2022-08-17第八组---MySQL查询案例练习

发布于:2023-01-13 ⋅ 阅读:(233) ⋅ 点赞:(0)

目录

一、构建数据库

(1)student表

(2)course表

(3)teacher表

(4)scores表

(5)给四张表插入数据

二、需求

-- 1.查询'01'号学生的姓名和各科成绩 **

-- 2.查询各个学科的平均成绩和最高成绩**

-- 3.查询每个同学的最高成绩和科目名称****(子查询)

-- 4.查询所有姓张的同学的各科成绩**

-- 5.查询每个课程的最高分的学生信息*****(子查询)

 -- 6.查询名字中含有'张'或'李'字的学生的信息和各科成绩。

-- 7.查询平均成绩及格的同学的信息。(子查询)

 -- 8.将学生按照总分数进行排名。(从高到低)

--- 9.查询数学成绩的最高分、最低分、平均分。

-- 10.将各科目按照平均分排序。

-- 11.查询老师的信息和他所带的科目的平均分

-- 12.查询被"Tom"和"Jerry"教的课程的最高分和最低分

-- 13.查询每个学生的最好成绩的科目名称(子查询)

-- 14.查询所有学生的课程及分数

-- 15.查询课程编号为1且课程成绩在60分以上的学生的学号和姓名(子查询)

-- 16. 查询平均成绩大于等于70的所有学生学号、姓名和平均成绩

-- 17.查询有不及格课程的学生信息

-- 18.查询每门课程有成绩的学生人数

-- 19.查询每门课程的平均成绩,结果按照平均成绩降序排列,如果平均成绩相同,再按照课程编号升序排列

-- 20.查询平均成绩大于60分的同学的学生编号和学生姓名和平均成绩

-- 21. 查询有且仅有一个课程成绩在80分以上的学生信息

-- 22 .查询出只有三门课程的学生的学号和姓名

-- 23.查询有不及格课程的课程信息

-- 24 .查询至少选择4门课程的学生信息

-- 25. 查询没有选全所有课程的同学的信息

-- 26.查询选全所有课程的同学的信息

-- 27.查询各学生都选了多少门课

-- 28.查询课程名称为"java",且分数低于60分的学生姓名和分数

-- 29.查询学过"Tom"老师授课的同学的信息

-- 30.查询没学过"Tom"老师授课的学生信息


一、构建数据库

 (1)student表

DROP TABLE IF EXISTS student;/*如果存在student表就删除*/
CREATE TABLE `student` (
	`id` INT ( 10 ) PRIMARY KEY,/*主键*/
	`name` VARCHAR ( 10 ),
	`age` INT ( 10 ) NOT NULL,
    `gender`VARCHAR(2)
);

(2)course表

DROP TABLE IF EXISTS course;
CREATE TABLE `course` (
	`id` INT ( 10 ) PRIMARY KEY,/*主键*/
	`name` VARCHAR ( 10 ),
	`t_id` INT(10)
);

(3)teacher表

DROP TABLE IF EXISTS teacher;
CREATE TABLE `teacher` (
	`id` INT ( 10 ) PRIMARY KEY,/*主键*/
	`name` VARCHAR ( 10 )
);

(4)scores表

DROP TABLE IF EXISTS scores;
CREATE TABLE `scores` (
	`s_id` INT ( 10 ),
	`score` INT ( 10 ),
	`c_id`int(10),
	PRIMARY KEY(s_id,c_id)
);

(5)给四张表插入数据

insert into  student (id,name,age,gender)VALUES(1,'张小明',19,'男'),(2,'李小红',19,'男'),(3,'小刚',24,'男'),(4,'小龙',11,'男'),(5,'小丽',18,'男'),(6,'张小军',18,'女'),(7,'小航',16,'男'),(8,'小亮',23,'男'),(9,'小杰',22,'女'),(10,'李小虎',21,'男');

insert into  course (id,name,t_id)VALUES(1,'数学',1),(2,'语文',2),(3,'c++',3),(4,'java',4),(5,'php',null);


insert into  teacher (id,name)VALUES(1,'Tom'),(2,'Jerry'),(3,'Tony'),(4,'Jack'),(5,'Rose');


