用简洁的sql 查询出平均值的最大值
表workload(empid,workdate,hours),用一条sql查询出平均每天工作小时数最多的员工id，试了很多，都感觉写的很累赘，
select empid
from (select empid, avg(hours) as avghours
from workload
group by empid) c
where c.avghours>= (select max(avghours)
from (select empid, avg(hours) as avghours
from workload
group by empid))
用了两次相同的子查询，
谁能用简洁的一条sql搞定呢，麻烦献上一条哦，感谢！！！
[最优解释]
select empid
from(
select empid,
avg(hours),
row_number()over(order by avg(hours) desc) as rn
from workload
group by empid
)where rn = 1 -- rn = 1 平均时长第一，rn=2 平均时长第二 通用不？
