巨简单题解 | #牛客直播各科目同时在线人数#

select t.course_id, course_name, max(num) as max_num
from(
select a1.course_id, count(distinct a2.user_id) as num
from attend_tb a1 join attend_tb a2 
on a1.course_id = a2.course_id and a1.out_datetime > a2.in_datetime and a1.out_datetime <= a2.out_datetime
group by 1, a1.user_id)t join course_tb c on t.course_id = c.course_id
group by 1,2
order by 1

找有时间交叉的就行了 然后计数