insert into  scores (s_id,score,c_id)VALUES(1,80,1);
insert into  scores (s_id,score,c_id)VALUES(1,56,2);
insert into  scores (s_id,score,c_id)VALUES(1,95,3);
insert into  scores (s_id,score,c_id)VALUES(1,30,4);
insert into  scores (s_id,score,c_id)VALUES(1,76,5);

insert into  scores (s_id,score,c_id)VALUES(2,35,1);
insert into  scores (s_id,score,c_id)VALUES(2,86,2);
insert into  scores (s_id,score,c_id)VALUES(2,45,3);
insert into  scores (s_id,score,c_id)VALUES(2,94,4);
insert into  scores (s_id,score,c_id)VALUES(2,79,5);

insert into  scores (s_id,score,c_id)VALUES(3,65,2);
insert into  scores (s_id,score,c_id)VALUES(3,85,3);
insert into  scores (s_id,score,c_id)VALUES(3,37,4);
insert into  scores (s_id,score,c_id)VALUES(3,79,5);

insert into  scores (s_id,score,c_id)VALUES(4,66,1);
insert into  scores (s_id,score,c_id)VALUES(4,39,2);
insert into  scores (s_id,score,c_id)VALUES(4,85,3);

insert into  scores (s_id,score,c_id)VALUES(5,66,2);
insert into  scores (s_id,score,c_id)VALUES(5,89,3);
insert into  scores (s_id,score,c_id)VALUES(5,74,4);


insert into  scores (s_id,score,c_id)VALUES(6,80,1);
insert into  scores (s_id,score,c_id)VALUES(6,56,2);
insert into  scores (s_id,score,c_id)VALUES(6,95,3);
insert into  scores (s_id,score,c_id)VALUES(6,30,4);
insert into  scores (s_id,score,c_id)VALUES(6,76,5);

insert into  scores (s_id,score,c_id)VALUES(7,35,1);
insert into  scores (s_id,score,c_id)VALUES(7,86,2);
insert into  scores (s_id,score,c_id)VALUES(7,45,3);
insert into  scores (s_id,score,c_id)VALUES(7,94,4);
insert into  scores (s_id,score,c_id)VALUES(7,79,5);

insert into  scores (s_id,score,c_id)VALUES(8,65,2);
insert into  scores (s_id,score,c_id)VALUES(8,85,3);
insert into  scores (s_id,score,c_id)VALUES(8,37,4);
insert into  scores (s_id,score,c_id)VALUES(8,79,5);

insert into  scores (s_id,score,c_id)VALUES(9,66,1);
insert into  scores (s_id,score,c_id)VALUES(9,39,2);
insert into  scores (s_id,score,c_id)VALUES(9,85,3);
insert into  scores (s_id,score,c_id)VALUES(9,79,5);

insert into  scores (s_id,score,c_id)VALUES(10,66,2);
insert into  scores (s_id,score,c_id)VALUES(10,89,3);
insert into  scores (s_id,score,c_id)VALUES(10,74,4);
insert into  scores (s_id,score,c_id)VALUES(10,79,5);

二、需求

-- 1.查询'01'号学生的姓名和各科成绩 **

SELECT
	s.id sid,
	s.`name` sname,
	c.`name` cname,
	sc.score 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON c.id = sc.c_id 
WHERE
	s.id = 1;

-- 2.查询各个学科的平均成绩和最高成绩**

SELECT
	c.id,
	c.`name`,
	AVG( sc.score ),
	max( sc.score ) 
FROM
	course c
	LEFT JOIN scores sc ON c.id = sc.c_id 
GROUP BY
	c.id,
	c.`name`;

-- 3.查询每个同学的最高成绩和科目名称****(子查询)

SELECT
	t.id 学生编号,
	t.`NAME` 学生姓名,
	c.id 课程编号,
	c.`NAME` 课程名称,
	r.score 最高成绩
FROM
	(
	SELECT
		s.id,
		s.`NAME`,(
		SELECT
			MAX( score ) 
		FROM
			scores r 
		WHERE
			r.s_id = s.id 
		) score 
	FROM
		student s 
	) t
	LEFT JOIN scores r ON r.s_id = t.id 
	AND r.score = t.score
	LEFT JOIN course c ON r.c_id = c.id;

