【MySQL】测试题03

文章目录

- - 1、创建数据库
  - 2、使用数据库
  - 3、创建数据表
  - - 【3.1】创建学生信息表Student
    - 【3.2】创建课程信息表Course
    - 【3.3】创建教师信息表Teacher
    - 【3.4】创建成绩信息表Score
  - 4、添加数据
  - - 【4.1】向学生student表添加数据
    - 【4.2】向课程course表添加数据
    - 【4.3】向教师信息teacher表添加数据
    - 【4.4】向成绩score表添加数据
  - 5、查询数据练习

1、创建数据库

CREATE DATABASE IF NOT EXISTS `exam`;

2、使用数据库

USE exam;

3、创建数据表

【3.1】创建学生信息表Student

CREATE TABLE IF NOT EXISTS `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`)
);

【3.2】创建课程信息表Course

CREATE TABLE IF NOT EXISTS `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);

【3.3】创建教师信息表Teacher

CREATE TABLE IF NOT EXISTS `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);

【3.4】创建成绩信息表Score

CREATE TABLE IF NOT EXISTS `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);

4、添加数据

【4.1】向学生student表添加数据

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');

在这里插入图片描述

【4.2】向课程course表添加数据

INSERT INTO Course VALUES('01' , '语文' , '02'); 
INSERT INTO Course VALUES('02' , '数学' , '01'); 
INSERT INTO Course VALUES('03' , '英语' , '03');

在这里插入图片描述

【4.3】向教师信息teacher表添加数据

INSERT INTO Teacher VALUES('01' , '张三'); 
INSERT INTO Teacher VALUES('02' , '李四'); 
INSERT INTO Teacher VALUES('03' , '王五');

在这里插入图片描述

【4.4】向成绩score表添加数据

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);

在这里插入图片描述

5、查询数据练习

01 查询“01”课程比“02”课程成绩高的学生的信息及课程分数

本质：行运算：比较运算
方法：1、直接比较：子查询 2、弄成两列：自关联（注意显式连接，提升查询速度）
备注：任一科成绩缺失不予比较

1】直接比较：子查询

SELECT s.*,sc.s_score
FROM (
SELECT sc1.s_id,sc1.s_score
FROM score sc1
WHERE sc1.c_id='01'
AND sc1.s_score>(SELECT s_score FROM score WHERE s_id=sc1.s_id AND c_id='02')) sc JOIN student s ON sc.s_id=s.s_id;

2】弄成两列：自关联（注意显式连接，提升查询速度）

SELECT s.*,sc3.s_score 
FROM
	(
	SELECT sc1.s_id,sc1.s_score 
	FROM
		( SELECT s_id, s_score FROM score WHERE c_id = '01' ) sc1
		JOIN ( SELECT s_id, s_score FROM score WHERE c_id = '02' ) sc2 ON sc1.s_id = sc2.s_id AND sc1.s_score > sc2.s_score 
	) sc3
	JOIN student s ON sc3.s_id = s.s_id

3】直接自关联

SELECT s.*,sc1.s_score
FROM student s 
JOIN score sc1 ON s.s_id=sc1.s_id AND sc1.c_id='01'
JOIN score sc2 ON s.s_id=sc2.s_id AND sc2.c_id='02' AND sc1.s_score>sc2.s_score

02 查询“01”课程比“02”课程成绩低的学生的信息及课程分数(题目 1 是成绩高)

本质：行运算：比较运算
方法：1、直接比较：子查询 2、弄成两列：自关联（注意显式连接，提升查询速度）
备注：任一科成绩缺失不予比较

1】直接比较：子查询

SELECT s.*,sc.s_score
FROM (
SELECT sc1.s_id,sc1.s_score
FROM score sc1
WHERE sc1.c_id='01'
AND sc1.s_score<(SELECT s_score FROM score WHERE s_id=sc1.s_id AND c_id='02')) sc JOIN student s ON sc.s_id=s.s_id

