题解 | #Where in 和Not in#
SQL类别高难度试卷得分的截断平均值
http://www.nowcoder.com/practice/a690f76a718242fd80757115d305be45
SELECT tag,difficulty,ROUND((SUM(score)-max(score)-MIN(score))/(COUNT(score)-2),1) AS clip_avg_socre
FROM exam_record as er
join examination_info AS ei
WHERE er.exam_id=ei.exam_id and difficulty='hard' and tag='SQL'
select tag,difficulty,round((sum(score) - max(score)-min(score)) / (count(score)-2),1) as clip_avg_score from exam_record as er join examination_info AS ei where ei.exam_id = er.exam_id and tag = 'SQL' and difficulty = 'hard';
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