select product_line,sum(defect_qty),sum(inspect_qty)
from (select product_line,defect_qty,inspect_qty from fr_badness group by product_line,defect_qty,inspect_qty)
group by product_line

------解决方案--------------------
SQL code

----------------------------
-- Author  :qiudf(day day up)
-- Date    :2011-10-12 9:10
-- Verstion:
----------------------------
--> 测试数据：[a]
with a as
(select 'a' as product_line , '1101' as barcode, '2' as defect_qty,
'10' as inspect_qty, 't1' as position, sysdate as create_date from dual
union all
select 'a' as product_line , '1101' as barcode, '2' as defect_qty,
'10' as inspect_qty, 't2' as position, sysdate as create_date from dual
union all
select 'a' as product_line , '1102' as barcode, '9' as defect_qty,
'20' as inspect_qty, 't3' as position, sysdate as create_date from dual
union all
select 'b' as product_line , '1103' as barcode, '3' as defect_qty,
'12' as inspect_qty, 't4' as position, sysdate as create_date from dual
union all
select 'b' as product_line , '1103' as barcode, '3' as defect_qty,
'12' as inspect_qty, 't5' as position, sysdate as create_date from dual
union all
select 'a' as product_line , '1101' as barcode, '2' as defect_qty,
'10' as inspect_qty, 't1' as position, sysdate as create_date from dual
) 
select product_line,sum(defect_qty),sum(inspect_qty)
from 
  (
         select product_line,barcode,defect_qty,inspect_qty,row_number() over(partition by product_line,barcode,defect_qty,inspect_qty order by product_line) rn  from a
  )
where rn=1 
group by product_line

--result:
a    11    30
b    3    12

------解决方案--------------------
#4楼的正解

------解决方案--------------------
select product_line,sum(distinct defect_qty),sum(distinct inspect_qty)
from 你的表

------解决方案--------------------
探讨

引用:

引用:
现在就是 sum(defect_qty),sum(inspect_qty)要过滤掉重复的条码，也就是重复的条码只算一次,相同条码的defect_qty和inspect_qty 一定是相同的，不同的分组一定没有……