题解 | #0级用户高难度试卷的平均用时和平均得分#

select uid
,round(sum(score)/count(*),0) as avg_score
,round(avg(case when score is not null then timestampdiff(minute,start_time,submit_time) else duration end),1) as avg_time_took 
from exam_record e1
join examination_info e2 
on e1.exam_id=e2.exam_id 
where uid in (select uid from user_info where level=0)
and e1.exam_id IN (select exam_id from examination_info where difficulty="hard")
group by uid