sql常见50道查询练习题

1. 表创建
- 1.1 表创建
- 1.2 数据插入
2. 简单查询例题(3题）
- 2.1 查询"李"姓老师的数量
- 2.2 查询男生、女生人数
- 2.3 查询名字中含有"风"字的学生信息
3. 日期相关例题(6题）
- 3.1 查询各学生的年龄
- 3.2 查询本周过生日的学生
- 3.3 查询下周过生日的学生
- 3.4 查询本月过生日的学生
- 3.5 查询下月过生日的学生
- 3.6 查询1990年出生的学生名单
4. 开窗函数查询（7题）
- 4.1 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
- 4.2 按各科成绩进行排序，并显示排名(实现不完全)
- 4.3 查询学生的总成绩并进行排名
- 4.4 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
- 4.5 查询学生平均成绩及其名次
- 4.6 查询各科成绩前三名的记录
- 4.7 查询每门功成绩最好的前两名
5. 表连接+子查询+聚合函数查询(34题）
- 5.1 查询"01"课程比"02"课程成绩高的学生的信息及课程分数
- 5.2 查询"01"课程比"02"课程成绩低的学生的信息及课程分数
- 5.3 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
- 5.4 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)
- 5.5 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
- 5.6 查询学过"张三"老师授课的同学的信息
- 5.7 查询没学过"张三"老师授课的同学的信息
- 5.8 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
- 5.9 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
- 5.10 查询没有学全所有课程的同学的信息
- 5.11 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
- 5.12 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
- 5.13 查询没学过"张三"老师讲授的任一门课程的学生姓名
- 5.14 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
- 5.15 检索"01"课程分数小于60，按分数降序排列的学生信息
- 5.16 查询各科成绩最高分、最低分和平均分
- 5.17 统计各科成绩各分数段人数
- 5.18 查询不同老师所教不同课程平均分从高到低显示
- 5.19 查询每门课程被选修的学生数
- 5.20 查询出只有两门课程的全部学生的学号和姓名
- 5.21 查询同名同性学生名单，并统计同名人数
- 5.22 查询每门课程的平均成绩
- 5.23 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
- 5.24 查询课程名称为"数学"，且分数低于60的学生姓名和分数
- 5.25 查询所有学生的课程及分数情况
- 5.26 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(学生的每门课都大于70)
- 5.27 查询不及格的课程
- 5.28 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
- 5.29 求每门课程的学生人数
- 5.30 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
- 5.31 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
- 5.32 统计每门课程的学生选修人数（超过5人的课程才统计）
- 5.33 检索至少选修两门课程的学生学号
- 5.34 查询选修了全部课程的学生信息

1. 表创建

在这里插入图片描述

1.1 表创建

#–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
CREATE TABLE `Student` (
    `s_id` VARCHAR(20),
    s_name VARCHAR(20) NOT NULL DEFAULT '',
    s_brith VARCHAR(20) NOT NULL DEFAULT '',
    s_sex VARCHAR(10) NOT NULL DEFAULT '',
    PRIMARY KEY(s_id)
);

#–2.课程表 
#Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号 
create table Course(
    c_id varchar(20),
    c_name VARCHAR(20) not null DEFAULT '',
    t_id VARCHAR(20) NOT NULL,
    PRIMARY KEY(c_id)
);

/*
–3.教师表 
Teacher(t_id,t_name) –教师编号,教师姓名 
*/
CREATE TABLE Teacher(
    t_id VARCHAR(20),
    t_name VARCHAR(20) NOT NULL DEFAULT '',
    PRIMARY KEY(t_id)
);

/*
–4.成绩表 
Score(s_id,c_id,s_score) –学生编号,课程编号,分数
*/
Create table Score(
    s_id VARCHAR(20),
    c_id VARCHAR(20) not null default '',
    s_score INT(3),
    primary key(`s_id`,`c_id`)
);

1.2 数据插入

#--插入学生表测试数据
#('01' , '赵雷' , '1990-01-01' , '男')
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');

#--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');

#--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

#--成绩表测试数据
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);

2. 简单查询例题(3题）

2.1 查询"李"姓老师的数量

SELECT
	count(1) as cnt
FROM
	teacher
WHERE
	t_name like "李%"

