题解 | #某乎问答高质量的回答中用户属于各级别的数量#

select
	level_cut,
    count(*) as num
from (
	select
		*,
		case 
			when author_level between 1 and 2 then '1-2级'
			when author_level between 3 and 4 then '3-4级'
			when author_level between 5 and 6 then '5-6级'
			else null
		end as level_cut
	from answer_tb
	join author_tb
		using (author_id)
) t
where t.char_len >= 100
group by level_cut
order by num desc