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！！！
[解决办法]

引用:

SQL code?123456789select 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，r……

正解，
分析函数，改成rank更好吧，可以查询出并列的。
[解决办法]
SELECT empid
FROM (SELECT AVG(hours), empid
FROM workload
GROUP BY empid
ORDER BY AVG(hours) DESC)
WHERE ROWNUM = 1