题解 | #某乎问答高质量的回答中用户属于各级别的数量#

1、按照等级分类，求不同等级下的创作者数量 with aaa as ( select count(x.author_id) as count_aid, y.author_level from answer_tb x join author_tb y on x.author_id = y.author_id where char_len >= 100 group by 2 ) 3、不同等级数量求和，用2个union就可以啦，虽然代码有点长，不过好理解，也有其他的做法~ select

from ( select '5-6级', sum(count_aid) as ww from aaa where aaa.author_level in (5, 6) union all select '1-2级', sum(count_aid) as ww from aaa where aaa.author_level in (1, 2) union all select '3-4级', sum(count_aid) as ww from aaa where aaa.author_level in (3, 4) order by 2 desc ) as gg where gg.ww!=0