题解 | #每个6/7级用户活跃情况#
每个6/7级用户活跃情况
http://www.nowcoder.com/practice/a32c7c8590324c96950417c57fa6ecd1
做法有些复杂。 总代码:
select qqq.uid, act_month_total, act_days_2021, act_days_2021_exam, act_days_2021_question
from
(
select ui.uid,
count(distinct if(year(er.submit_time)=2021,date(er.submit_time),null)) as act_days_2021_exam,#date(er.submit_time)天数不含秒
count(distinct if(year(pr.submit_time)=2021,date(pr.submit_time),null)) as act_days_2021_question
from exam_record as er
left join practice_record as pr on er.uid = pr.uid
right join user_info as ui on ui.uid = er.uid
where `level` in (6,7)
group by ui.uid
) as qqq
left join
(
select uid,
count(distinct date_format(time,'%y%m')) as act_month_total,
count(distinct if(year(time)=2021,date(time),null)) as act_days_2021
from(
select ui.uid,start_time as time
from exam_record as er right join user_info as ui on er.uid = ui.uid
union all
select uid,submit_time as time
from practice_record
) as q
group by uid
) as qq
on qqq.uid = qq.uid
order by act_month_total desc, act_days_2021 desc;
1、统计总的月份和2021年总的天数
统计总的月份和总的天数,最容易想到的就是把试卷时间和练习时间放在同一列,即纵向联结,然后去重做统计。
select uid,
count(distinct date_format(time,'%y%m')) as act_month_total,
count(distinct if(year(time)=2021,date(time),null)) as act_days_2021
from(
select ui.uid,start_time as time
from exam_record as er right join user_info as ui on er.uid = ui.uid
union all
select uid,submit_time as time
from practice_record
) as q
group by uid
这里需要注意:在试卷表和练习表中,没有1005用户,但是输出我们需要将它输出,因为没有记录所以总的月份数和天数全为0。所以需要将它的记录加上。 这样就得到了表的前两列。
2、统计试卷和练习在2021年的天数
统计试卷和练习的天数,把试卷表和练习表横向合并,然后去重统计。
select ui.uid,
count(distinct if(year(er.submit_time)=2021,date(er.submit_time),null)) as act_days_2021_exam,#date(er.submit_time)天数不含秒
count(distinct if(year(pr.submit_time)=2021,date(pr.submit_time),null)) as act_days_2021_question
from exam_record as er
left join practice_record as pr on er.uid = pr.uid
right join user_info as ui on ui.uid = er.uid
where `level` in (6,7)
group by ui.uid
3、将步骤1和步骤2的表合并
因为步骤1和步骤2都将用户信息表合并在一起了,所以包含了所有用户uid。所以联结方式不影响。