select er.uid as uid, avg(if(score is not null,score,0)as avg_score, avg(if(submit_time is not null, timestampdiff(minute,start_time,submit_time),duration)) as avg_time_took from exam_record as er join user_info as ui on ui.uid=er.uid join examination_info as ei on ei.exam_id=er.exam_id where level=0 and difficulty='hard' group by uid; 这个老报错,说有数组越界等异常,到底哪里错了?
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不愿透露姓名的神秘牛友
10-29 21:14
疯犬丨哈士奇:喜欢你的人会主动表白,对你有想法的人会很主动,所以要你的公司不会吊着你所以懂了吧
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