之前联系了一些MySQL的查询相关知识，现在补充作为一个记录，免得自己忘记。
致谢博主：https://blog.csdn.net/dehu_zhou/article/details/52881587
#建表语句与测试数据
--创建测试数据
create table Student(
S varchar(10),
Sname varchar(10),
Sage datetime,
Ssex nvarchar(10)
) ;
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
create table SC(
Sid varchar(10),
Cid varchar(10),
score decimal(18,1)
);
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);
create table Course(
Cid varchar(10),
Cname varchar(10),
Tid varchar(10)
);
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
create table Teacher(
Tid varchar(10),
Tname varchar(10)
);
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
练习题：
题一：查询"01"课程比"02"课程成绩高的学生的信息及课程分数。
--分析过程：
--基表：学生的信息（Student），课程分数（SC） --条件： #1.成绩自己比自己； #2.查询学生的信息； #方法1：使用内连接，并作为子查询条件进行查询 SELECT t2.*, t1.`01分数`, t1.`02分数` FROM ( SELECT a.Sid, a.score AS '01分数', b.score AS '02分数' FROM SC a INNER JOIN SC b ON a.Sid = b.Sid and a.Cid='01' and b.Cid='02' WHERE a.score > b.score ) t1, Student t2 where t1.Sid=t2.Sid; #方法2：使用三层内连接连续使用 #考点：三层嵌套内连接 SELECT a.*, b.score AS '01分数', c.score AS '02分数' FROM Student a INNER JOIN SC b ON a.Sid = b.Sid AND b.Cid = '01' INNER JOIN SC c ON a.Sid = c.Sid AND c.Cid = '02' WHERE b.score > c.score;
2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
#与题目1一致，修改成绩比较条件 SELECT a.*, b.score AS '01分数', c.score AS '02分数' FROM Student a INNER JOIN SC b ON a.Sid = b.Sid AND b.Cid = '01' INNER JOIN SC c ON a.Sid = c.Sid AND c.Cid = '02' WHERE b.score < c.score;
3、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
#考点，group by 子句以及having的条件删选 SELECT a.*, b.avg_score FROM Student a, ( SELECT t.Sid, round(avg(t.score), 2) AS avg_score FROM SC t GROUP BY t.Sid HAVING avg(t.score) > 60 ) b WHERE a.Sid = b.Sid; #+外连接版本 SELECT t1.*,t2.avg_score FROM ( SELECT t.Sid, round(avg(t.score), 2) AS avg_score FROM SC t GROUP BY t.Sid HAVING avg(t.score) > 60 ) t2 LEFT JOIN Student t1 ON t2.Sid = t1.Sid; #外连接凝练版本 SELECT t1.*, round(avg(t.score), 2) AS avg_score FROM SC t RIGHT JOIN Student t1 ON t.Sid = t1.Sid GROUP BY t.Sid HAVING avg(t.score) > 60;
4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
SELECT t1.*, round(avg(t.score), 2) AS avg_score FROM SC t RIGHT JOIN Student t1 ON t.Sid = t1.Sid GROUP BY t.Sid HAVING avg(t.score) < 60;
5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
SELECT b.Sname, a.Sid, count(DISTINCT a.Cid) as '选课总数', sum(a.score) as '总成绩' FROM SC a LEFT JOIN Student b ON a.Sid = b.Sid GROUP BY a.Sid;
6、查询"李"姓老师的数量
select count(*) from Teacher a where a.Tname like '李%';
7、查询学过"张三"老师授课的同学的信息
