题解 | #SQL类别高难度试卷得分的截断平均值#

select
    "SQL" AS tag,
    "hard" as difficulty,
    
       format((sum(a.score) - max(a.score) - min(a.score)) / ((count(1)) - 2),1) 
     as clip_avg_score
from
    exam_record a
    join examination_info b on a.exam_id = b.exam_id
    and b.difficulty = "hard"
    and a.submit_time is not null and b.tag = "SQL"

33%.......什么情况