题解 | #得分不小于平均分的最低分#

select min(er.score) as min_score_over_avg
# 获取符合列最小的数
from exam_record er 
join examination_info ei using(exam_id)
where tag = 'SQL' AND score >=
(
    select avg(e1.score)
    from exam_record e1 
    join examination_info e2 using(exam_id)
    where tag = 'SQL'
)
# 选中sql类，分数大于平均分，这里的平均分用子查询获取直接使用