SELECT a.Tname, b.Cid, d.* FROM Teacher a, Course b, SC c, Student d WHERE a.Tname = '张三' AND a.Tid = b.Tid AND b.Cid = c.Cid AND c.Sid = d.Sid; #方法2：使用三层内连接连续使用 SELECT a.* FROM Student a INNER JOIN SC b ON a.Sid = b.Sid INNER JOIN Course c ON b.Cid = c.Cid INNER JOIN Teacher d ON c.Tid = d.Tid WHERE d.Tname = '张三' GROUP BY 1,2,3,4; #按照查询（select后面的）的1,2,3,4字段来分组 #方法3：方法2基础上补充 SELECT a.* FROM Student a LEFT JOIN ( SELECT a.* FROM Student a INNER JOIN SC b ON a.Sid = b.Sid INNER JOIN Course c ON b.Cid = c.Cid INNER JOIN Teacher d ON c.Tid = d.Tid WHERE d.Tname = '张三' GROUP BY 1,2,3,4 ) t ON a.Sid = t.Sid WHERE t.Sid IS NOT NULL;
8、查询没学过"张三"老师授课的同学的信息
#方法1：直接对原来学过张三老师课程的学生信息进行not in SELECT * FROM Student t WHERE t.Sid NOT IN ( SELECT d.Sid FROM Teacher a, Course b, SC c, Student d WHERE a.Tname = '张三' AND a.Tid = b.Tid AND b.Cid = c.Cid AND c.Sid = d.Sid ); #NOT IN 与 NOT EXISTS的用法区别 SELECT * FROM Student t WHERE not EXISTS ( SELECT * #此处返回的是整个结果集，如果单挑sid，则返回的结果与预期结果不一致 FROM Teacher a, Course b, SC c, Student d WHERE a.Tname = '张三' AND a.Tid = b.Tid AND b.Cid = c.Cid AND c.Sid = d.Sid AND d.Sid = t.Sid #此处为结果集的判断 );
9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
SELECT c.*,a.score as '01分数',b.score as '02分数' FROM Student c INNER JOIN SC a ON c.Sid = a.Sid INNER JOIN SC b ON a.Sid = b.Sid AND a.Cid = '01' AND b.Cid = '02';
10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
#重要，我没有做出来 ##方法：连续的左连接，筛选出差异结果 SELECT * FROM Student a LEFT JOIN SC b ON a.Sid = b.Sid AND b.Cid = '01' LEFT JOIN SC c ON a.Sid = c.Sid AND c.Cid = '02' WHERE b.Cid = '01' AND c.Cid IS NULL;
11、查询没有学全所有课程的同学的信息
SELECT * FROM Student b LEFT JOIN SC a ON a.Sid = b.Sid GROUP BY a.Sid HAVING count(a.cid) < 3; #完善方法 SELECT a.* FROM Student a LEFT JOIN SC b ON a.Sid = b.Sid LEFT JOIN ( SELECT COUNT(1) anum FROM Course ) c ON 1 = 1 #通过子查询和恒真条件筛选 GROUP BY 1, 2, 3, 4 HAVING MAX(c.anum) > COUNT(b.Cid);
12、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 DISTINCT b.* FROM Student b LEFT JOIN SC c ON b.Sid = c.Sid AND c.Cid IN ( SELECT a.Cid FROM SC a WHERE a.Sid = '01' ); #方法2： SELECT a.* FROM Student a INNER JOIN SC b ON a.Sid = b.Sid WHERE EXISTS ( SELECT 1 FROM SC WHERE Sid = '01' AND Cid = b.Cid ) GROUP BY 1, 2, 3, 4; #方法3：：通过组装cid字段比对进行筛选 #重点：GROUP_CONCAT(cid ORDER BY cid) 函数的用法 #1.需要配合group by 子句使用 #2.内部字段的排序的用法 SELECT DISTINCT t.* FROM Student t INNER JOIN SC t1 ON t.Sid = t1.Sid WHERE t.Sid IN ( SELECT t3.Sid FROM ( SELECT sid, GROUP_CONCAT(cid ORDER BY cid) AS cid_list FROM SC a WHERE a.sid = '01' GROUP BY sid ) t2 RIGHT JOIN ( SELECT sid, GROUP_CONCAT(cid ORDER BY cid) AS cid_list FROM SC a GROUP BY sid ) t3 ON t2.cid_list = t3.cid_list WHERE t2.sid IS NOT NULL );
