--1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

SELECT st.*, class1, class2 FROM student st JOIN(SELECT * FROM(SELECT SId s1, score class1 FROM sc WHERE CId = '01') c1,

(SELECT SId s2, score class2 FROM sc WHERE CId = '02') c2WHERE s1 = s2 AND class1 >class2) cON st.SId =c.s1;--1.1 查询同时存在" 01 "课程和" 02 "课程的情况

SELECT * FROM((SELECT * FROM sc WHERE CId = '01') t1JOIN(SELECT * FROM sc WHERE CId = '02') t2ONt1.Sid=t2.Sid);--1.2 查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )

SELECT * FROM((SELECT * FROM sc WHERE CId = '01') t1left JOIN(SELECT * FROM sc WHERE CId = '02') t2ONt1.Sid=t2.Sid);--1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

select * fromscwhere sc.CId = '02' ANDsc.SIdNOT IN(SELECT SId FROM sc WHERE sc.CId = '01');--2.查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩

SELECT st.SId, Sname, AVG(score) 'avg' FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING avg >= 60;--3.查询在 SC 表存在成绩的学生信息

SELECT * FROM student WHERE SId IN(SELECT DISTINCT SId FROMsc);--4.查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和

SELECT st.sid, st.Sname, COUNT(sc.CId) 'nums', SUM(sc.score) 'sum' FROMstudent stJOIN sc on st.SId =sc.SIdGROUP BYst.SId;--4.1显示没选课的学生(显示为NULL)

SELECT st.sid, st.Sname, CASE WHEN COUNT(sc.SId) > 0 THEN COUNT(sc.SId) ELSE NULL END 'nums', SUM(sc.score) 'sum' FROMstudent stleft JOIN sc on st.SId =sc.SIdGROUP BYst.SId;--4.2查有成绩的学生信息

SELECT * FROM student WHERE SId IN (SELECT DISTINCT SId FROM sc); --适用于右表小

SELECT * FROM student st WHERE EXISTS (SELECT * FROM sc WHERE sc.SId = st.SId); --适用于右表大

--5.查询「李」姓老师的数量

SELECT COUNT(*) FROM teacher t WHERE t.Tname LIKE '李%';--6.查询学过「张三」老师授课的同学的信息

SELECT * FROM student st WHERE st.SId IN(SELECT SId from sc WHERE sc.CId =(SELECT CId FROM course WHERE course.TId =(SELECT TId FROM teacher WHERE teacher.Tname = '张三')));--7.查询没有学全所有课程的同学的信息

SELECT * FROM student WHERE SId NOT IN(SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROMcourse));--8.查询至少有一门课与学号为" 01 "的同学所学相同的同学的信息

SELECT * FROM student WHERE SId IN (SELECT SId FROM sc WHERE CId IN (SELECT CId FROM sc WHERE SId = '01'));--9.查询和" 01 "号的同学学习的课程完全相同的其他同学的信息--解法一

SELECT * FROM student WHERE SId IN(SELECT SId FROM sc GROUP BY SId HAVING SId <> '01' AND GROUP_CONCAT(CId ORDER BY CId) =(SELECT GROUP_CONCAT(CId ORDER BY CId) FROM sc WHERE SId = '01'));--解法二

SELECT * FROM student WHERE SId IN(SELECT SId FROM sc WHERE CId IN (SELECT CId FROM sc WHERE SId = '01') AND SId <> '01'

GROUP BYSIdHAVING COUNT(*) = (SELECT COUNT(*) FROM sc WHERE SId = '01'));--10.查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT student.Sname FROM student WHERE SId NOT IN(SELECT SId FROM sc WHERE CId in(SELECT course.CId FROM course WHERE TId =(SELECT teacher.TId FROM teacher WHERE tname = '张三')));--11.查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

SELECT st.SId, sname, avg(score) FROM student st JOIN sc ON st.sid = sc.SId GROUP BYst.SIdHAVING COUNT(score <= 60 OR NULL) >= 2;--错误解法：

select student.sid, student.sname, AVG(sc.score) fromstudent,scwherestudent.sid= sc.sid and sc.score< 60

group bysc.sidhaving count(*)>1;--12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT st.*, scoreFROM student st JOINscON st.SId = sc.SId AND sc.CId = '01' AND sc.score < 60

