select A.part,A.avgvalue,A.sumvalue from (
select t.*,sum(sal) over(partition by t.part) as sumvalue
,avg(sal) over(partition by t.part) as avgvalue from ATEST t)A
group by (part,avgvalue)
发现原来这个有点麻烦,下面这个就成
select t.part, sum(sal),avg(sal) from ATEST t group by t.part
还有这种方法
select distinct t.part,sum(sal) over(partition by t.part) as sumvalue from ATEST t