--------------------------表结构--------------------------
student(StuId,StuName,StuAge,StuSex) 学生表
teacher(TId,Tname) 教师表
course(CId,Cname,C_TId) 课程表
sc(SId,S_CId,Score) 成绩表
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问题二十二：查询如下课程成绩第3名到第6名的学生成绩单：企业管理（001），马克思（002），UML（003），数据库（004）格式：[学生ID]，[学生姓名]，企业管理，马克思，UML，数据库，平均成绩
SELECT stu.StuId,stu.StuName, BM.Score AS BM,Marx.Score AS Marx,UML.Score AS UML,DB.Score AS DB, (IFNULL(BM.Score,0)+IFNULL(Marx.Score,0)+IFNULL(UML.Score,0)+IFNULL(DB.Score,0))/ (SUM(CASE WHEN IFNULL(BM.Score,0)=0 THEN 0 ELSE 1 END)+ SUM(CASE WHEN IFNULL(Marx.Score,0)=0 THEN 0 ELSE 1 END)+ SUM(CASE WHEN IFNULL(UML.Score,0)=0 THEN 0 ELSE 1 END)+ SUM(CASE WHEN IFNULL(DB.Score,0)=0 THEN 0 ELSE 1 END)) AS AvgScore FROM student stu LEFT JOIN sc AS BM ON BM.SId=stu.StuId AND BM.S_CId=‘001‘ LEFT JOIN sc AS Marx ON Marx.SId=stu.StuId AND Marx.S_CId=‘002‘ LEFT JOIN sc AS UML ON UML.SId=stu.StuId AND UML.S_CId=‘003‘ LEFT JOIN sc AS DB ON DB.SId=stu.StuId AND DB.S_CId=‘004‘ GROUP BY stu.StuId ORDER BY AvgScore DESC LIMIT 2,4;
答案仅供参考，不一定完全正确，若发现错误或有更好的，欢迎评论，互相交流，一起成长！！！
原文：https://www.cnblogs.com/love-ziji/p/13044950.html