select t2.level
,count( t.uid) as level_cnt
from exam_record t inner join examination_info t1 on t.exam_id =t1.exam_id 
inner join user_info t2 using(uid)
where t.score>80 and t1.tag = "SQL"
group by t2.level
order by level_cnt desc,level desc

为了保险起见，可以加入 level desc,不然可能会有别的报错

这个题比上一道更简单，这个适合放到非技术入门部分。

[[联合查询]]