MySQL--关于MySQL的那些练习题
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);
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