declare @test table(n int, item nvarchar(10))
insert @test select 0,'A001' union all
select 100,'B001' union all
select 100,'C001' union all
select 100,'D001' union all
select 0,'E001' union all
select 0,'F001' union all
select -1,'I001' union all
select -1,'J001' 

n的值只有100,0和-1三种情况
需要sum(n<>-1)/count(n<>-1)
只汇总n不等于-1的值然后求平均值
然后放在第三列
结果如下
n    item     result
0    A001     50
100    B001     50
100    C001     50
100    D001     50
0    E001     50
0    F001     50
-1    I001     50
-1    J001     50

declare @test table(n int, item nvarchar(10))
insert @test select 0,'A001' union all
select 100,'B001' union all
select 100,'C001' union all
select 100,'D001' union all
select 0,'E001' union all
select 0,'F001' union all
select -1,'I001' union all
select -1,'J001' 

select *,result=(select sum(n) from @test where n<>-1)/(select count(*) from @test where n<>-1) from @test
/*
n           item       result
----------- ---------- -----------
0           A001       50
100         B001       50
100         C001       50
100         D001       50
0           E001       50
0           F001       50
-1          I001       50
-1          J001       50

(8 行受影响)
*/

------解决方案--------------------
SQL code
declare @test table(n int, item nvarchar(10))
 
insert @test select 0,'A001' union all
select 100,'B001' union all
select 100,'C001' union all
select 100,'D001' union all
select 0,'E001' union all
select 0,'F001' union all
select -1,'I001' union all
select -1,'J001' 

select * ,(select sum(n) from @test where n<>-1)/(select count(*) from @test where n<>-1)as result from  @test group by n,item



(8 行受影响)
n           item       result
----------- ---------- -----------
-1          I001       50
-1          J001       50
0           A001       50
0           E001       50
0           F001       50
100         B001       50
100         C001       50
100         D001       50

(8 行受影响)

------解决方案--------------------
SQL code
select
 *,
 (select sum(n) from @test where n<>-1)/(select count(1) from @test where n<>-1) as result
from
 @test