98.SQL语句面试50题

# 01.课程1比课程2分数高的

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] 筛选出课程号01这门课的成绩比02这门课成绩高的学生，输出这些学生的信息及课程分（表：Student，Score)——表联结
select a.*, b.s_score as 01_score, c.s_score as 02_score
from Student as a inner join Score as b on a.s_id = b.s_id and b.c_id = '01'
inner join Score as c on a.s_id = c.s_id and c.c_id = '02' 
where b.s_score > c.s_score;

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Student表和Score表中学生s_id相同的数据

MySQL> SELECT * FROM Student as a INNER JOIN Score as b ON a.s_id=b.s_id;
+------+--------+------------+-------+------+------+---------+
| s_id  | s_name  | s_birth   | s_sex | s_id  | c_id  | s_score |
+------+--------+------------+-------+------+------+----  -----+
| 01   | 赵雷   | 1990-01-01 | 男    | 01   | 01   |    80 |
| 01   | 赵雷   | 1990-01-01 | 男    | 01   | 02   |    90 |

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# 02.平均成绩大于等于60分

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] 要求平均成绩，就需要用group by对学生分组，然后利用聚合函数avg求出平均成绩，由于where语句中不能包含聚合函数，故再利用having语句和60分比较。
-- 输出的结果学生编号和学生姓名在student表中，成绩在score表中，故需要用到内联结。
SELECT a.s_id, a.s_name, AVG(s_score) as avg_score
FROM Student as a INNER JOIN Score as b ON a.s_id=b.s_id
GROUP BY a.s_id            # 以Student表中的学生id分组
HAVING AVG(b.s_score) >= 60;

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# 03.所有课程的总成绩

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]学生编号、学生姓名在表student中；选课数可通过在表score中通过group by 学生编号后利用count(c_id)算出；所有课程的总成绩则用sum可以算出。
-- 因为要用到表student和表score，故需要对表进行联结。
SELECT a.s_id, a.s_name, count(b.c_id), SUM(b.s_score)
FROM Student as a INNER JOIN Score as b ON a.s_id = b.s_id
GROUP BY a.s_id,a.s_name;

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查询 Student表和Score表中数据

MySQL> SELECT * FROM Student as a INNER JOIN Score as b on a.s_id=b.s_id;
+------+-------  -+------------+------ -+------+------+---------+
| s_id | s_name  |  s_birth   | s_sex  | s_id | c_id | s_score |
+------+------  --+------------+-------+------+------+---------+
| 01   | 赵雷   | 1990-01-01 | 男    | 01   | 01   |    80 |
| 01   | 赵雷   | 1990-01-01 | 男    | 01   | 02   |    90 |
| 01   | 赵雷   | 1990-01-01 | 男    | 01   | 03   |    99 |
| 02   | 钱电   | 1990-12-21 | 男    | 02   | 01   |    70 |
| 02   | 钱电   | 1990-12-21 | 男    | 02   | 02   |    60 |

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# 04.查询"李"姓老师的数量

-- [分析] 这里用到知识点字符串模糊查询，如：where 姓名 like '猴%' 即可查到猴什么什么。
-- 关于字符串模糊查询的知识点可查看我的另一篇文章《SQL学习笔记——汇总分析》
select count(t_name) as number
from Teacher
where t_name like '李%';

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# 05.学过"张三"老师课的同学

-- [分析] 老师的信息在表teacher中，可通过teacher.t_id与表course联结，
-- 再通过course.c_id与表score联结，
-- 再再通过score.s_id与表student联结，查找出上过张三老师课的同学的信息。
SELECT s.*,a.t_name
FROM Teacher as a INNER JOIN Course as b on a.t_id = b.t_id
INNER JOIN Score as c ON b.c_id = c.c_id 
INNER JOIN Student as s ON c.s_id = s.s_id
WHERE a.t_name = '张三'

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# 06.学过01和02两门课程同学

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]由于既要学过01课程的，又要学过02课程的，所以共需要联结两次才能达到筛选效果
SELECT * FROM Student  # 显示这些学生的信息
WHERE s_id in (
	SELECT a.s_id FROM Student as a INNER JOIN Score as b ON a.s_id = b.s_id AND b.c_id = '01'  # 查询所有选择课程01的学生s_id
	INNER JOIN Score as c ON b.s_id = c.s_id AND c.c_id='02'  # 过滤出选择 02课程的学生并且也选择了 课程01
)

