题解 | #得分不小于平均分的最低分#

近三个月未完成试卷数为0的用户完成情况

http://www.nowcoder.com/practice/4a3acb02b34a4ecf9045cefbc05453fa

select uid, count(score) as exam_complete_cnt from ( select uid, start_time, score, dense_rank() over(PARTITION by uid order by date_format(start_time, '%Y%m') DESC) as ranking from exam_record ) table1 where ranking <= 3 group by uid having count(start_time) = count(score) order by exam_complete_cnt desc, uid desc

全部评论

相关推荐

点赞 评论 收藏
分享
06-10 23:36
已编辑
首都经济贸易大学 C++
点赞 评论 收藏
分享
评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客网在线编程
牛客网题解
牛客企业服务