2.2 查询男生、女生人数

SELECT
	s.s_sex,
	count(1) as 人数
FROM
	student s
group by
	s.s_sex

2.3 查询名字中含有"风"字的学生信息

SELECT
	*
FROM
	student
WHERE
	s_name like "%风%"

3. 日期相关例题(6题）

3.1 查询各学生的年龄

(按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一)

-- if函数

select
	a.*,
	year(NOW())-year(a.s_brith)-if(DATE_FORMAT(now(),"%m%d") >DATE_FORMAT(a.s_brith,"%m%d"),0,1) as age
FROM
	student a

-- case函数
 
select s_brith,
	(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_brith,'%Y') - (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_brith,'%m%d') then 0 else 1 end)) as age
from student;

3.2 查询本周过生日的学生

SELECT
	*
FROM
	student
WHERE
	WEEKOFYEAR(STR_TO_DATE(concat(year(NOW()),DATE_FORMAT(s_brith,'%m%d')),"%Y%m%d"))=WEEKOFYEAR(NOW())
-- 	WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)

3.3 查询下周过生日的学生

SELECT
	*
FROM
	student
WHERE
	WEEKOFYEAR(STR_TO_DATE(concat(year(NOW()),DATE_FORMAT(s_brith,'%m%d')),"%Y%m%d"))=WEEKOFYEAR(NOW()+interval "7" day)
-- 	WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1=WEEK(s_birth)

3.4 查询本月过生日的学生

SELECT
	*
FROM
	student
WHERE
	MONTH(now())=month(s_brith)

3.5 查询下月过生日的学生

SELECT
	*
FROM
	student
WHERE
	MONTH(now()+interval "1" month)=month(s_brith)

3.6 查询1990年出生的学生名单

SELECT
	*
FROM
	student
WHERE
	s_brith like "1990%"
-- 	left(s_brith,4)="1990"
-- 	year(s_brith)="1990"

4. 开窗函数查询（7题）

4.1 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

方法一:开窗函数

select
	a.*,
	avg(a.s_score) over(PARTITION by a.s_id) as avg_score
FROM
	score a

方法二：临时表连接

SELECT
	a.*,
	t.avg_score
FROM
	score a,
	(SELECT
		a.s_id,
		round(avg(a.s_score),2) as avg_score
	FROM
		score a
	group by
		a.s_id) t
WHERE
	a.s_id=t.s_id
order by
	t.avg_score desc

方法三:长型数据转为宽型数据

SELECT
	a.s_id,
	ifnull((select s_score from score where s_id=a.s_id and c_id="01"),0) as "语文",
	ifnull((select s_score from score where s_id=a.s_id and c_id="02"),0) as "数学",
	ifnull((select s_score from score where s_id=a.s_id and c_id="03"),0) as "英语",
	ifnull(round(avg(a.s_score),2),0) as avg_score
FROM
	score a
group by
	a.s_id
order by
	ifnull(round(avg(a.s_score),2),0) desc

4.2 按各科成绩进行排序，并显示排名(实现不完全)

方法一：开窗函数

SELECT
	a.*,
	rank() over(PARTITION by c_id order by s_score desc) rank排名,
	row_number() over(PARTITION by c_id order by s_score desc) row_number排名,
	dense_rank() over(PARTITION by c_id order by s_score desc) dense_rank排名
FROM
	score a

方法二：子查询

SELECT	
	a.*,
	(select count(s_score) from score b where a.c_id=b.c_id and  a.s_score<b.s_score)+1 rk,
	(select count(distinct s_score) from score b where a.c_id=b.c_id and  a.s_score<=b.s_score) den_rk
FROM
	score a
order by
	c_id,s_score desc

4.3 查询学生的总成绩并进行排名

方法一：开窗函数

SELECT
	t.*,
	rank() over(order by sum_score desc) rank排名
FROM
	(SELECT
		s_id,
		sum(s_score) as sum_score
	FROM
		score
	group by
		s_id) t

4.4 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

方法一：子查询+开窗函数

SELECT
	a.*,
	t.c_id,
	t.rk,
	t.s_score
FROM
	student a,
	(SELECT
		a.s_id,
		a.c_id,
		a.s_score,
		dense_rank() over(PARTITION by a.c_id order by a.s_score desc) as rk
	FROM
		score a) t
WHERE
	t.rk in (2,3)
AND
	a.s_id=t.s_id

