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题解 | 牛客每个人最近的登录日期(四)

select distinct date, sum(case when date = first then 1 else 0 end ) as new from login l left join ( select user_id,min(date) as first from login ...

2026-04-25

1 16

题解 | 使用子查询的方式找出属于Action分类的所有电影对应的title,description

select t.title,t.description from(select f.title,f.description from film f join film_category fc on f.film_id=fc.film_id join category c o...

2026-04-11

0 23

题解 | 统计每个用户的平均刷题数

select up.university, qd.difficult_level, round(count(qpd.question_id)/count(distinct qpd.device_id),4) as avg_answer_cnt from user_profile up inn...

2026-04-03

0 25