题解 | #SQL类别高难度试卷得分的截断平均值#

select 
    tag,
    difficulty,
    round((sum(score)-max(score)-min(score))/(count(score)-2),1) as clip_avg_score #注意这里如果是
	#count(exam_id)会算出错误结果
from
    exam_record as main 
join examination_info as b 
on b.exam_id = main.exam_id
where tag = 'SQL' and difficulty = 'hard'
group by tag,difficulty