题解 | #大小写混乱时的筛选统计#

with q as 
(select 
		tag,
		count(start_time) as answer_cnt
from exam_record er right join examination_info ei
on er.exam_id = ei.exam_id
where start_time is not null
group by tag),
qq as
(
select 
		tag
from exam_record er right join examination_info ei
on er.exam_id = ei.exam_id
where start_time is not null
group by tag
having count(start_time)<3
)
select qq.tag, q.answer_cnt
from q,qq
where q.tag = upper(qq.tag) and q.tag != qq.tag;#如果转换后tag并没有发生变化，不输出该条结果。