-- 4.查询所有姓张的同学的各科成绩**

SELECT
	s.id,
	s.`name`,
	c.`name` cname,
	sc.score 
FROM
SELECT
	s.id,
	s.`name`,
	c.`name` cname,
	sc.score 
FROM
	student s
	LEFT JOIN scores sc ON sc.s_id = s.id
	LEFT JOIN course c ON c.id = sc.c_id 
WHERE
	s.`name` LIKE '张%';

-- 5.查询每个课程的最高分的学生信息*****(子查询)

SELECT
	* 
FROM
	student s 
WHERE
	id IN (
	SELECT DISTINCT
		r.s_id 
	FROM
		(
		SELECT
			c.id,
			c.NAME,
			max( score ) score 
		FROM
			student s
			LEFT JOIN scores r ON r.s_id = s.id
			LEFT JOIN course c ON c.id = r.c_id 
		GROUP BY
			c.id,
			c.NAME 
		) t
		LEFT JOIN scores r ON r.c_id = t.id 
	AND t.score = r.score 
	)

 -- 6.查询名字中含有'张'或'李'字的学生的信息和各科成绩。

SELECT
s.id AS 学生编号,
s.NAME AS 学生姓名,
c.NAME 课程名称,
sc.score 成绩
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON c.id = sc.c_id 
WHERE
	( s.`name` LIKE '李%' OR s.`name` LIKE '张%' );

-- 7.查询平均成绩及格的同学的信息。(子查询)

/*方法1*/
SELECT
	a.s_id AS 学生编号,
	AVG( a.score ) AS 平均成绩
FROM
	scores AS a 
GROUP BY
	a.s_id 
HAVING
	avg( a.score )>=60
/*方法2*/
SELECT
	* 
FROM
	student 
WHERE
	id IN (
	SELECT
		sc.s_id 
	FROM
		scores sc 
	GROUP BY
		sc.s_id 
	HAVING
	avg( sc.score ) >= 60 
	)

 -- 8.将学生按照总分数进行排名。(从高到低)

SELECT
	s.id AS 学生编号,
	s.NAME 学生姓名,
	sum( sc.score ) 学生总分数 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON c.id = sc.c_id 
GROUP BY
	s.NAME 
ORDER BY
	score DESC;

--- 9.查询数学成绩的最高分、最低分、平均分。

SELECT
	c.id AS 科目编号,
	c.`name` AS 科目名称,
	max( sc.score ) AS 最高分,
	min( sc.score ) AS 最低分,
	AVG( sc.score ) AS 平均分 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON c.id = sc.c_id 
WHERE
	c.id = 1;

-- 10.将各科目按照平均分排序。

SELECT
	c_id AS 科目编号,
	c.name AS 科目名称,
	AVG( sc.score ) AS 平均分 
FROM
	student s
	LEFT JOIN scores sc ON s.id = sc.s_id
	LEFT JOIN course c ON c.id = sc.c_id 
GROUP BY
	c_id 
ORDER BY
	sc.score DESC,
	c_id

-- 11.查询老师的信息和他所带的科目的平均分

SELECT
t.id 老师编号,
t.`name` 老师姓名,
AVG(sc.score) 科目平均成绩,
c.`name` 科目名称
FROM
teacher t
LEFT JOIN course c ON t.id = c.t_id 
LEFT JOIN scores sc ON c.id = sc.c_id
GROUP BY
  t.`name`,
  t.id;

-- 12.查询被"Tom"和"Jerry"教的课程的最高分和最低分

SELECT

t.id 老师编号,

t.`name` 老师名字,

c.`name` 课程名称,

max(sc.score) 课程最高分,

MIN(sc.score) 课程最低分

FROM

teacher t

LEFT JOIN course c ON t.id = c.t_id

LEFT JOIN scores sc ON c.id = sc.c_id

GROUP BY

t.id,

t.`name`;