4.5 查询学生平均成绩及其名次

方法一：开窗函数

SELECT
	t.*,
	rank() over(order by t.avg_score desc) 排名
FROM
	(SELECT
		a.s_id,
		round(avg(a.s_score),2) as avg_score
	FROM
		score a
	group by
		a.s_id) t

4.6 查询各科成绩前三名的记录

方法一：开窗函数

SELECT
	t.*
from
	(SELECT
		a.c_id,
		a.c_name,
		b.s_score,
		rank() over(PARTITION by a.c_id order by b.s_score desc) rk
	FROM
		course a
	LEFT JOIN
		score b
	ON
		a.c_id=b.c_id) t
WHERE
	t.rk<=3;

方法二：子查询

SELECT
	*
from (
	SELECT
		a.c_id,
		a.c_name,
		b.s_score,
		(select count(c.s_score) from score c where a.c_id=c.c_id and b.s_score<c.s_score)+1 as rk
	FROM
		course a
	LEFT JOIN
		score b
	ON
		a.c_id=b.c_id) t
WHERE
	t.rk<=3
order by
	t.c_name,t.rk asc;

4.7 查询每门功成绩最好的前两名

方法一：开窗函数

SELECT
	t.s_id,
	t.c_id,
	t.s_score
FROM
	(SELECT
		*,
		rank() over(PARTITION by b.c_id order by b.s_score desc) rk
	FROM
		score b) t
WHERE
	t.rk<=2;

方法二：自连接

SELECT
	t.s_id,
	t.c_id,
	t.s_score
FROM
	(SELECT
		a.*,
		(select count(1) from score b where b.c_id=a.c_id and a.s_score<b.s_score)+1 as rk
	FROM
		score a
	order by
		a.c_id,rk) t
WHERE	
	t.rk<=2

方法三：条件查询+子查询

SELECT
	a.*
FROM
	score a
WHERE
	(select count(1) from score b where b.c_id=a.c_id and a.s_score<b.s_score)+1<=2
order by
	a.c_id

5. 表连接+子查询+聚合函数查询(34题）

5.1 查询"01"课程比"02"课程成绩高的学生的信息及课程分数

方法一：自连接，同列比较，使用自查询

思路：先找出查询条件的学生信息及分数，根据子查询得到最终结果

SELECT
st.*,t1.sc1,t1.sc2
FROM
 student st,
	(SELECT
		s1.s_id,s1.s_score as sc1,s2.s_score as sc2
	FROM
		score s1,score s2
	WHERE	
		s1.c_id="01"
	AND	
		s2.c_id="02"
	AND
		s1.s_id=s2.s_id
	AND
		s1.s_score>s2.s_score) t1
WHERE
	st.s_id=t1.s_id;

方法二：表连接

SELECT
st.*,s1.s_score as sc1,s2.s_score as sc2
FROM
	student st
left JOIN
	score s1
ON
	s1.s_id=st.s_id
left JOIN
	score s2
ON
	s2.s_id=st.s_id
WHERE	
	s1.c_id="01"
AND	
	s2.c_id="02"
AND
	s1.s_id=s2.s_id
AND
	s1.s_score>s2.s_score

数据长型数据变为宽型数据

-- IF函数或case函数
SELECT
	a.*,
	t.s01,
	t.s02
from
	student a,
	(SELECT
		a.s_id,
		max(case when a.c_id="01" then a.s_score end) as s01,
		max(case when a.c_id="02" then a.s_score end) as s02
--  max(if(a.c_id="01",a.s_score,null)) as s01,
--  max(if(a.c_id="02",a.s_score,null)) as s02
	from
		score a
	group by
		a.s_id) t
WHERE
	a.s_id=t.s_id
AND
	t.s01>t.s02

5.2 查询"01"课程比"02"课程成绩低的学生的信息及课程分数

与上一题思路一致，条件大于变小于

方法一：自连接

SELECT
st.*,t1.sc1,t1.sc2
FROM
 student st,
	(SELECT
		s1.s_id,s1.s_score as sc1,s2.s_score as sc2
	FROM
		score s1,score s2
	WHERE	
		s1.c_id="01"
	AND	
		s2.c_id="02"
	AND
		s1.s_id=s2.s_id
	AND
		s1.s_score<s2.s_score) t1
WHERE
	st.s_id=t1.s_id;