14、查询没学过"张三"老师讲授的任一门课程的学生姓名
SELECT t.*, t1.Tid FROM Student t LEFT JOIN ( SELECT a.*, d.Tid FROM Student a LEFT JOIN SC b ON a.sid = b.sid LEFT JOIN Course c ON c.Cid = b.Cid LEFT JOIN Teacher d ON c.Tid = d.Tid AND d.Tname = '张三' WHERE d.Tid IS NOT NULL ) t1 ON t.Sid = t1.sid WHERE t1.tid IS NULL;
15、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 * FROM Student t LEFT JOIN ( SELECT sid, round(avg(score), 1) AS avg_score FROM SC a WHERE a.score < 60 GROUP BY sid HAVING count(*) > 1 ) t1 ON t.Sid = t1.sid WHERE t1.avg_score IS NOT NULL;
#优化 SELECT a.Sid , a.Sname , round(avg(score), 1) AS avg_score FROM Student a INNER JOIN SC b ON a.Sid = b.Sid GROUP BY 1, 2 HAVING SUM( CASE WHEN b.Score >= 60 THEN 0 ELSE 1 END ) >= 2;
16、检索"01"课程分数小于60，按分数降序排列的学生信息
SELECT a.* FROM Student a LEFT JOIN SC b ON a.Sid = b.Sid AND b.Cid = '01' WHERE b.score < 60 ORDER BY b.Cid DESC;
17、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 t.*, round((s01+s02+s03)/3,2) as avs from ( SELECT a.* ,SUM(CASE WHEN b.Cid='01' THEN b.score ELSE 0 END) AS s01 ,SUM(CASE WHEN b.Cid='02' THEN b.score ELSE 0 END) AS s02 ,SUM(CASE WHEN b.Cid='03' THEN b.score ELSE 0 END) AS s03 FROM Student a LEFT JOIN SC b ON a.Sid=b.Sid GROUP BY 1,2,3,4 ) t ORDER BY avs desc; select * from Student t LEFT JOIN ( select a.Sid ,max(case when a.Cid='01' THEN a.score ELSE 0 END) as s01 ,max(case when a.Cid='02' THEN a.score ELSE 0 END) as s02 ,max(case when a.Cid='03' THEN a.score ELSE 0 END) as s03 ,round(avg(case when a.score is null then 0 else a.score end),2) as avs from SC a group BY a.Sid ) t1 on t.Sid = t1.Sid ;
18、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 #考察重点再分组 select a.cid,b.Cname,max(a.score),min(a.score),avg(a.score) ,concat(ROUND(count(IF(a.score >=60 ,1,null))/count(1) * 100,2),'%') as '及格率' ,concat(ROUND(sum(case when a.score >= 70 and a.score < 80 then 1 else 0 end)/count(1) * 100,2),'%') as '中等率' ,concat(ROUND(sum(case when a.score >= 80 and a.score < 90 then 1 else 0 end)/count(1) * 100,2),'%') as '优良率' ,concat(ROUND(sum(case when a.score >= 90 then 1 else 0 end)/count(1) * 100,2),'%') as '优秀率' from SC a INNER JOIN Course b on a.Cid = b.Cid group by 1,2;
19、按各科成绩进行排序，并显示排名
#单科排序方法，自定义变量 SET @rn:=0; SELECT a.*,@rn:=@rn+1 as '排名' FROM( select a.*,b.score from Course a INNER JOIN SC b on a.Cid = b.Cid where a.Cid='01' ORDER BY a.Cid,b.score desc )a ; #按科目排序方法，通过自连接比对排序规则 SELECT a.*, count(b.Cid) + 1 AS tp FROM SC a LEFT JOIN SC b ON a.Cid = b.Cid AND a.score < b.score GROUP BY a.Sid, a.Cid ORDER BY a.Cid, a.score DESC;
