select tag,difficulty,
round((sum(score)-(max(score)+min(score)))/(count(score)-2),1) as clip_avg_score
# 计算分数之和减去最大值和最小值然后除以得分数量减去2
from examination_info ei
join exam_record er
on ei.exam_id = er.exam_id
where tag = 'SQL' and difficulty = 'hard'
and score is not null
# 根据题目要求选择科目SQL,难度hard，并且去除无效试卷（分数为null的）