2】弄成两列：自关联（注意显式连接，提升查询速度）

SELECT s.*,sc3.s_score 
FROM
	(
	SELECT sc1.s_id,sc1.s_score 
	FROM
		( SELECT s_id, s_score FROM score WHERE c_id = '01' ) sc1
		JOIN ( SELECT s_id, s_score FROM score WHERE c_id = '02' ) sc2 ON sc1.s_id = sc2.s_id AND sc1.s_score < sc2.s_score 
	) sc3
	JOIN student s ON sc3.s_id = s.s_id

3】直接自关联

SELECT s.*,sc1.s_score
FROM student s 
JOIN score sc1 ON s.s_id=sc1.s_id AND sc1.c_id='01'
JOIN score sc2 ON s.s_id=sc2.s_id AND sc2.c_id='02' AND sc1.s_score<sc2.s_score

03 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

本质：行聚合，后having筛选
方法：直接聚合，后having对聚合函数进行筛选

SELECT sc.s_id,s.s_name,ROUND(AVG(sc.s_score),0) AS score
FROM score sc JOIN student s ON sc.s_id=s.s_id
GROUP BY sc.s_id
HAVING score>=60

1】附加题:总分超过200分的同学

SELECT sc.s_id,s.s_name,SUM(sc.s_score) AS score
FROM score sc JOIN student s ON sc.s_id=s.s_id
GROUP BY sc.s_id
HAVING score>200

04 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)

本质：行聚合，后having筛选
方法：直接聚合，聚合后用having筛选，注意主表
备注：1、主表是学生表，没有成绩的需要包含在内，故left join 2、having可看作聚合函数的where：为空判断，比较运算，子查询 3、用ifnull处理空/left join情况

1】null 判断

SELECT s.s_id,s.s_name,ROUND(AVG(sc.s_score),0) AS score
FROM student s
LEFT JOIN score sc ON s.s_id=sc.s_id
GROUP BY s.s_id
HAVING score IS NULL OR score < 60

2】ifnull 函数：更兼容

SELECT s.s_id,s.s_name,ROUND(AVG(IFNULL(sc.s_score,0)),0) AS score
FROM student s
LEFT JOIN score sc ON s.s_id=sc.s_id
GROUP BY s.s_id
HAVING score < 60

05 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

本质：行聚合
方法：直接聚合，注意对象范围
备注：主表是学生表，需考虑没有选课or没有成绩的情况

SELECT s.s_id,s.s_name,
COUNT(DISTINCT sc.c_id) AS c_num,
SUM(sc.s_score) AS score
FROM student s LEFT JOIN score sc ON s.s_id=sc.s_id
GROUP BY s.s_id

如果需要聚合后筛选，则使用if和ifnull函数
if和ifnull是好函数，处理null情况非常方便。
注：ifnull搭配值运算函数使用，if函数搭配count函数

SELECT s.s_id,s.s_name,
IF(sc.s_id IS NULL,0,COUNT(DISTINCT sc.c_id)) AS c_num,
SUM(IFNULL(sc.s_score,0)) AS score
FROM student s LEFT JOIN score sc ON s.s_id=sc.s_id
GROUP BY s.s_id

06 查询“李”姓老师的数量

本质：行筛选聚合
方法：筛选计数
问题：like是最优的吗

1】like通配符

SELECT COUNT(DISTINCT t_id) AS t_num
FROM Teacher
WHERE t_name LIKE ‘李%’

2】截取姓氏，然后判断

SELECT COUNT(DISTINCT t_id) AS t_num
FROM Teacher
WHERE SUBSTR(t_name,1,1)=‘李’

07 查询学过张三老师授课的同学的信息

本质：行筛选：逐级
方法：表关联or子查询
1】逐级子查询，这样更好理解

SELECT s.*
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
WHERE c_id IN(SELECT c_id FROM Course WHERE t_id IN(SELECT t_id FROM Teacher WHERE t_name=‘张三’))

2】表关联

SELECT s.*
FROM score sc
JOIN course c ON c.c_id=sc.c_id
JOIN teacher t ON t.t_id=c.t_id AND t.t_name=‘张三’
JOIN student s ON sc.s_id=s.s_id

