题解 | #某乎问答高质量的回答中用户属于各级别的数量#

select case when author_level in (1,2) then '1-2级'
when author_level in (3,4) then '3-4级'
else '5-6级' end as 'level_cut',
count(at.author_id) as num
from answer_tb at
left join author_tb ab
on at.author_id=ab.author_id
where char_len>=100
group by case when author_level in (1,2) then '1-2级'
when author_level in (3,4) then '3-4级'
else '5-6级' end 
order by num desc