with a1 as(select a.pmonth, sum(a.amount) sum1 from ( select month (submit_time) pmonth, uid, count(distinct (day (submit_time))) amount from exam_record where year (submit_time) = 2021 group by pmonth, uid ) a group by a.pmonth ) select month,round(a1.sum1 / mau, 2) avg_active_days,mau from (select...
牛客141649392号：我收到了科技规划岗的
投递人保科技等公司6个岗位 >