方法二：表连接

SELECT
	st.*,s1.s_score as sc1,s2.s_score as sc2
FROM
	student st
left JOIN
	score s1
ON
	s1.s_id=st.s_id
left JOIN
	score s2
ON
	s2.s_id=st.s_id
WHERE	
	s1.c_id="01"
AND	
	s2.c_id="02"
AND
	s1.s_id=s2.s_id
AND
	s1.s_score<s2.s_score
	
-- 方法二
SELECT
	st.*,s1.s_score as sc1,s2.s_score as sc2
FROM
	student st
left JOIN
	score s1
ON
	s1.s_id=st.s_id
AND
	s1.c_id="01"
left JOIN
	score s2
ON
	s2.s_id=st.s_id
AND
	s2.c_id="02"
AND
	s1.s_id=s2.s_id
WHERE
	s1.s_score<s2.s_score

方法三:数据长型数据变为宽型数据

-- IF函数或case函数
SELECT
	a.*,
	t.s01,
	t.s02
from
	student a,
	(SELECT
		a.s_id,
		max(case when a.c_id="01" then a.s_score end) as s01,
		max(case when a.c_id="02" then a.s_score end) as s02
--  max(if(a.c_id="01",a.s_score,null)) as s01,
--  max(if(a.c_id="02",a.s_score,null)) as s02
	from
		score a
	group by
		a.s_id) t
WHERE
	t.s01<t.s02
AND
	a.s_id=t.s_id

5.3 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

方法一：子查询

-- 子查询一
SELECT
st.s_id,st.s_name,t.avg_s
FROM
	student ST,
	(SELECT
		s.s_id,round(avg(s.s_score),2) as avg_s
	FROM	
		score s
	GROUP BY
		s.s_id
	HAVING
		round(avg(s.s_score),2)>=60) t
WHERE
	st.s_id=t.s_id

-- 方法二：子查询二

SELECT
	s.s_id,
	(select s_name from student where s_id=s.s_id) as s_name,
	round(avg(s.s_score),2) as avg_s
FROM	
	score s
GROUP BY
	s.s_id
HAVING
	avg_s>=60

方法二：表连接

SELECT
	a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
GROUP BY
	a.s_id
HAVING
	round(avg(b.s_score),2)>=60;

5.4 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)

方法一：子查询

-- 有成绩的

SELECT
	a.s_id,a.s_name,t.avg_acore
FROM
	student a,
	(SELECT
		a.s_id,round(avg(a.s_score),2) as avg_acore
	FROM
		score a
	GROUP BY
		a.s_id
	HAVING
		round(avg(a.s_score),2)<60) t
WHERE
	a.s_id=t.s_id
 
UNION
-- 没有成绩的：没有成绩的s_id不存在
SELECT
	a.s_id,a.s_name,0 as avg_acore
FROM
	student a
WHERE
	a.s_id not in (SELECT DISTINCT s_id FROM score);

方法二：表连接

SELECT
	a.s_id,a.s_name,ifnull(round(avg(b.s_score),2),0) as avg_score
FROM
	student a
LEFT JOIN
	score b
on 
	a.s_id=b.s_id
GROUP BY
	a.s_id
HAVING
	avg_score<60

5.5 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
	a.s_id,
	a.s_name,
	count(b.c_id) as cnt_course,
	ifnull(sum(b.s_score),0) as sum_score
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
group by
	a.s_id

5.6 查询学过"张三"老师授课的同学的信息

方法一：表连接+子查询单层嵌套

SELECT
	a.*
FROM
	student a
LEFT JOIN
	score b
on 
	a.s_id=b.s_id
LEFT JOIN
	course c
ON
	b.c_id=c.c_id
where c.t_id in(SELECT t_id FROM teacher WHERE t_name = "张三")

方法二：表连接+子查询多层嵌套

SELECT
	a.*
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
WHERE
	b.c_id in (
SELECT
	c_id
FROM
	course where t_id in(SELECT t_id from teacher where t_name="张三")
);

方法三：多表连接

select
	a.*
from
	student a,score b,course c,teacher d
WHERE
	a.s_id=b.s_id
AND
	b.c_id=c.c_id
AND
	c.t_id=d.t_id
AND
	d.t_name="张三"