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# 07.学过01没学过02

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] 在上述语句中用and联结，注意不能直接将“=”换成“<>”
SELECT * FROM Student
WHERE s_id IN
	(SELECT a.s_id FROM Student as a INNER JOIN Score as b ON a.s_id = b.s_id and b.c_id='01')
and s_id NOT IN 
	(SELECT a.s_id FROM Student as a INNER JOIN Score as c ON a.s_id = c.s_id and c.c_id='02')

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# 08.没有学全所有课程的同学

查询没有学全所有课程的同学的信息

-- [分析]没学全所有课程的情况很多，如学0门，学1门，学2门。所以这里先筛选出学全所有课程的同学再取反，即没有学全所有课程的同学。
SELECT * FROM Student
WHERE s_id NOT IN
(
	SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(c_id)   # 聚合统计所有同学学习的课程数量
	= (select count(c_id) from Course)  # 学习数量 等于 课程总数量
)

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# 09.没学过"张三"课程学生

查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT s_name FROM Student
WHERE  s_id NOT In (
	SELECT s_id FROM Score WHERE c_id IN 
	(
		# 找到 张三 老师教授的所有课程
		SELECT c_id FROM Course WHERE t_id = (SELECT t_id FROM Teacher WHERE t_name='张三')
	)
)

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# 10.两门不及格课程同学

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 a.s_id,a.s_name,avg(b.s_score)
FROM Student as a INNER JOIN Score AS b ON a.s_id=b.s_id and b.s_score < 60
GROUP BY a.s_id HAVING COUNT(b.s_score)>= 2;

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# 11."01"课程分数小于60

检索"01"课程分数小于60，按分数降序排列的学生信息

-- [分析] 内联结+子查询筛选+排序
SELECT a.s_id,a.s_name,b.s_score,b.c_id
FROM Student as a INNER JOIN Score as b 
ON a.s_id = b.s_id and b.c_id = 01 AND b.s_score <= 60 ORDER BY b.s_score DESC;

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# 12 按平均成绩排序

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（重要！）

select a.s_id,(select s_score from Score where s_id = a.s_id and c_id = '01') as 语文,
              (select s_score from Score where s_id = a.s_id and c_id = '02') as 数学,
              (select s_score from Score where s_id = a.s_id and c_id = '03') as 英语,
              round(avg(s_score),2) as 平均分 
from Score as a 
group by a.s_id 
order by 平均分 desc;

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# 13.查询各科成绩最高分、最低分和平均分

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（重要！）

-- [分析] 用到case语句，后面记得按课程号分组来显示全部结果
select a.c_id,a.c_name,max(b.s_score),min(b.s_score),avg(b.s_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 between 70 and 80 then 1 else 0 end)/sum(case when b.s_score then 1 end)),2) as 中等率,
		   round(100*(sum(case when b.s_score between 80 and 90 then 1 else 0 end)/sum(case when b.s_score then 1 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 end)),2) as 优秀率
from Course as a inner join Score as b on a.c_id = b.c_id
group by a.c_id,a.c_name;

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# 14.学生的总成绩并进行排名

SELECT s_id, SUM(s_score) FROM Score GROUP BY s_id ORDER BY SUM(s_score) DESC

# 15.不同老师不同课程平均分

查询不同老师所教不同课程平均分从高到低显示

SELECT a.t_id,avg(b.s_score) 
FROM Course as a INNER JOIN Score as b ON a.c_id=b.c_id 
GROUP BY a.t_id ORDER BY avg(b.s_score) desc;

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# 16.总成绩第2名到第3名

查询所有课程的总成绩第2名到第3名的学生信息及该课程成绩

SELECT a.s_id,SUM(b.s_score) 
FROM Student as a INNER JOIN Score as b ON a.s_id=b.s_id 
GROUP BY a.s_id ORDER BY SUM(b.s_score) DESC
LIMIT 2 OFFSET 1

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# 17.各科成绩前三名的记录

查询各科成绩前三名的记录（重要！！！！！）

(select * from Score where c_id = '01' order by s_score desc limit 3)
UNION
(select * from Score where c_id = '02' order by s_score desc limit 3)
UNION
(select * from Score where c_id = '03' order by s_score desc limit 3);

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# 18.每门课程被选修的学生数

-- [分析] 有点简单，写完甚至有点难以置信，还亲自去数了一遍。。。。。。
select c_id, count(s_id)
from Score
group by c_id;

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# 19.只有两门课程学生

查询出只有两门课程的全部学生的学号和姓名

-- [分析] 内联结+筛选
select a.s_id,a.s_name
from student as a inner join score as b on a.s_id = b.s_id
group by a.s_id
having count(b.c_id) = 2;

