Problem: 1211. 查询结果的质量和占比

👨‍🏫 参考题解

Code

select query_name,round(avg(rating/position),2) as quality,round(100 * avg(rating < 3), 2) as poor_query_percentage
from Queries
group by query_name -- 到此结束过不了最后一个极端样例
having query_name is not null -- 得加上查询名 非空判断才行