08 找出没有学过张三老师课程的学生

本质：反向行筛选
方法：按照学过筛选，最后取相反情况left join 或right join，且is null

1】表关联

SELECT s.*
FROM score sc
JOIN course c ON c.c_id=sc.c_id
JOIN teacher t ON t.t_id=c.t_id AND t.t_name=‘张三’
RIGHT JOIN student s ON sc.s_id=s.s_id
WHERE sc.s_id IS NULL

1】not in，这个好理解

SELECT *
FROM student
WHERE s_id NOT IN(
SELECT sc.s_id
FROM score sc
JOIN course c ON c.c_id=sc.c_id
JOIN teacher t ON t.t_id=c.t_id AND t.t_name=‘张三’
)

09 查询学过编号为 01，并且学过编号为 02 课程的学生信息

本质：行筛选：取交集
方法：自关联 or 子查询

1】自关联

SELECT s.*
FROM Score sc1
JOIN Score sc2 ON sc1.s_id=sc2.s_id AND sc1.c_id=‘01’ AND sc2.c_id=‘02’
JOIN Student s ON sc1.s_id=s.s_id

1】子查询

SELECT s.*
FROM Score sc1
JOIN Student s ON sc1.s_id=s.s_id
WHERE sc1.c_id=‘01’
AND sc1.s_id IN(SELECT s_id FROM Score sc2 WHERE sc1.s_id=sc2.s_id AND sc2.c_id=‘02’)

10 查询学过 01 课程，但是没有学过 02 课程的学生信息(注意和上面9题目的区别)

本质行筛选：取交集
方法子查询，不能使用自关联，因为一条cid一条记录，不能用否判断

SELECT s.*
FROM Score sc1
JOIN Student s ON sc1.s_id=s.s_id
WHERE sc1.c_id=‘01’
AND sc1.s_id NOT IN(SELECT s_id FROM Score sc2 WHERE sc1.s_id=sc2.s_id AND sc2.c_id=‘02’)

11 查询没有学完全部课程的同学的信息

本质：行聚合，后having筛选
方法直接聚合，注意表对象范围
备注 having可看作聚合函数的where：为空判断，比较运算，子查询

SELECT s.*
FROM student s
LEFT JOIN score sc ON sc.s_id=s.s_id
GROUP BY s.s_id
HAVING COUNT(DISTINCT sc.c_id)<(SELECT COUNT(*) FROM course)

发现筛选，先筛选出学完全部课程的同学

SELECT *
FROM Student
WHERE s_id NOT IN(SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(c_id)=(SELECT COUNT(*) FROM Course))

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

本质行筛选：子查询
方法 1、学号01所学课程c_id01 2、学号不等于01且c_id in c_id01

SELECT s.*
FROM score sc
JOIN student s ON sc.s_id=s.s_id AND sc.s_id<>‘01’
WHERE sc.c_id IN(SELECT c_id FROM score WHERE s_id=‘01’)
GROUP BY s.s_id

13 查询和 01 同学学习的课程完全相同的同学的信息

#本质行筛选：子查询
#方法 1、学号01所学课程 2、和01所学相同课程且课程数等于01课程数
#备注关键是思考方法
#group_concat()函数用法
#group_concat([DISTINCT] 字段 [Order BY ASC/DESC 排序字段] [Separator ‘分隔符’])

1】方法1 课程相同的数量相同

SELECT s.*
FROM score sc
JOIN student s ON sc.s_id=s.s_id

课程数量相等

WHERE sc.s_id IN(SELECT s_id FROM Score WHERE s_id <>‘01’ GROUP BY s_id HAVING COUNT(DISTINCT c_id)=(SELECT COUNT(DISTINCT c_id) FROM score WHERE s_id=‘01’))

课程相同的数量相同

AND c_id IN(SELECT c_id FROM score WHERE s_id=‘01’)
GROUP BY s.s_id
HAVING COUNT(DISTINCT sc.c_id)=(SELECT COUNT(DISTINCT c_id) FROM score WHERE s_id=‘01’)