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# 20.查询男生、女生人数

# 方法1：
select s_sex,count(s_sex) as 人数
from Student
GROUP BY s_sex

# 方法2：
(select s_sex,count(s_sex) as 人数 from Student where s_sex = '男')
union
(select s_sex,count(s_sex) as 人数 from Student where s_sex = '女')

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# 21.名字中含有"风"字的学生

SELECT * FROM Student WHERE  s_name LIKE'%风%'

# 22.姓名性别相同的学生数量

查询同名同性学生名单，并统计同名人数（重要！！！！！）

-- [分析] 将表student定义两次，然后进行内联结，并给出判断条件
SELECT a.s_name,a.s_sex,COUNT(*)
FROM Student AS a INNER JOIN Student as b on 
		a.s_id <> b.s_id           # 用户id不同
		AND a.s_name = b.s_name      # 名字和姓名都相同
		and a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex;

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# 23.查询1990年出生的学生

-- [分析] 字符串模糊查询
SELECT *
FROM Student
WHERE s_birth like "1990%"

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# 24.课程平均成绩排列

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

-- [分析] 考察order by 语句的升降序及先后次序
SELECT c_id, AVG(s_score)
FROM Score
GROUP BY c_id
ORDER BY AVG(s_score) DESC,c_id ASC

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# 25.平均成绩大于等于85学生

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.s_id,a.s_name,AVG(b.s_score) 
FROM Student as a INNER JOIN Score AS bON a.s_id = b.s_id 
GROUP BY a.s_id,a.s_name 
HAVING AVG(b.s_score) > 85

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# 26.课程为数学且分数低于60

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.s_name,b.s_score,b.c_id,c.c_name
FROM Student AS a INNER JOIN Score AS b ON a.s_id=b.s_id
INNER JOIN Course as c ON b.c_id = c.c_id AND c.c_name = '数学'

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# 27.任一门成绩在70以上

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 a.s_name,b.s_score,c.c_name
FROM Student AS a INNER JOIN Score AS b ON a.s_id = b.s_id
INNER JOIN Course as c ON b.c_id = c.c_id
WHERE b.s_score >= 70

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# 28.查询所有不及格学生

SELECT a.s_id,a.s_name,c.c_id,c.c_name,b.s_score
FROM Student AS a INNER JOIN Score AS b ON a.s_id = b.s_id
INNER JOIN Course as c ON b.c_id = c.c_id
WHERE b.s_score < 60

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# 29.课程为01成绩在80分上

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 a.s_id,a.s_name,b.c_id,b.s_score
FROM Student as a INNER JOIN Score AS b ON a.s_id = b.s_id
WHERE b.c_id = '01' AND b.s_score >= 80

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# 30.求每门课程的学生人数

SELECT a.c_name,COUNT(b.s_id)
FROM Course AS a INNER JOIN Score as b ON a.c_id = b.c_id GROUP BY a.c_name

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# 31."张三"老师课程最高学生

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

SELECT *
FROM Student AS a INNER JOIN Score as b ON a.s_id = b.s_id
INNER JOIN Course AS c ON c.c_id = b.c_id 
INNER JOIN Teacher as d ON c.t_id = d.t_id
WHERE d.t_name = '张三' 
ORDER BY b.s_score DESC LIMIT 1

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# 32.不同课程成绩相同的学生

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（重要！！！！）

SELECT distinct b.s_id,b.c_id,b.s_score
FROM Score AS a, Score AS b
WHERE a.c_id !=b.c_id AND a.s_score = b.s_score

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# 33.每门功成绩最好的前两名

查询每门功成绩最好的前两名

(select c_id,s_score from Score where c_id = '01' order by s_score desc limit 2)
UNION
(select c_id,s_score from Score where c_id = '02' order by s_score desc limit 2)
UNION
(select c_id,s_score from Score where c_id = '03' order by s_score desc limit 2)

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# 34.选修学生超过5人

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 c_id,COUNT(s_id) 
FROM Score GROUP BY c_id
HAVING COUNT(s_id) >= 5
ORDER BY COUNT(s_id) DESC,c_id ASC

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# 35.选修全部课程学生

查询选修了全部课程的学生信息

SELECT a.* ,COUNT(c_id)
FROM Student AS a INNER JOIN Score AS b on a.s_id = b.s_id
GROUP BY s_id
HAVING COUNT(b.c_id) = (SELECT COUNT(c_id) FROM Course)

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# 36.计算学生年龄

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

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

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# 37.下周过生日的学生

select * from Student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1=WEEK(s_birth)

# 38.每门成绩大于80学生信息

SELECT * FROM Student WHERE s_id IN (
	select s_id from Score group by s_id  having min(s_score)>=80
)

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