5.7 查询没学过"张三"老师授课的同学的信息

注意：一个学生有几门课程包含张三课程，不是张三课程的，根据没学过的查询不出来，因为一个人有多个老师的课程

方法一：多层嵌套子查询

SELECT
	s.*
FROM
	student s
WHERE
	s.s_id NOT IN (
		-- 查找学的学生
		SELECT DISTINCT
			a.s_id
		FROM
			student a
		LEFT JOIN score b ON a.s_id = b.s_id
		WHERE
			b.c_id IN (
				-- 查找学过的课程
				SELECT c_id
				FROM course
				WHERE t_id IN ( SELECT t_id FROM teacher WHERE t_name = "张三")
			)
	)

方法二:条件查询+子表连接

SELECT
	*
FROM
	student s
WHERE
	s.s_id not in (
				select
					a.s_id
				from
					score a,
					course b,
					teacher c
				WHERE
					a.c_id=b.c_id
				AND
					b.t_id=c.t_id
				AND
					c.t_name="张三")

5.8 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

方法一：子查询+自连接,同列对比可以用自连接

SELECT
	*
FROM
	student s
WHERE
	s.s_id in(
					SELECT
						a.s_id
					FROM
						score a,score b
					WHERE
						a.c_id="01" 
					AND
						b.c_id="02"
					AND
						a.s_id=b.s_id)

方法二：连表+自连接,同列对比可以用自连接

SELECT
	s.*
FROM
	student s
LEFT JOIN 
	score a
ON
	s.s_id=a.s_id
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
WHERE
	a.c_id="01" 
AND
	b.c_id="02"

方法三：条件查询+子查询

SELECT
	*
FROM
	student
WHERE
	s_id in (SELECT
					s_id
				FROM
					score
				where
					c_id="01" or c_id="02"
				GROUP BY
					s_id
				HAVING
					count(1)=2)

方法四：自连接，条件连接

SELECT
	s.*
FROM
	student s,score a,score b
WHERE
	s.s_id=a.s_id
AND
	a.s_id=b.s_id
AND
	a.c_id="01" 
AND
	b.c_id="02"

方法五：子查询+数据长型数据变为宽型数据

SELECT
	a.*
FROM
	student a,
	(select
		a.s_id,
		max(if(a.c_id="01",a.s_score,0)) as s01,
		max(if(a.c_id="02",a.s_score,0)) as s02
	from
		score a
	group by
		a.s_id) t
WHERE
	a.s_id=t.s_id
AND
	t.s01>0
AND
	t.s02>0

5.9 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

方法一：条件查询+子查询

select
	a.*
from
	student a
WHERE	
	a.s_id in(select s_id from score where c_id="01") 
AND
	a.s_id not in (select s_id from score where c_id="02")

方法二：子查询+分组聚合

SELECT
	s.*
FROM
	student s,
	(
		SELECT
			a.s_id,
			max(case when a.c_id="01" then a.s_score end) s01,
			max(case when a.c_id="02" then a.s_score end) s02
		FROM
			score a
		group by
			a.s_id) t
WHERE
	s.s_id=t.s_id
AND
	t.s01 is not NULL
AND
	t.s02 is null

方法三：数据长型数据变为宽型数据

SELECT
	a.*
FROM
	student a,
	(select
		a.s_id,
		max(if(a.c_id="01",a.s_score,null)) as s01,
		max(if(a.c_id="02",a.s_score,null)) as s02
	from
		score a
	group by
		a.s_id) t
WHERE
	a.s_id=t.s_id
AND
	t.s01 is not null
AND
	t.s02 is null

5.10 查询没有学全所有课程的同学的信息

方法一:条件查询+子查询

SELECT
	s.*
FROM
	student s
WHERE
	s.s_id in(
			SELECT
				a.s_id
			FROM
				score a
			group by
				a.s_id
			having
				count(1)<(select count(1) from course))

方法二:表连接

SELECT
	s.*,
	count(a.c_id) cnt
FROM
	student s
LEFT JOIN
	score a
ON
	a.s_id=s.s_id
group by
	s.s_id
HAVING
	count(a.c_id)<(select count(1) from course)

5.11 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

方法一:子查询

SELECT
	s.*
FROM
	student s
WHERE
	s.s_id in(
			SELECT
				distinct a.s_id
			FROM
				score a
			WHERE
				a.c_id in(
						SELECT
							b.c_id
						FROM
							score b
						WHERE
							b.s_id="01"))
AND
	s.s_id!='01'