20、查询学生的总成绩并进行排名
SET @ct := 0; SELECT (@ct := @ct + 1) AS '排名', b.Sname, ifnull(sum(a.score), 0) AS '总成绩' FROM SC a RIGHT JOIN Student b ON a.Sid = b.Sid GROUP BY a.Sid ORDER BY sum(a.score) DESC;
21、查询不同老师所教不同课程平均分从高到低显示
##没有统计未参考学生的平均分 select d.Tname,c.Cname,ROUND(sum(b.score)/count(a.Sid),2) as '科目平均分' from Student a LEFT JOIN SC b on a.Sid = b.Sid LEFT JOIN Course c on c.Cid = b.Cid LEFT JOIN Teacher d on c.Tid = d.Tid where b.Cid is not null group by b.Cid ORDER BY sum(b.score)/count(a.Sid) desc; SELECT a.*,b.Cname,AVG(c.Score) ascore FROM Teacher a INNER JOIN Course b ON a.Tid=b.Tid INNER JOIN SC c ON b.Cid=c.Cid GROUP BY 1,2,3 ORDER BY ascore DESC ; #统计未参考学生的平均分 select sum(case when b.cid = '01' then b.score else 0 end)/count(distinct a.Sid) as avg_01, sum(case when b.cid = '02' then b.score else 0 end)/count(distinct a.Sid) as avg_02, sum(case when b.cid = '03' then b.score else 0 end)/count(distinct a.Sid) as avg_03 from Student a LEFT JOIN SC b on a.Sid = b.Sid;
22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
#核心为题19按科目的成绩排序 SELECT c.Sname, d.cid, d.score, d.tp FROM Student c INNER JOIN ( SELECT a.*, count(b.Cid) + 1 AS tp FROM SC a LEFT JOIN SC b ON a.Cid = b.Cid AND a.score < b.score GROUP BY a.Sid, a.Cid HAVING count(b.Cid) + 1 IN (2, 3) ) d ON c.Sid = d.sid;
23、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],[0-60]及所占百分比
select t.*,c.Cname ,concat(round(t.bjg/sum(t.bjg+t.pt+t.lh+t.yx)* 100 ,2),'%') as bl_bjg ,concat(round(t.pt/sum(t.bjg+t.pt+t.lh+t.yx)* 100 ,2),'%') as bl_pt ,concat(round(t.lh/sum(t.bjg+t.pt+t.lh+t.yx)* 100 ,2),'%') as bl_lh ,concat(round(t.yx/sum(t.bjg+t.pt+t.lh+t.yx)* 100 ,2),'%') as bl_yx from ( select cid ,sum(case when a.score < 60 then 1 else 0 end) as 'bjg' ,sum(case when a.score > 60 and a.score <= 70 then 1 else 0 end) as 'pt' ,sum(case when a.score > 70 and a.score <= 80 then 1 else 0 end) as 'lh' ,sum(case when a.score > 80 then 1 else 0 end) as 'yx' from SC a group by a.cid ) t LEFT JOIN Course c on t.cid =c.cid GROUP BY t.cid ; #直接计算 SELECT a.Cid ,a.cname ,SUM(CASE WHEN b.score<=100 AND b.score>85 THEN 1 ELSE 0 END) AS 'yx' ,SUM(CASE WHEN b.score<=85 AND b.score>70 THEN 1 ELSE 0 END) AS 'lh' ,SUM(CASE WHEN b.score<=70 AND b.score>60 THEN 1 ELSE 0 END) AS 'jg' ,SUM(CASE WHEN b.score<=60 AND b.score>0 THEN 1 ELSE 0 END) AS 'bjg' ,SUM(CASE WHEN b.score<=100 AND b.score>85 THEN 1 ELSE 0 END)/COUNT(1) AS 'yx%' ,SUM(CASE WHEN b.score<=85 AND b.score>70 THEN 1 ELSE 0 END)/COUNT(1) AS 'lh%' ,SUM(CASE WHEN b.score<=70 AND b.score>60 THEN 1 ELSE 0 END)/COUNT(1) AS 'jg%' ,SUM(CASE WHEN b.score<=60 AND b.score>0 THEN 1 ELSE 0 END)/COUNT(1) AS 'bjg%' FROM Course a INNER JOIN SC b ON a.Cid=b.Cid GROUP BY 1,2 ;
