通过代码

SELECT
    exam_record.uid,
    round(avg(IF(score is NULL,0,score)),0) avg_score,
    ROUND((sum(TIMESTAMPDIFF(MINUTE,start_time,submit_time))
    +sum(if(submit_time is null ,duration,0))) / count(*),1) avg_time_took
FROM
    exam_record
RIGHT JOIN
    examination_info
ON
    exam_record.exam_id = examination_info.exam_id
WHERE
    exam_record.uid in 
    (select
        user_info.uid
     from
        user_info
    left join
        exam_record
    on
        user_info.uid = exam_record.uid
    where
        level = 0)
    and difficulty = 'hard'
GROUP BY
    exam_record.uid

思路

这道题要求没多少啊，所以我们就直接简单粗暴来

请输出每个0级用户所有的高难度试卷考试平均用时和平均得分，未完成的默认试卷最大考试时长和0分处理

1. 0级用户筛选，我们选择用where in

select
  user_info.uid
from
  user_info
left join
  exam_record
on
  user_info.uid = exam_record.uid
where
  level = 0

2.考试平均用时和得分

用时保留一位小数，得分保留整数，if round， timesatamp都是老熟人啊就不多说了

round(avg(IF(score is NULL,0,score)),0) avg_score,
ROUND((sum(TIMESTAMPDIFF(MINUTE,start_time,submit_time))
+sum(if(submit_time is null ,duration,0))) / count(*),1) avg_time_took

还有最后的高难度跟group by 很简单的题啊。