
select     a.[user],b.rating from      u_near a join     u1_base b on a.near5=b.userid and b.itemid='?'
)T
group by [user]

--> 已知u_near表中的user值
declare @user int
set @user=???

--> 还已知itemid
declare @itemid int
set @itemid=???

--> 要得到rating的平均值并返回
select avg(rating) from u1_base where itemid=@itemid and userid in
(
    --> 选择出 near1...near5 这五个分别作为u1_base表中的 userid
    select near1 from u_near where [user]=@user
    union all
    select near2 from u_near where [user]=@user
    union all
    select near3 from u_near where [user]=@user
    union all
    select near4 from u_near where [user]=@user
    union all
    select near5 from u_near where [user]=@user
)