方法二:表连接+子查询

SELECT
	a.*
FROM
	student a
LEFT JOIN
	score b
on 
	a.s_id=b.s_id
WHERE
	b.c_id in (
			SELECT
				b.c_id
			FROM
				score b
			WHERE
				b.s_id="01")
group by 1,2,3,4

5.12 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

筛选课程与01号一样的数据，计算课程数与01一致的

SELECT
	s.*
FROM
	student s
WHERE
	s.s_id in(
					SELECT distinct 
						a.s_id
					FROM
						score a
					WHERE
						a.c_id in(
								SELECT
									a.c_id
								FROM
									score a
								WHERE
									a.s_id="01")
					AND
						a.s_id!="01"
					group by 
						a.s_id
					HAVING
						count(distinct a.c_id)=(select count(1) from score a where a.s_id="01")
					)

5.13 查询没学过"张三"老师讲授的任一门课程的学生姓名

查询学过张三老师的学生，在学生表中反向查询

SELECT
	s.s_name
FROM
	student s
WHERE
	s.s_id not in(
				SELECT
					a.s_id
				FROM
					score a
				WHERE
					a.c_id in (
							SELECT
								a.c_id
							FROM
								course a
							WHERE
								a.t_id in (SELECT t.t_id FROM teacher t WHERE t.t_name="张三")))

5.14 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

方法一：表连接+分组+having条件

SELECT
	a.s_id,
	a.s_name,
	round(avg(b.s_score),2) as avg_score
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
group by
	a.s_id
having
	sum(if(b.s_score>=60,0,1))>=2

方法二：自连接+子查询

select
	a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
FROM
	student a,score b
WHERE
	a.s_id=b.s_id
AND
	a.s_id in(
			SELECT
				a.s_id
			FROM
				score a
			WHERE
				a.s_score<60
			group by
				a.s_id
			HAVING
				count(1)>=2)
group by
	a.s_id

方法三：表连接+子查询

select
	a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
FROM
	student a
LEFT JOIN
	score b
on
	a.s_id=b.s_id
where
	a.s_id in(
			SELECT
				a.s_id
			FROM
				score a
			WHERE
				a.s_score<60
			group by
				a.s_id
			HAVING
				count(1)>=2)
group by
	a.s_id

5.15 检索"01"课程分数小于60，按分数降序排列的学生信息

方法一：表连接

SELECT
	a.*,b.c_id,b.s_score
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
WHERE
	b.c_id="01" and b.s_score<60
order by
	b.s_score desc

5.16 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

方法一：if语句

SELECT
	a.c_id,
	a.c_name,
	max(b.s_score) as max_score,
	min(b.s_score) as min_score,
	round(avg(b.s_score),2) as avg_score,
	round(100*sum(if(b.s_score>=60,1,0))/count(1),2) as "及格率",
	round(100*sum(if(b.s_score>=70 and b.s_score<80,1,0))/count(1),2) as "中等率",
	round(100*sum(if(b.s_score>=80 and b.s_score<90,1,0))/count(1),2) as "优良率",
	round(100*sum(if(b.s_score>=90,1,0))/count(1),2) as "优秀率"
FROM
	course a,
	score b
WHERE
	a.c_id=b.c_id
group by
	a.c_id

方法二：case when

SELECT
	a.c_id,
	a.c_name,
	max(b.s_score) as max_score,
	min(b.s_score) as min_score,
	round(avg(b.s_score),2) as avg_score,
	round(100*sum(case when b.s_score>=60 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as "及格率",
	round(100*sum(case when b.s_score>=70 and b.s_score<80 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as "中等率",
	round(100*sum(case when b.s_score>=80 and b.s_score<90 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as "优良率",
	round(100*sum(case when b.s_score>=90 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as "优秀率"
FROM
	course a,
	score b
WHERE
	a.c_id=b.c_id
group by
	a.c_id