24、查询学生平均成绩及其名次
set @mc := 0;
select t.* ,@mc := @mc + 1 as '名次' from (
select c.Sid,c.Sname,sum(b.score)/3 as pjf from Course a LEFT JOIN SC b on a.Cid = b.Cid
RIGHT JOIN Student c on b.Sid =c.Sid group by b.Sid ORDER BY pjf desc ) t ;
#方法2
SELECT a.*
,COUNT(b.Sid)+1
FROM (
SELECT a.*,AVG(CASE WHEN b.SCore IS NULL THEN 0 ELSE b.SCore END) AS aSCore
FROM Student a
LEFT JOIN SC b
ON a.Sid=b.Sid
GROUP BY 1,2,3,4
)a
LEFT JOIN(
SELECT a.*,AVG(CASE WHEN b.SCore IS NULL THEN 0 ELSE b.SCore END) AS aSCore
FROM Student a
LEFT JOIN SC b
ON a.Sid=b.Sid
GROUP BY 1,2,3,4
)b
ON a.aSCore<b.aSCore
GROUP BY 1,2,3,4,5 ;
25、查询各科成绩前三名的记录
##题22 方法的同种类型 SELECT c.Sname, d.cid, d.score, d.tp FROM Student c INNER JOIN ( SELECT a.*, count(b.Cid) + 1 AS tp FROM SC a LEFT JOIN SC b ON a.Cid = b.Cid AND a.score < b.score GROUP BY a.Sid, a.Cid HAVING count(b.Cid) + 1 <4 ) d ON c.Sid = d.sid order by d.cid; SELECT a.*,COUNT(b.cid)+1 AS ascore FROM SC a LEFT JOIN SC b ON a.cid=b.cid AND a.score<b.score GROUP BY 1,2,3 HAVING ascore<=3 ORDER BY a.cid,ascore ;
26查询每门课程被选修的学生数
SELECT a.* ,COUNT(b.sid) FROM Course a LEFT JOIN SC b ON a.cid=b.cid GROUP BY 1,2,3 ;
27、查询出只有两门课程的全部学生的学号和姓名
SELECT a.* FROM Student a LEFT JOIN SC b ON a.Sid = b.Sid GROUP BY b.Sid HAVING count(b.cid) = 2;
28、查询男生、女生人数
SELECT a.Ssex, count(a.Sid) FROM Student a GROUP BY a.Ssex;
#29、查询名字中含有"风"字的学生信息 select * from Student a where a.Sname like '%风%';
#30、查询同名同性学生名单，并统计同名人数 select * from Student a group by a.Sname HAVING count(a.Sname)>1; #31、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime) SELECT * FROM Student WHERE YEAR(sage)=1990 ;
#32、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
SELECT a.*,AVG(b.Score) as ore FROM Course a LEFT JOIN SC b ON a.cid=b.cid GROUP BY 1,2,3 ORDER BY ore DESC,a.cid ;
33、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 a.Sid, a.Sname, sum( CASE WHEN b.score IS NULL THEN 0 ELSE b.score END ) / 3 AS avgsc FROM Student a LEFT JOIN SC b ON a.Sid = b.Sid GROUP BY a.Sid HAVING avgsc > 85;
34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
SELECT a.*, b.score FROM Student a LEFT JOIN SC b ON a.Sid = b.sid LEFT JOIN Course c ON b.Cid = c.Cid WHERE b.score < 60 AND c.Cname = '数学' or c.Cname is null;
35、查询所有学生的课程及分数情况；
SELECT c.cname,a.*,b.score FROM Student a LEFT JOIN SC b ON a.Sid = b.Sid LEFT JOIN Course c on b.Cid = c.Cid ORDER BY c.cname, a.Sid, b.Cid;