2】使用group_concat()函数，如果记录数比较大，建议采用方法1

SELECT s.*
FROM student s
JOIN score sc ON s.s_id=sc.s_id AND sc.s_id<>‘01’
GROUP BY s.s_id
HAVING GROUP_CONCAT(sc.c_id ORDER BY sc.c_id)=
(SELECT GROUP_CONCAT(c_id ORDER BY c_id) FROM score WHERE s_id=‘01’ GROUP BY s_id)

14 查询没有修过张三老师讲授的任何一门课程的学生姓名

#本质反向行筛选
#方法修过张三老师任一门课即不符合要求
#备注注意表对象是全体学生，主表是学生表，因对s_id筛选，故不用再和score关联

SELECT s_name
FROM student
WHERE s_id NOT IN(
SELECT s_id FROM score
WHERE c_id IN(SELECT c_id FROM course WHERE t_id=(SELECT t_id FROM teacher WHERE t_name=‘张三’)))

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

#本质行筛选并聚合
#方法 1、行筛选：及格与否 2、求不及格课程数 3、having筛选
#备注修过才会有及格与否之说，故score为主表

1】方式1

SELECT sc.s_id,s.s_name,ROUND(AVG(sc.s_score),0) AS avg_score
FROM score sc
JOIN student s ON sc.s_id=s.s_id
GROUP BY sc.s_id
HAVING SUM(IF(sc.s_score<60,1,0))>=2

2】方式2 查询表更小

SELECT sc.s_id,s.s_name,ROUND(AVG(sc.s_score),0) AS avg_score
FROM score sc
JOIN student s ON sc.s_id=s.s_id AND sc.s_score<60
GROUP BY sc.s_id
HAVING COUNT(*)>=2

16 检索 01 课程分数小于 60，按分数降序排列的学生信息

#本质行筛选并排序
#方法分数<60、分数降序
#备注排序函数区别 1、row_number()over：123 2、rank()over()：113 3、dense_rank()over()：112

SELECT s.*
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
WHERE c_id=‘01’
AND s_score<60
ORDER BY s_score DESC

17 按平均成绩从高到低(降序)显示所有学生的所有课程的成绩以及平均成绩

#本质表关联并排序
#方法找到主表，找到排序字段
#备注如何将排序字段和其他字段关联是关键点,这里的平均成绩是所选课程的平均成绩

1】纵表

SELECT s.s_name,c.c_name,sc.s_score,sc2.avg_score
FROM Student s
JOIN Course c
LEFT JOIN Score sc ON s.s_id=sc.s_id AND c.c_id=sc.c_id
LEFT JOIN (SELECT s_id,ROUND(AVG(s_score),0) AS avg_score FROM Score GROUP BY s_id) sc2 ON s.s_id=sc2.s_id
ORDER BY sc2.avg_score DESC

2】横表

SELECT s.s_name,
MAX(CASE sc.c_id WHEN ‘01’ THEN sc.s_score END) AS ‘语文’,
MAX(CASE sc.c_id WHEN ‘02’ THEN sc.s_score END) AS ‘数学’,
MAX(CASE sc.c_id WHEN ‘03’ THEN sc.s_score END) AS ‘英语’,
sc2.avg_score AS ‘平均成绩’
FROM Student s
JOIN Course c
LEFT JOIN Score sc ON s.s_id=sc.s_id AND c.c_id=sc.c_id
LEFT JOIN (SELECT s_id,ROUND(AVG(s_score),0) AS avg_score FROM Score GROUP BY s_id) sc2 ON s.s_id=sc2.s_id
GROUP BY s.s_name,sc2.avg_score
ORDER BY sc2.avg_score DESC

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:>=60，中等为:70-80，优良为:80-90，优秀为:>=90