ORDER BY score DESC;--13.按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

SELECT * from sc JOIN(SELECT SId, avg(score) avg FROM sc GROUP BYSId) avgsON sc.SId =avgs.SIdORDER BY avgs.avg desc;--14.查询各科成绩最高分、最低分和平均分--以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率--及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90--要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列

SELECTsc.CId'课程 ID',count(*) '选修人数',

course.Cname'课程 name',MAX(score) '最高分',MIN(score) '最低分',AVG(score) '平均分',COUNT(score >= 60 OR NULL) / COUNT(*) '及格率',COUNT(score >= 70 AND score < 80 OR NULL) / COUNT(*) '中等率',COUNT(score >= 80 AND score < 90 OR NULL) / COUNT(*) '优良率',COUNT(score >= 90 OR NULL) / COUNT(*) '优秀率'

FROM sc join course ON sc.CId = course.CId GROUP BYsc.CIdORDER BY COUNT(*) desc, sc.CId;--15.按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

SELECT CId, SId, score , (SELECT COUNT(*) FROM sc WHERE CId = o.CId AND score > o.score) + 1rankFROM sc o ORDER BY CId, score DESC;--16.查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

SET @rank = 0;SELECT q.sid, q.sum, (@rank := @rank + 1) rank FROM(SELECT sid, sum(score) sum FROM sc GROUP BY SId ORDER BY sum) q;--17.统计各科成绩各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0] 及所占百分比

SELECTs.CId, c.Cname,

CONCAT(COUNT(score >= 85 OR NULL), ',',COUNT(score >= 85 OR NULL) / COUNT(*)) '[100-85]',COUNT(score < 85 AND score >= 70 OR NULL) '[85-70]',COUNT(score < 70 AND score >= 60 OR NULL) '[70-60]',COUNT(score < 60 OR NULL) '[60-0]'

FROM sc s JOIN course c ON s.CId = c.CId GROUP BYs.CId;--18. 查询各科成绩前三名的记录

SELECT o.* FROM sc o HAVING(SELECT COUNT(*) FROM sc WHERE o.CId = CId AND o.score < score) + 1 <= 3

ORDER BY CId, score DESC, SId;--19.查询每门课程被选修的学生数

SELECT CId, COUNT(*) FROM sc GROUP BYCId;--20.查询出只选修两门课程的学生学号和姓名

SELECTst.SId, st.SnameFROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING COUNT(*) = 2;--21.查询男生、女生人数

select ssex, count(*) fromstudentgroup byssex;--22.查询名字中含有「风」字的学生信息

SELECT * FROM student WHERE Sname LIKE '%风%';--23.查询同名学生名单，并统计同名人数

SELECT st.*, (SELECT COUNT(*) FROM student WHERE Sname = st.Sname) 'same' FROM student st WHERE st.SId IN(SELECT s1.SId FROM student s1 JOIN student s2 ON s1.SId != s2.SId AND s1.Sname =s2.Sname);--24.查询 1990 年出生的学生名单

SELECT * FROM student WHERE YEAR(Sage) = 1990;--25.查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列

SELECT CId, avg(score) avg

FROM sc GROUP BY CId ORDER BY avg DESC, CId;--26.查询平均成绩大于等于 85 的所有学生的学号、姓名和平均成绩

SELECT st.SId, Sname, AVG(score) avg

FROM student st JOIN sc s ON st.SId = s.SId GROUP BY st.SId HAVING avg >= 85;--27.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

SELECTst.Sname, s.scoreFROMstudent st, sc s, course cWHERE st.SId = s.SId AND s.CId = c.CId AND s.score < 60 AND c.Cname = '数学';--28.查询所有学生的课程及分数情况(存在学生没成绩，没选课的情况)

SELECTSname, st.SId, scoreFROM student st LEFT JOIN sc s ON st.SId =s.SId;--29.查询任何一门课程成绩在 70 分以上的姓名、课程名称和分数--这个我没太理解，理解一是任意一门成绩均在70分以上

SELECTst.Sname, c.Cname, s.scoreFROM student st, sc s, course c WHERE st.SId = s.SId AND s.cid = c.cid AND st.SId IN(SELECT sid FROM sc GROUP BY SId HAVING MIN(score) > 70);--理解二是存在一门成绩在70分以上即可满足条件