5.17 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],[0-60]个数及所占百分比

方法一：if函数

SELECT
	b.c_id,
	a.c_name,
    round(100*sum(if(b.s_score>85 and b.s_score<=100,1,0))/count(1),2) as "[100-85]百分比",
	sum(if(b.s_score>85 and b.s_score<=100,1,0)) as "[100-85]",
	round(100*sum(if(b.s_score>70 and b.s_score<=85,1,0))/count(1),2) as "[85-70]百分比",
	sum(if(b.s_score>70 and b.s_score<=85,1,0)) as "[85-70]",
	round(100*sum(if(b.s_score>60 and b.s_score<=70,1,0))/count(1),2) as "[70-60]百分比",
	sum(if(b.s_score>60 and b.s_score<=70,1,0)) as "[70-60]",
	round(100*sum(if(b.s_score>0 and b.s_score<=60,1,0))/count(1),2) as "[0-60]百分比",
	sum(if(b.s_score>=0 and b.s_score<=60,1,0)) as "[0-60]"
FROM
	course a,
	score b
WHERE
	a.c_id=b.c_id
group by
	b.c_id

5.18 查询不同老师所教不同课程平均分从高到低显示

方法一：表连接

SELECT
	c.t_name,
	a.c_name,
	round(avg(b.s_score),2) as avg_score
FROM
	course a
left JOIN
	score b
ON
	a.c_id=b.c_id
LEFT JOIN
	teacher c
ON
	a.t_id=c.t_id
group by
	c.t_name,a.c_name
order by
	avg_score DESC

5.19 查询每门课程被选修的学生数

SELECT
	a.c_id,
	a.c_name,
	count(1) as cnt
FROM
	course a
LEFT JOIN
	score b
ON	
	a.c_id=b.c_id
group by
	a.c_id

5.20 查询出只有两门课程的全部学生的学号和姓名

方法一:连表

SELECT
	distinct a.s_id,a.s_name
FROM
	student a,
	score b
WHERE
	a.s_id=b.s_id
group by
	a.s_id
HAVING
	count(b.c_id)=2

方法二：条件查询

select 
	s_id,
	s_name 
from 
	student 
where 
	s_id in (select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);

5.21 查询同名同性学生名单，并统计同名人数

方法一:分组条件查询

SELECT
	s_name,
	count(1) as "人数"
FROM
	student
group by
	s_name,s_sex
HAVING
	count(1)>1

方法二:自连接(同列比较可以用自连接）

select 
	a.s_name,
	a.s_sex,
	count(*) 
from 
	student a  
JOIN 
	student b 
on 
	a.s_id !=b.s_id 
and 
	a.s_name = b.s_name 
and 
	a.s_sex = b.s_sex
GROUP BY 
	a.s_name,a.s_sex

5.22 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
	a.c_id,
	round(avg(a.s_score),2) as avg_score
FROM
	score a
group by
	a.c_id
order by
	avg_score desc,a.c_id asc

5.23 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

方法一：子查询

SELECT
	a.s_id,
	a.s_name,
	t.avg_score
FROM
	student a,
	(SELECT
		a.s_id,
		round(avg(a.s_score),2) as avg_score
	FROM
		score a
	group by
		a.s_id
	HAVING
		avg_score>=85) t
WHERE	
	a.s_id=t.s_id
AND
	t.avg_score is not null

方法二：表连接

select
	a.s_id,
	b.s_name,
	ifnull(round(avg(a.s_score),2),0) as avg_score
FROM
	score a
LEFT JOIN
	student b
ON
	a.s_id=b.s_id
GROUP BY
	a.s_id
HAVING
	avg_score>=85

5.24 查询课程名称为"数学"，且分数低于60的学生姓名和分数

方法一：条件查询+子查询

SELECT
	b.s_name,
	a.s_score
FROM
	score a
LEFT JOIN
	student b
ON
	a.s_id=b.s_id
WHERE
	c_id in (SELECT c_id FROM course where c_name="数学")
AND
	a.s_score<60

方法二：多表连接

SELECT
	b.s_name,
	a.s_score
FROM
	score a
LEFT JOIN
	student b
ON
	a.s_id=b.s_id
LEFT JOIN
	course c
ON
	a.c_id=c.c_id
WHERE
	c.c_name="数学"
AND
	a.s_score<60

5.25 查询所有学生的课程及分数情况

方法一：表连接

SELECT
	a.s_name,c.c_name,b.s_score
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
LEFT JOIN
	course c
ON
	c.c_id=b.c_id

