题解 | #月均完成试卷数不小于3的用户爱作答的类别#

select 
    tag,
    count(tag) as tag_cnt
from
    exam_record as c
inner join examination_info as d 
on c.exam_id = d.exam_id
where uid in(
    select 
        uid
    from
        exam_record 
    where submit_time is not null
    group by uid
    having count(exam_id)/count(distinct date_format(start_time,'%Y%m')) >= 3
)
group by tag
order by tag_cnt desc;

注意，使用子查询时，in()括号里的子查询只能返回一列，否则会报错