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,sum(cnt) num
        from 
         (
            select
                case when a.author_level>=1 and a.author_level<=2 then '1-2级'
                when a.author_level>=3 and a.author_level<=4 then '3-4级'
                when a.author_level>=5 and a.author_level<=6 then '5-6级'
                else '0' end as level_cut,
                count(1) cnt
            from
                author_tb a,
                answer_tb b
            where
                a.author_id = b.author_id
                and char_len > 100
            group by
                a.author_level
        ) c
        group by c.level_cut
        order by sum(cnt) desc