【牛客】SQL135 每个6/7级用户活跃情况

描述

现有用户信息表user_info（uid用户ID，nick_name昵称, achievement成就值, level等级, job职业方向, register_time注册时间）：

试卷信息表examination_info（exam_id试卷ID, tag试卷类别, difficulty试卷难度, duration考试时长, release_time发布时间）：

试卷作答记录表exam_record（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分）：

题目练习记录表practice_record（uid用户ID, question_id题目ID, submit_time提交时间, score得分）：

请统计每个6/7级用户总活跃月份数、2021年活跃天数、2021年试卷作答活跃天数、2021年答题活跃天数，按照总活跃月份数、2021年活跃天数降序排序。由示例数据结果输出如下：

解释：6/7级用户共有5个，其中1006在202109、202108、202008共3个月活跃过，2021年活跃的日期有20210907、20210804、20210803、20210802共4天，2021年在试卷作答区20210907活跃1天，在题目练习区活跃了3天。

代码1：

with cte1 as(
    select uid
    from
    user_info
    where level=6 or level=7
),
cte2 as(
    select uid,start_time as act_time,'exam' as label from exam_record
    union
    select uid,submit_time as act_time,'question' as lebel from practice_record
),
cte3 as(
    select uid,
    count(distinct left(act_time,7)) as act_month_total
    from cte1 left join cte2 using(uid)
    group by uid
    order by act_month_total desc
),
cte4 as(
    select uid,
    count(distinct left(act_time,10)) as act_days_2021
    from cte1 left join cte2 using(uid)
    where left(act_time,4)=2021
    group by uid
),
cte5 as(
    select uid,
    count(distinct left(act_time,10)) as act_days_2021_exam
    from cte1 left join cte2 using(uid)
    where left(act_time,4)=2021 and label='exam'
    group by uid
),
cte6 as(
    select uid,
    count(distinct left(act_time,10)) as act_days_2021_question
    from cte1 left join cte2 using(uid)
    where left(act_time,4)=2021 and label='question'
    group by uid
)
 
select uid,
ifnull(act_month_total,0),ifnull(act_days_2021,0), 
ifnull(act_days_2021_exam,0),ifnull(act_days_2021_question,0)
from cte1 left join cte3 using(uid)
left join cte4 using(uid)
left join cte5 using(uid)
left join cte6 using(uid)
order by act_month_total desc,act_days_2021 desc

代码2（来源于讨论区的机智写法）：

SELECT uid,
COUNT(DISTINCT DATE_FORMAT(act_time,'%Y%m')) as act_month_total,
COUNT(DISTINCT IF(YEAR(act_time)=2021,act_time,null)) as act_days_2021,
COUNT(DISTINCT IF(YEAR(act_time)=2021 and tag='exam',act_time,null)) as act_days_2021_exam,
COUNT(DISTINCT IF(YEAR(act_time)=2021 and tag='question',act_time,null)) as act_days_2021_question
FROM user_info
LEFT JOIN  (
        SELECT uid,DATE(start_time) as act_time,'exam' as tag
        FROM exam_record
        UNION
        SELECT uid,DATE(submit_time) as act_time,'question' as tag
        FROM practice_record
)t
using(uid)
WHERE level>=6
GROUP BY uid
ORDER BY act_month_total DESC,act_days_2021 DESC;