SELECT username, (SUM(rating)/count(*)) as TheAverage, count(*) as TheCount WHERE month ='Aug' AND TheCount > 1 ORDER BY TheAverage DESC, TheCount DESC该表如下:我试图计算出每个用户的平均评分,然后根据平均评分和评分数量对结
我想查询所有cno及其对应的平均学位,该学位至少有5名学生(sno表示学生数量),从3开始。我尝试了以下查询语句: select cno,avg(degree) from score where cno in (select cno from score group by cno HAVING我的MySql版本是8.0.18 Expression #1 of SELECT list is not in GROUP BY clause and contains nonaggregated column