#本质表关联并聚合
#方法主表course，left join 成绩表score，求最值、均值及分组
#备注注意小数位数及百分率符号

SELECT c.c_id AS ‘课程id’,c.c_name AS ‘课程name’,
MAX(sc.s_score) AS ‘最高分’,
MIN(sc.s_score) AS ‘最低分’,
ROUND(AVG(sc.s_score),2) AS ‘平均分’,
CONCAT(ROUND(SUM(IF(sc.s_score>=60,1,0))/COUNT(sc.s_id)*100,2),‘%’) AS ‘及格率’,
CONCAT(ROUND(SUM(IF(sc.s_score>=70 AND sc.s_score<80,1,0))/COUNT(sc.s_id)*100,2),‘%’) AS ‘中等率’,
CONCAT(ROUND(SUM(IF(sc.s_score>=80 AND sc.s_score<90,1,0))/COUNT(sc.s_id)*100,2),‘%’) AS ‘优良率’,
CONCAT(ROUND(SUM(IF(sc.s_score>=90,1,0))/COUNT(sc.s_id)*100,2),‘%’) AS ‘优秀率’
FROM Course c
LEFT JOIN Score sc ON c.c_id=sc.c_id
GROUP BY c.c_id

19 按照各科成绩进行排序，并且显示排名

#本质分组排序
#方法选择排序依据并排名

SELECT c.c_name,s.s_name,sc.s_score,
ROW_NUMBER() OVER(PARTITION BY c.c_name ORDER BY sc.s_score DESC) AS ‘排名’
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
JOIN Course c ON sc.c_id=c.c_id

20 查询学生的总成绩，并进行排名

#本质行聚合后排序
#方法 sum、排序

SELECT s.s_name,SUM(sc.s_score) AS ‘总成绩’,
ROW_NUMBER()OVER(ORDER BY SUM(sc.s_score) DESC) AS ‘排名’
FROM Student s
LEFT JOIN Score sc ON s.s_id=sc.s_id
GROUP BY s.s_name

21 查询不同老师所教不同课程平均分从高到低显示

#本质表关联后聚合再排序
#方法主表是教师表teacher和课程表course笛卡尔积，left join 成绩表score，求均值后order by

SELECT t.t_name,c.c_name,ROUND(AVG(sc.s_score),2) AS ‘平均分’
FROM Teacher t
JOIN Course c ON t.t_id=c.t_id
LEFT JOIN Score sc ON c.c_id=sc.c_id
GROUP BY t.t_name,c.c_name
ORDER BY AVG(sc.s_score) DESC

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

#本质行排序后筛选
#方法所有课程成绩排名，再取2和3名

SELECT r.c_name,r.rank_num,s.s_name,r.s_score
FROM
(SELECT c.c_name,sc.s_id,sc.s_score,
ROW_NUMBER()OVER(PARTITION BY c.c_name ORDER BY sc.s_score DESC) AS rank_num
FROM Course c
LEFT JOIN Score sc ON c.c_id=sc.c_id)r
JOIN Student s ON r.s_id=s.s_id AND r.rank_num IN(2,3)

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

#本质行分组聚合
#方法先列基本：科目、成绩，后分组计数

SELECT sc.c_id,c.c_name,
SUM(IF(sc.s_score>=85,1,0)) AS ‘[100-85]人数’,
SUM(IF(sc.s_score>=70 AND sc.s_score<85,1,0)) AS ‘[85-70]人数’,
SUM(IF(sc.s_score>=60 AND sc.s_score<70,1,0)) AS ‘[70-60]人数’,
SUM(IF(sc.s_score<60,1,0)) AS ‘[0-60]人数’,
CONCAT(ROUND(SUM(IF(sc.s_score>=85,1,0))/COUNT(*)100,2),‘%’) AS ‘[100-85]百分比’,
CONCAT(ROUND(SUM(IF(sc.s_score>=70 AND sc.s_score<85,1,0))/COUNT()100,2),‘%’) AS ‘[85-70]百分比’,
CONCAT(ROUND(SUM(IF(sc.s_score>=60 AND sc.s_score<70,1,0))/COUNT()100,2),‘%’) AS ‘[70-60]百分比’,
CONCAT(ROUND(SUM(IF(sc.s_score<60,1,0))/COUNT()*100,2),‘%’) AS ‘[0-60]百分比’
FROM Score sc
JOIN Course c ON sc.c_id=c.c_id
GROUP BY sc.c_id

