SQL经典题型解析:1-10题

已知是以下4张表：

–1.学生表

Student(s_id, s_name, s_birth, s_sex) –学生编号,学生姓名, 出生年月,学生性别

–2.课程表

Course(c_id, c_name, t_id) – –课程编号, 课程名称, 教师编号

–3.教师表

Teacher(t_id, t_name) –教师编号,教师姓名

–4.成绩表

Score(s_id, c_id, s_score) –学生编号,课程编号,分数

首先，我们需要搞清楚这4张表的关联关系，这样有助于编写对应的SQL语句。下面是我做的关于4个表的关联图：

关联关系

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搞清表之间的关联关系后，开始创建数据库和表，我用的是navicat客户端。

1.创建表：

(1)学生表：

CREATE TABLE `Student`(

`s_id` VARCHAR(20),

`s_name` VARCHAR(20) NOT NULL DEFAULT ' ',

`s_birth` VARCHAR(20) NOT NULL DEFAULT ' ',

`s_sex` VARCHAR(10) NOT NULL DEFAULT ' ',

PRIMARY KEY(`s_id`)

);

(2)成绩表：

CREATE TABLE `Score`(

`s_id` VARCHAR(20),

`c_id` VARCHAR(20),

`s_score` INT(3),

PRIMARY KEY(`s_id`,`c_id`)

);

(3)课程表：

CREATE TABLE `Course`(

`c_id` VARCHAR(20),

`c_name` VARCHAR(20) NOT NULL DEFAULT ' ',

`t_id` VARCHAR(20) NOT NULL,

PRIMARY KEY(`c_id`)

);

(4)教师表：

CREATE TABLE `Teacher`(

`t_id` VARCHAR(20),

`t_name` VARCHAR(20) NOT NULL DEFAULT ' ',

PRIMARY KEY(`t_id`)

);

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2.插入数据

(1)插入学生表数据：

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');

(2)插入成绩表数据：

insert into Score values('01','01',80);

insert into Score values('01','02',90);

insert into Score values('01','03',99);

insert into Score values('02','01',70);

insert into Score values('02','02',60);

insert into Score values('02','03',80);

insert into Score values('03','01',80);

insert into Score values('03','02',80);

insert into Score values('03','03',80);

insert into Score values('04','01',50);

insert into Score values('04','02',30);

insert into Score values('04','03',20);

insert into Score values('05','01',76);

insert into Score values('05','02',87);

insert into Score values('06','01',31);

insert into Score values('06','03',34);

insert into Score values('07','02',89);

insert into Score values('07','03',98);

(3)插入课程表数据：

insert into Course values('01','语文','02');

insert into Course values('02','数学','01');

insert into Course values('03','英语','03');

(4)插入教师表数据：

insert into Teacher values('01','张三');

insert into Teacher values('02','李四');

insert into Teacher values('03','王五');

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以上4张表已创建完成并插入数据，下面详细解析经典题型

****-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数

理解题意：查询学生信息及01、02课程的成绩在'01'课程分数大于'02'课程条件下

分析思路：由于要查询学生信息和分数，那么需要用2个表，即"Student"和"Score"表，多表查询需要用到Join语句连接，要使用子查询查询各课程的成绩。

方法一：

#查询'01'课程的成绩

SELECT s_id, s_score as s1 From score where c_id='01';

#查询'02'课程的成绩

SELECT s_id, s_score as s2 From score where c_id='02';

（注：s1,s2 分别是'01'和'02'课程对应分数的别名，方便作比较）

#查询'01'课程分数>'02'课程分数的s_id,s1,s2

SELECT r1.s_id, s1,s2 From 

(SELECT s_id, s_score as s1 From score where c_id='01')r1,

(SELECT s_id, s_score as s2 From score where c_id='02')r2 

where r1.s_id=r2.s_id and s1>s2;

得到的结果集如下：

#使用JOIN连接"student"表,得到学生信息及课程成绩

SELECT * FROM Student s Right Join

(SELECT r1.s_id, s1,s2 From 

(SELECT s_id, s_score as s1 From score where c_id='01')r1,

(SELECT s_id, s_score as s2 From score where c_id='02')r2 

where r1.s_id=r2.s_id and s1>s2)R

On s.s_id=r.s_id;

最终结果应该是这样的：

方法二：

#使用INNER JOIN 连接子查询，将满足条件的s_id, s1,s2查询出来

SELECT r1.s_id, s1,s2 from

(SELECT s_id, s_score as s1 from score where c_id='01')r1

INNER JOIN

(SELECT s_id, s_score as s2 from score where c_id='02')r2

on r1.s_id=r2.s_id and s1>s2

#再使用INNER JOIN 连接Student表和上面的临时表R

SELECT s.* ,R.s1,R.s2 from student s

INNER JOIN

(SELECT r1.s_id, s1,s2 from

(SELECT s_id, s_score as s1 from score where c_id='01')r1

INNER JOIN

(SELECT s_id, s_score as s2 from score where c_id='02')r2

on r1.s_id=r2.s_id and s1>s2)R

on s.s_id=R.s_id;

得到如下结果：

与方法一一样

总结：方法一和方法二都能得到正确的结果，相较于方法二，方法一逻辑更清晰，语句较短，比较推荐。

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-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数