-- 13.查询每个学生的最好成绩的科目名称(子查询)

SELECT

t.id,

t.sname,

r.c_id,

c.NAME,

t.score 

FROM

(

SELECT

s.id,

s.NAME sname,

max( r.score ) score 

FROM

student s

LEFT JOIN scores r ON r.s_id = s.id 

GROUP BY

s.id,

s.NAME 

) t

LEFT JOIN scores r ON r.s_id = t.id 

AND r.score = t.score

LEFT JOIN course c ON r.c_id = c.id;

-- 14.查询所有学生的课程及分数

SELECT

s.id,

s.NAME,

c.id,

c.NAME,

r.score 

FROM

student s

LEFT JOIN scores r ON s.id = r.s_id

LEFT JOIN course c ON c.id = r.c_id;

-- 15.查询课程编号为1且课程成绩在60分以上的学生的学号和姓名(子查询)

/*方法一*/

SELECT

* 

FROM

student s 

WHERE

s.id IN (

SELECT

r.s_id 

FROM

scores r 

WHERE

r.c_id = 1 

AND r.score > 60)

--------------------------------------------------------

/*方法二---外连接*/

SELECT

s.*,

r.* 

FROM

student s

LEFT JOIN scores r ON s.id = r.s_id 

WHERE

r.c_id = 1 

AND r.score > 60

-- 16. 查询平均成绩大于等于70的所有学生学号、姓名和平均成绩

/*方法一*/

SELECT

s.id 学生学号,

s.`name` 学生姓名,

AVG(sc.score) 平均成绩

FROM

student s

LEFT JOIN scores sc ON sc.s_id = s.id

GROUP BY

s.id

HAVING

AVG(sc.score) >=70

----------------------------------------------------

/*方法二*/

SELECT

s.id 学生学号,

s.`name` 学生姓名,

t.score 平均成绩

FROM

student s

LEFT JOIN 

( SELECT 

r.s_id, AVG( r.score ) score 

FROM

scores r 

GROUP BY 

r.s_id ) t 

ON s.id = t.s_id

WHERE

t.score >=70

-- 17.查询有不及格课程的学生信息

SELECT 

*

FROM

student S 

WHERE

id IN ( 

SELECT 

r.s_id 

FROM 

scores r 

GROUP BY 

r.s_id 

HAVING 

min(r.score) <60 );

-- 18.查询每门课程有成绩的学生人数

/*方法一*/

SELECT

c.id,

c.NAME,

t.number 

FROM

course c

LEFT JOIN ( SELECT r.c_id, count(*) number FROM scores r GROUP BY r.c_id ) t ON c.id = t.c_id;

----------------------------------------------------

/*方法二*/

SELECT

c.id,

c.NAME,

count(*) 

FROM

course c

LEFT JOIN scores r ON c.id = r.c_id 

GROUP BY

c.id,

c.NAME;

-- 19.查询每门课程的平均成绩,结果按照平均成绩降序排列,如果平均成绩相同,再按照课程编号升序排列

SELECT

c.id 课程编号,

c.`name` 课程名称,

AVG(score) 平均成绩

FROM

course c

LEFT JOIN scores r ON r.c_id = c.id 

GROUP BY

c.id,

c.`name` 

ORDER BY

AVG(score) DESC,

c.id ASC;

-- 20.查询平均成绩大于60分的同学的学生编号和学生姓名和平均成绩

SELECT

s.id,

s.NAME sname,

avg( r.score ) score 

FROM

student s

LEFT JOIN scores r ON r.s_id = s.id

LEFT JOIN course c ON c.id = r.c_id 

GROUP BY

s.id,

s.NAME 

HAVING

score > 65;

-- 21. 查询有且仅有一个课程成绩在80分以上的学生信息

/*方法一*/

SELECT

s.age,

s.gender,

s.id,

s.`name`

FROM

student s

LEFT JOIN scores sc on sc.s_id =s.id

LEFT JOIN course c ON c.id = sc.c_id

WHERE

sc.score >80

GROUP BY

s.id

HAVING COUNT(c.id)=1;

