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 between 5 and 6 then '5-6级' 
            when author_level between 3 and 4 then '3-4级'
            else '1-2级' end 
            as level_cut,
            count(case when char_len >= 100 then issue_id else null end) as 高质量评论数
            from author_tb join answer_tb on author_tb.author_id = answer_tb.author_id
            group by level_cut
            having 高质量评论数 > 0
            order by 高质量评论数 desc;