解题思路与上题一致。

只需将上题的查询语句稍加改变即可，如下：

SELECT * FROM Student s Right Join

(SELECT r1.s_id, s1,s2 From

(SELECT s_id, s_score as s1 From score where c_id='01')r1,

(SELECT s_id, s_score as s2 From score where c_id='02')r2 

where r1.s_id=r2.s_id and s1

On s.s_id=r.s_id;

结果集：

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-- 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

分析思路：要使用AVG(),那么就要用到group by，结果要有学生姓名，那么需要用的join连接student表

方法一：子查询

#查询平均成绩大于等于60的学生编码和平均成绩

SELECT s_id,AVG(s_score) avs from score GROUP BY s_id HAVING avs>=60;

结果集

#把学生姓名加进来

SELECT s.s_id, s_name, r.avs from student s right join

(SELECT s_id,AVG(s_score) avs from score GROUP BY s_id HAVING avs>=60)r 

on s.s_id=r.s_id ;

结果集

方法二：不使用join语句

SELECT s.s_id, s_name, r.avs from student s ，

(SELECT s_id,AVG(s_score) avs from score GROUP BY s_id HAVING avs>=60)r 

where s.s_id=r.s_id ;

结果和上面一样。

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-- 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(包括有成绩的和无成绩的)

分析思路：有成绩的且平均分小于60的学生和没有成绩的同学，分2部再连接在一起

#没成绩的学生编码，姓名和平均成绩（null）

SELECT s_id, s_name, null from student where s_id not in(SELECT DISTINCT s_id from score) ;  

结果集

#查询平均成绩低于60分的学生编码，姓名和平均成绩

SELECT r.s_id, s_name, r.avs from student s,

(SELECT s_id, ROUND(avg(s_score),2) avs from score 

GROUP BY s_id HAVING avs<60)r 

where s.s_id=r.s_id;

结果集

用了ROUND函数，使平均成绩保留2位小数。

#最后使用UNION合并到一起

SELECT r.s_id, s_name, r.avs from student s, (SELECT s_id,ROUND(avg(s_score),2) avs from score GROUP BY s_id HAVING avs<60)r where s.s_id=r.s_id

UNION

SELECT s_id,s_name,null from student where s_id not in(SELECT DISTINCT s_id from score)

最终结果集

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-- 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

分析思路：要用到COUNT(),SUM(), group by 分组，student表和score表，join连接，注意是所有学生，也就是包括没成绩的，所以要以student表为基准表

SELECT r.s_id, s.s_name, r.count_course, r.total_score from student s

LEFT JOIN

(SELECT s_id, count(*) as count_course, sum(s_score) as total_score from score GROUP BY s_id)r on s.s_id=r.s_id;

结果集

还有另一种写法：

select s.s_id, s.s_name, count(sc.c_id) as count_course, sum(sc.s_score) as sum_score from student s

left join score sc on s.s_id=sc.s_id  GROUP BY s.s_id, s.s_name;

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-- 6、查询"李"姓老师的数量

分析思路:想到用LIKE，count()函数，teacher表

SELECT count(*) from teacher where t_name like "李%";

结果集

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-- 7、查询学过"张三"老师授课的同学的信息

分析思路：4张表都要用到，需要从teacher先找到张三老师的编号，在course表中找到教的课程，在从score表找选修这门课程的学生编码，在连接student表查询出学生信息。

SELECT * from student where s_id in

(SELECT s_id from score where c_id=

(SELECT c_id from course where t_id=

(SELECT T_id from teacher where t_name='张三')))

这还真是一层一层的查（嵌套查询）

结果集

还有一种方法:（联合查询）

SELECT student.* from student where s_id in

(SELECT s_id from score sc,course c,teacher t

where sc.c_id=c.c_id

and c.t_id=t.t_id

and t_name='张三');

结果是一样的。

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-- 8、查询没学过"张三"老师授课的同学的信息

分析思路：首先要知道张三老师教的哪门课程，然后逆向思维，先找到学过张三老师课程的学生，用not in找到没学过的学生信息

SELECT * from student where s_id not in

(SELECT s_id from score where c_id in

(SELECT c_id from course where t_id in

(SELECT t_id from teacher where t_name='张三')));

还有一种方法：

SELECT * from student where s_id not in

(SELECT s_id from score sc, course c, teacher t

where sc.c_id = c.c_id 

and c.t_id=t.t_id 

and t_name='张三')

相较第一种，第二种更简洁些。和7题相反，逻辑相似。

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-- 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

分析思路：分别查询学习2门课程的学生编号，再找到相同的，最后连接student表查询。

SELECT r.s_id,s_name,s_birth,s_sex from student

right join (SELECT s1.s_id from (SELECT s_id from score where c_id='01')s1

INNER JOIN (SELECT s_id from score where c_id='02')s2 on s1.s_id=s2.s_id)r

on student.s_id=r.s_id;

还有一种简洁的查询语句：

SELECT s.* from student s, score a, score b

where s.s_id = a.s_id  and s.s_id = b.s_id and a.c_id='01' and b.c_id='02';

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-- 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

分析思路：查询学生编号在学习01课程的学生编号但不在学习02课程的编号里。

SELECT s.* from student s where s.s_id in

(SELECT s_id from score where c_id='01') 

and s.s_id not in (SELECT s_id from score where c_id='02');

结果集

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以上10个题，其中1，3，4，8，9，10题很重要，也有些难度，需要搞清逻辑，当然方法不是唯一的，我相信还有其他解法的，学习sql一定要会做这几道题哦，大家相互参考，共同学习。