-----------------------------------------------------

/*方法二*/
SELECT
* 
FROM
student 
WHERE
id IN ( SELECT r.s_id FROM scores r WHERE r.score > 80 GROUP BY r.s_id HAVING COUNT(*)= 1 );

-- 22 .查询出只有三门课程的学生的学号和姓名

/*方法一*/

SELECT

s.id 学生学号,

s.`name` 学生姓名

FROM

student s

LEFT JOIN scores sc ON sc.s_id = s.id

GROUP BY

s.id,

s.`name`

HAVING

COUNT(*)=3;

-------------------------------------

/*方法二*/

SELECT

* 

FROM

student s 

WHERE

id IN ( SELECT r.s_id FROM scores r GROUP BY r.s_id HAVING count(*) = 3 );

-- 23.查询有不及格课程的课程信息

/*方法一*/
SELECT
c.id 课程编号,
c.NAME 课程名称
FROM
course c
LEFT JOIN scores sc ON c.id = sc.c_id 
GROUP BY
c.id,
c.NAME 
HAVING
min( sc.score ) < 60;

/*方法二*/
SELECT
* 
FROM
course c 
WHERE
id IN (
SELECT
r.c_id 
FROM
scores r 
GROUP BY
r.c_id 
HAVING
min( r.score )< 60
 )

-- 24 .查询至少选择4门课程的学生信息

/*方法一*/
SELECT
s.id 学生学号,
s.`name` 学生姓名
FROM
student s
LEFT JOIN scores sc ON sc.s_id = s.id
GROUP BY
s.id,
s.`name`
HAVING
COUNT(*)>=4;





/*方法二*/
SELECT
* 
FROM
student 
WHERE
id IN (
SELECT
r.s_id 
FROM
scores r 
GROUP BY
r.s_id 
HAVING
count(*) >= 4 
)

-- 25. 查询没有选全所有课程的同学的信息

SELECT

* 

FROM

student 

WHERE

id IN (

SELECT

r.s_id 

FROM

scores r 

GROUP BY

r.s_id 

HAVING

count(*) != 5

)

-- 26.查询选全所有课程的同学的信息

SELECT

s.id,

s.NAME,

count(*) number 

FROM

student s

LEFT JOIN scores r ON s.id = r.s_id 

GROUP BY

s.id,

s.NAME 

HAVING

number = ( SELECT count(*) FROM course );

-- 27.查询各学生都选了多少门课

/*方法一*/
SELECT
s.id 学生学号,
s.`name` 学生姓名,
COUNT(c.id) 选课数量
FROM
student s
LEFT JOIN scores sc ON sc.s_id = s.id
LEFT JOIN course c ON  c.id = sc.c_id
GROUP BY
s.id
HAVING
COUNT(c.id)

/*方法二*/
SELECT
s.id,
s.NAME,
count(*) number 
FROM
student s
LEFT JOIN scores r ON s.id = r.s_id 
GROUP BY
s.id,
s.NAME

-- 28.查询课程名称为"java",且分数低于60分的学生姓名和分数

SELECT
s.id 学生学号,
s.`name` 学生姓名,
sc.score 学生成绩,
c.`name` 课程名称
FROM
student s
LEFT JOIN scores sc ON sc.s_id = s.id
LEFT JOIN course c ON  c.id = sc.c_id
WHERE
sc.score < 60 AND c.name = 'Java'

-- 29.查询学过"Tom"老师授课的同学的信息

SELECT
s.id 学生学号,
s.`name` 学生姓名
FROM
student s
LEFT JOIN scores sc ON sc.s_id = s.id
LEFT JOIN course c ON  c.id = sc.c_id
LEFT JOIN teacher t ON t.id = c.t_id
WHERE
t.`name` = 'Tom'

-- 30.查询没学过"Tom"老师授课的学生信息

SELECT
* 
FROM
student 
WHERE
id NOT IN (
SELECT DISTINCT
s.id 
FROM
student s
LEFT JOIN scores r ON r.s_id = s.id
LEFT JOIN course c ON c.id = r.c_id
LEFT JOIN teacher t ON t.id = c.t_id 
WHERE
t.NAME = 'Tom' 
)


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