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题解 | #试卷发布当天作答人数和平均分#
select level, count(distinct uid) as level_cnt from exam_record join examination_info using (exam_id) join user_info using (uid) where tag = 'SQL' and...
Mysql
2021-12-07
0
304
题解 | #试卷发布当天作答人数和平均分#
select exam_id, count(distinct uid) as uv, round(sum(score)/count(uid),1) as avg_score from exam_record join examination_info using(exam_id) join user...
Mysql
2021-12-05
0
207