SELECTst.Sname, c.Cname, s.scoreFROM student st, sc s, course c WHERE st.SId = s.SId AND s.cid = c.cid AND st.SId IN(SELECT sid FROM sc GROUP BY SId HAVING MAX(score) > 70);--理解三就是找出所有大于70分的得分。

select student.sname, course.cname,sc.score fromstudent,course,scwhere sc.score>70

and student.sid =sc.sidand sc.cid =course.cid;SELECT * FROM sc WHERE score > 70

--30.查询存在不及格的课程

SELECT cid FROM sc GROUP BY CId HAVING MIN(score) < 60;select cid fromscwhere score< 60

group bycid;SELECT DISTINCT cid FROM sc WHERE score < 60;--31.查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名

SELECT sid, sname FROM student WHERE sid IN(SELECT SId FROM sc WHERE CId = '01' AND score >= 80);--32.求每门课程的学生人数

SELECT cid, count(*) FROM sc GROUP BYcid;--33.成绩不重复，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

SELECT st.*, score FROM student st JOIN sc s ON st.SId = s.SId AND s.CId =(SELECT CId FROM course WHERE TId =(SELECT TId FROM teacher WHERE Tname = '张三')) ORDER BY score DESC LIMIT 1;--34.成绩有重复的情况下，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

SELECT st.*, score, CId FROM student st JOIN sc s ON st.SId = s.SId AND s.CId =(SELECT CId FROM course WHERE TId =(SELECT TId FROM teacher WHERE Tname = '张三'))WHERE score = (SELECT max(score) FROM sc WHERE CId =s.CId);--35.查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩--这个问题其实一开始没太明白啥意思，后来理解为某个人的几科分数是一样的，需要把这个人找出来

select a.cid, a.sid, a.score from sc asainner joinscasbon a.sid =b.sidand a.cid !=b.cidand a.score =b.scoregroup bycid, sid;--36.查询每门功成绩最好的前两名

SELECT o.* FROM sc o HAVING (SELECT COUNT(*) FROM sc WHERE CId = o.CId AND score > o.score) + 1 <= 2 ORDER BYo.cid, sid;--37.统计每门课程的学生选修人数(超过 5 人的课程才统计)

SELECT CId, COUNT(*) sum FROM sc GROUP BY CId HAVING sum > 5;--38.检索至少选修两门课程的学生学号

SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) >= 2;--39.查询选修了全部课程的学生信息

SELECT * FROM student WHERE sid IN(SELECT SId FROM sc GROUP BY SId HAVING COUNT(*) = (SELECT COUNT(*) FROMcourse));--40.查询各学生的年龄，只按年份来算

SELECT SId, Sname, Sage, YEAR(NOW()) - YEAR(Sage)FROMstudent;--41.按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

SELECT SId '学生编号', Sname '学生姓名', TIMESTAMPDIFF(YEAR, Sage, NOW()) '学生年龄'

FROMstudent;--42.查询本周过生日的学生--有点复杂，需要拼接出本周的起止日期

SELECT * FROMstudentWHERE DATE(CONCAT(YEAR(NOW()),'-',MONTH(Sage), '-', DAY(Sage)))BETWEEN DATE(DATE_SUB(NOW(),INTERVAL WEEKDAY(NOW()) DAY))AND DATE(DATE_ADD(NOW(),INTERVAL 6 - WEEKDAY(NOW()) DAY));SELECT * FROMstudent--43. 查询下周过生日的学生--同42

SELECT * FROMstudentWHERE DATE(CONCAT(YEAR(NOW()),'-',MONTH(Sage), '-', DAY(Sage)))BETWEEN DATE(DATE_SUB(NOW(),INTERVAL WEEKDAY(NOW()) - 7 DAY))AND DATE(DATE_ADD(NOW(),INTERVAL 13 - WEEKDAY(NOW()) DAY));--44.查询本月过生日的学生

SELECT * FROM student WHERE month(sage) = month(NOW())--45.查询下月过生日的学生--注意本月是12月的话，下一个月份是1即可

SELECT * FROM student WHERE month(Sage) = (CASE WHEN month(NOW()) = 12 THEN 1 ELSE MONTH(NOW()) + 1 END);