24 查询学生的平均成绩及名次

#本质行聚合后排序
#方法求平均成绩，后排序

SELECT s.s_name,ROUND(AVG(sc.s_score),2) AS ‘总成绩’,
ROW_NUMBER()OVER(ORDER BY AVG(sc.s_score) DESC) AS ‘排名’
FROM Student s
LEFT JOIN Score sc ON s.s_id=sc.s_id
GROUP BY s.s_name

25 查询各科成绩前三名的记录

#本质行排序后筛选
#方法各科成绩排序，取前3

26 查询每门课被选修的学生数

#本质行聚合
#方法成绩表按科目对s_id计数

SELECT c.c_name,COUNT(DISTINCT sc.s_id) AS s_num
FROM Score sc
JOIN Course c ON sc.c_id=c.c_id
GROUP BY c.c_name

27 查询出只有两门课程的全部学生的学号和姓名

#本质行聚合
#方法成绩表按学生对c_id计数

SELECT s.s_id,s.s_name
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
GROUP BY s.s_id
HAVING COUNT(DISTINCT sc.c_id)=2

28 查询男女生人数

#本质行筛选聚合
#方法学生表student按性别分组计数

SELECT s_sex,COUNT(DISTINCT s_id) AS ‘人数’
FROM Student
GROUP BY s_sex

29 查询名字中含有风字的学生信息

#本质模糊筛选
#方法

SELECT *
FROM Student
WHERE s_name LIKE ‘%风%’

30 查询同名同性的学生名单，并统计同名人数

#本质分组聚合
#方法分组计数后having筛选

SELECT s1.s_name,s1.s_sex,s2.num AS ‘同名人数’
FROM Student s1
JOIN (SELECT s_name,COUNT() AS num FROM Student GROUP BY s_name HAVING COUNT()>=2)s2 ON s1.s_name=s2.s_name
GROUP BY s1.s_name,s1.s_sex
HAVING COUNT(*)>=2

31 查询 1990 年出生的学生信息

#本质行筛选
#方法获取出生年份，并筛选
#备注 DATE()可以将varchar转成日期型 YEAR()函数返回一个指定日期or时间的年份值，范围为1000到9999，如果日期为零，YEAR()函数返回0

SELECT *
FROM Student
WHERE YEAR(DATE(s_birth))=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 c_id 升序排列

#本质行聚合后排序
#方法对课程求均值，后排序
#备注 order by后可跟聚合函数

SELECT c.c_name,ROUND(AVG(sc.s_score),2) AS ‘平均分’
FROM Score sc
JOIN Course c ON sc.c_id=c.c_id
GROUP BY c.c_id
ORDER BY AVG(sc.s_score) DESC,c.c_id ASC

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

#本质行聚合后筛选
#方法对学生求成绩均值，后筛选

SELECT s.s_id,s.s_name,ROUND(AVG(sc.s_score),2) AS ‘平均成绩’
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
GROUP BY s.s_id
HAVING AVG(sc.s_score)>=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

#本质行筛选
#方法按要求筛选，用到子查询

SELECT s.s_name,sc.s_score
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
WHERE c_id=(SELECT c_id FROM Course WHERE c_name=‘数学’)
AND s_score<60

35 查询所有学生的课程及分数情况

#本质表关联后聚合
#方法学生表与课程表笛卡尔积，获取所有学生及所有课程，然后关联成绩表获取成绩，最后聚合注：成绩表为窄表，故外层需要聚合

SELECT s.s_name,
SUM(CASE c.c_name WHEN ‘语文’ THEN sc.s_score ELSE 0 END) AS ‘语文’,
SUM(CASE c.c_name WHEN ‘数学’ THEN sc.s_score ELSE 0 END) AS ‘数学’,
SUM(CASE c.c_name WHEN ‘英语’ THEN sc.s_score ELSE 0 END) AS ‘英语’,
SUM(sc.s_score) AS ‘总分’
FROM Student s
JOIN Course c
LEFT JOIN Score sc ON s.s_id=sc.s_id AND c.c_id=sc.c_id
GROUP BY s.s_name