36、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；
SELECT DISTINCT b.Sname, c.Cname, a.score FROM Student b LEFT JOIN SC a ON a.Sid = b.Sid LEFT JOIN Course c ON c.Cid = a.Cid WHERE a.score > 70;
#37 不及格的
SELECT DISTINCT b.Sname, c.Cname, a.score FROM Student b LEFT JOIN SC a ON a.Sid = b.Sid LEFT JOIN Course c ON c.Cid = a.Cid WHERE a.score < 60 or a.score is null;
#38查询课程编号为01且课程成绩在80分以上的学生的学号和姓名；
SELECT DISTINCT b.Sname, c.Cname, a.score FROM Student b LEFT JOIN SC a ON a.Sid = b.Sid LEFT JOIN Course c ON c.Cid = a.Cid WHERE a.score > 80 and a.Cid='01';
39、求每门课程的学生人数
SELECT a.Cid, count(a.Sid) FROM SC a GROUP BY a.Cid;
40、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
#方法1：排序获取结果
SELECT
*
FROM
Student a
LEFT JOIN SC b ON a.Sid = b.Sid
LEFT JOIN Course c ON b.Cid = c.Cid
LEFT JOIN Teacher d ON c.Tid = d.Tid
WHERE
d.Tname = '张三'
ORDER BY
b.score DESC
LIMIT 1;
#方法2： 通过内连接最大值匹配
SELECT a.*,b.score
FROM Student a
INNER JOIN SC b
ON a.Sid=b.Sid
INNER JOIN(
SELECT c.Cid ,MAX(c.score) AS maxscore FROM Teacher a
INNER JOIN Course b
ON a.Tid=b.Tid
INNER JOIN SC c
ON b.Cid=c.Cid
WHERE a.Tname='张三'
GROUP BY c.Cid )c
ON b.Cid=c.Cid AND b.score=c.maxscore ;
41、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
SELECT DISTINCT
c.Sname,
c.Sid,
a.Cid,
d.cname,
a.score
FROM
Student c
LEFT JOIN SC a ON a.Sid = c.Sid
INNER JOIN SC b ON a.sid = b.Sid
LEFT JOIN Course d ON a.Cid = d.Cid
WHERE
a.Cid != b.Cid
AND a.score = b.score;
#方法2
SELECT a.sid
,a.cid
,a.SCore
FROM SC a
INNER JOIN (
SELECT a.SCore
,b.sid
,COUNT(1)
FROM SC a
INNER JOIN Student b
ON a.sid=b.sid
GROUP BY a.SCore,b.sid
HAVING COUNT(1)>1
)b
ON a.sid=b.sid AND a.SCore=b.SCore ;
42、查询每门功成绩最好的前两名 ，同22
SELECT c.Sname, d.cid, d.score, d.tp FROM Student c INNER JOIN ( SELECT a.*, count(b.Cid) + 1 AS tp FROM SC a LEFT JOIN SC b ON a.Cid = b.Cid AND a.score < b.score GROUP BY a.Sid, a.Cid HAVING count(b.Cid) + 1 <3 ) d ON c.Sid = d.sid order by d.cid;
43、统计每门课程的学生选修人数（超过5人的课程才统计）。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
SELECT a.Cid, a.Cname, count(b.Sid) FROM Course a LEFT JOIN SC b ON a.Cid = b.Cid GROUP BY b.Cid HAVING count(b.Sid) > 5 ORDER BY count(b.Sid) DESC, b.Cid;
#方法3：：通过组装cid字段比对进行筛选#重点：GROUP_CONCAT(cid ORDER BY cid) 函数的用法#1.需要配合group by 子句使用#2.内部字段的排序的用法
SELECT DISTINCTt.*FROMStudent tINNER JOIN SC t1 ON t.Sid = t1.SidWHEREt.Sid IN (SELECTt3.SidFROM(SELECTsid,GROUP_CONCAT(cid ORDER BY cid) AS cid_listFROMSC aWHEREa.sid = '01'GROUP BYsid) t2RIGHT JOIN (SELECTsid,GROUP_CONCAT(cid ORDER BY cid) AS cid_listFROMSC aGROUP BYsid) t3 ON t2.cid_list = t3.cid_listWHEREt2.sid IS NOT NULL);