题解 | #SQL类别高难度试卷得分的截断平均值#

SELECT e.tag, e.difficulty, round( (SUM(er.score) - max(er.score) - min(er.score)) / (count(er.score) - 2), 1)  as clip_avg_score
from examination_info e join   exam_record er  using(exam_id)
where tag = 'SQL' and difficulty = 'hard';