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

#本质行筛选
#方法成绩表筛选>70

SELECT s.s_name,c.c_name,sc.s_score
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
JOIN Course c ON sc.s_id=c.c_id
WHERE s_score>70

37 查询不及格的课程

#本质行筛选

SELECT sc.c_id,c.c_name,sc.s_score
FROM Score sc
JOIN Course c ON sc.c_id=c.c_id
WHERE sc.s_score<60

38 查询课程编号为 01 且课程成绩大于等于 80 的学生的学号和姓名

#本质行筛选

SELECT s.s_id,s.s_name
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
WHERE c_id=‘01’
AND s_score>=80

39 每门课程的学生人数

#本质行聚合

SELECT c.c_name,COUNT(DISTINCT sc.s_id) AS ‘人数’
FROM Score sc
JOIN Course c ON sc.c_id=c.c_id
GROUP BY c.c_name

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

#本质行筛选后聚合
#方法张三老师所授课程中，成绩最高（max、order by、row_number()over）的学生 1、max需要关联表匹配 2、order by和limit 1配合使用，推荐 3、row_number()over需要外层表限制rank_num=1

SELECT s.*,s_score
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
WHERE c_id=(SELECT c_id FROM Course WHERE t_id=(SELECT t_id FROM Teacher WHERE t_name=‘张三’))
ORDER BY s_score DESC
LIMIT 1

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

#本质行筛选
#方法符合条件的成绩：按成绩分组，对课程id计数且>2

SELECT *
FROM Score
WHERE s_score IN(SELECT s_score FROM Score GROUP BY s_score HAVING COUNT(DISTINCT c_id)>=2)

42 查询每门功成绩最好的前两名

#本质行排序后筛选
#方法这里分组取前两名，需窗口函数row_number()over

SELECT r.*
FROM(
SELECT c_name,s_id,s_score,
ROW_NUMBER()OVER(PARTITION BY c_name ORDER BY s_score DESC) AS rank_num
FROM Score sc
JOIN Course c ON sc.c_id=c.c_id)r
WHERE r.rank_num<=2

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

#本质行聚合后排序

SELECT c_id,COUNT(DISTINCT s_id) AS ‘人数’
FROM Score
GROUP BY c_id
ORDER BY ‘人数’ DESC,c_id ASC

44 检索至少选修两门课程的学生学号

#本质行聚合后筛选

SELECT s_id
FROM Score
GROUP BY s_id
HAVING COUNT(DISTINCT c_id)>=2

45 查询选修了全部课程的学生信息

#本质行聚合后筛选

SELECT s.*
FROM Score sc
JOIN Student s ON sc.s_id=s.s_id
GROUP BY s.s_id
HAVING COUNT(DISTINCT c_id)=(SELECT COUNT(*) FROM Course)

46 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 1

#本质行筛选：日期-年月日
#方法年月日获取year()、month()、day()

SELECT *,
CASE
WHEN MONTH(NOW())<MONTH(DATE(s_birth)) THEN
YEAR(NOW())-YEAR(DATE(s_birth))-1
WHEN MONTH(NOW())=MONTH(DATE(s_birth)) AND DAY(NOW())<DAY(DATE(s_birth)) THEN
YEAR(NOW())-YEAR(DATE(s_birth))-1
ELSE
YEAR(NOW())-YEAR(DATE(s_birth))
END AS age
FROM Student

47 查询本周过生日的学生

#本质行筛选：日期-周
#方法周获取week()

SELECT *
FROM Student
WHERE WEEK(DATE(s_birth))=WEEK(NOW())

48 查询下周过生日的学生

#本质行筛选：日期-周
#方法周获取week()

SELECT *
FROM Student
WHERE WEEK(DATE(s_birth))=WEEK(NOW())+1

49 查询本月过生的同学

#本质行筛选：日期-月

SELECT *
FROM Student
WHERE MONTH(DATE(s_birth))=MONTH(NOW())

50 查询下月过生的同学

#本质行筛选：日期-月

SELECT *
FROM Student
WHERE MONTH(DATE(s_birth))=MONTH(NOW())+1