题解 | #根据指定记录是否存在输出不同情况#

select ui.uid,count(start_time)-count(submit_time) incomplete_cnt,round((count(start_time)-count(submit_time))/count(*),3) incomplete_rate
from user_info ui 
left join exam_record er 
on ui.uid = er.uid
where case when (
select uid from
(
select ui.uid,level,count(start_time)-count(submit_time) incomplete_cnt 
from user_info ui 
left join exam_record er 
on ui.uid = er.uid
where level = 0
group by ui.uid
having incomplete_cnt >2
) a
) is not null 
then level = 0
else 
ui.uid in (select uid
from exam_record) end
group by ui.uid
order by incomplete_rate;