方法二：if函数

SELECT
	a.s_id,
	a.s_name,
	sum(if(c.c_name="语文",b.s_score,0)) as "语文",
	sum(if(c.c_name="数学",b.s_score,0)) as "数学",
	sum(if(c.c_name="英语",b.s_score,0)) as "英语",
	sum(b.s_score) as "总分"
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
LEFT JOIN
	course c
ON
	c.c_id=b.c_id
group by
	a.s_id,a.s_name

方法三：case函数

select
	a.s_id,
	a.s_name,
	sum(case when c.c_name="语文" then b.s_score else 0 end) as "语文",
	sum(case when c.c_name="数学" then b.s_score else 0 end) as "数学",
	sum(case when c.c_name="英语" then b.s_score else 0 end) as "英语",
	sum(b.s_score) as "总分"
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
LEFT JOIN
	course c
ON
	c.c_id=b.c_id
group by
	a.s_id,a.s_name

5.26 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(学生的每门课都大于70)

方法一：表连接+子查询

SELECT
	a.s_name,
	c.c_name,
	b.s_score

FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
LEFT JOIN
	course c
ON
	c.c_id=b.c_id
WHERE
	a.s_id in (select s_id from score group by s_id having min(s_score)>70);

5.27 查询不及格的课程

方法一：表连接

SELECT
distinct
	b.s_id,
	b.c_id,
	a.c_name,
	b.s_score
from
	course a
LEFT JOIN
	score b
ON
	a.c_id=b.c_id
WHERE
	b.s_score<60

5.28 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

方法一：子查询

SELECT
	t.s_id,
	t.s_name
FROM
	student t
WHERE
	t.s_id in(
			SELECT
				a.s_id
			FROM
				score a
			WHERE
				a.c_id="01" 
			AND
				a.s_score>80)

方法二：表连接

select
	a.s_id,
	a.s_name
from
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
WHERE
	b.c_id="01"
AND
	b.s_score>80

5.29 求每门课程的学生人数

SELECT
	a.c_name,
	count(1) as "人数"
FROM
	course a
LEFT JOIN
	score b
ON
	a.c_id=b.c_id
group by
	a.c_id

5.30 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

方法一：表连接+子查询

SELECT
	a.*,
	b.s_score as max_score,
	b.c_id,
	c.c_name
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
LEFT JOIN
	course c
ON
	c.c_id=b.c_id
WHERE
-- 查询id
	b.c_id in (
						SELECT
							c_id
						FROM
							course 
						WHERE
							t_id in (select t_id from teacher where t_name="张三")
						)
AND
-- 查询最大分数
	b.s_score=(select distinct max(s_score) from score where c_id="02")

方法二：表连接

 SELECT
	a.*,
	b.s_score as max_score,
	b.c_id,
	c.c_name
FROM
	student a
LEFT JOIN
	score b
ON
	a.s_id=b.s_id
LEFT JOIN
	course c
ON
	c.c_id=b.c_id
LEFT JOIN
	teacher d
ON
	d.t_id=c.t_id
WHERE
	d.t_name="张三"
order by
	max_score desc
limit 1;

5.31 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
	a.*
FROM
	score a,
	score b
WHERE
	a.c_id!=b.c_id
AND
	a.s_score=b.s_score

5.32 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

方法一：分组聚合

SELECT
	c_id,
	count(1) as "选修人数"
FROM
	score
group by
	c_id
HAVING
	count(1) >5
order by
	"选修人数" desc,c_id asc

方法二：连表+分组聚合

SELECT
	a.c_id,
	count(b.s_id) cnt
FROM
	course a
LEFT JOIN
	score b
ON
	a.c_id=b.c_id
group by
	a.c_id
HAVING
	count(b.s_id)>5
order by
	cnt desc,a.c_id asc

5.33 检索至少选修两门课程的学生学号

SELECT
	s_id
FROM
	score
group by
	s_id
HAVING
	count(c_id)>=2;

5.34 查询选修了全部课程的学生信息

方法一：连表查询

SELECT
	a.*
FROM
	student a,
	score b
WHERE
	a.s_id=b.s_id
group by
	s_id
HAVING
	count(1)=(select count(1) from course)

方法二：子查询

SELECT
	*
FROM
	student a
WHERE
	a.s_id in(
		   select 
				s_id
			FROM
				score
			group by
				s_id
			HAVING	
				count(1)=(select count(1) from course))