sql50题带建表语句带详细答案（sql高阶函数详解）

末蓝、 2022-03-01 16:50 612阅读 0赞

## 1. sql基础 ##

![1545658351688.png][])

### 建表语句 ###

--建表
    --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`)
    );
    --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`)
    );
    --教师表
    CREATE TABLE `Teacher`(
        `t_id` VARCHAR(20),
        `t_name` VARCHAR(20) NOT NULL DEFAULT '',
        PRIMARY KEY(`t_id`)
    );
    --成绩表
    CREATE TABLE `Score`(
        `s_id` VARCHAR(20),
        `c_id`  VARCHAR(20),
        `s_score` INT(3),
        PRIMARY KEY(`s_id`,`c_id`)
    );
    --插入学生表测试数据
    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');
    --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);

### 1.1 主表&从表关系 ###

*  主表中有的主键对应的列值,从表中可以没有和这个列值对应的数据,
    
    主表中没有的主键对应的列值,从表中一定不能有这个列值对应的数据
 *  join连接关系  
    ![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjY0MTkwOQ_size_16_color_FFFFFF_t_70]
 *  先分组,再排序  
    ![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjY0MTkwOQ_size_16_color_FFFFFF_t_70 1]

### 1.2 mysql语句顺序 ###

*  查询书写格式
    
        select...(avg(),sum())from...left join...on...
        	where...group by...
        	order by...asc/desc
        	limit ...
 *  执行顺序
    
        form...on...left join...where...
        	group by...
        	avg(),sum()...having...
        	select...
        	order by...asc/desc
        	limit...
    
     *  列别名, 不要在条件中使用, 因为类别名在select后面
     *  表别名可以在条件中使用, from后面是表别名

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjY0MTkwOQ_size_16_color_FFFFFF_t_70 2]

### 1.3 window系统mysql运行远程访问 ###

*  截断表 truncate table xxx;
 *  window系统mysql修改权限
    
        mysql -uroot -proot
        use mysql;
        create

# 2. sql50题详细答案 #

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

-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数 
    SELECT a.*, b.s_score AS 01_score, c.s_score AS 02_score FROM 
    	student a JOIN score b ON a.s_id = b.s_id AND b.c_id = '01'
    	LEFT JOIN score c ON a.s_id = c.s_id AND c.c_id = '02' OR c.c_id = NULL WHERE b.s_score > c.s_score;
    	
    	-- 或者
    	SELECT * (SELECT a.*, b.s_score AS 01_score, c.s_score AS 02_score FROM 
    	student a JOIN score b ON a.s_id = b.s_id AND b.c_id = '01'
    	LEFT JOIN score c ON a.s_id = c.s_id AND c.c_id = '02' OR c.c_id = NULL) d WHERE d.01_01_score > d.02_score;

### 1.1 多表查询规律 ###

*  要先把学生表,成绩表关联起来—关键就是s\_id 和 c\_id
 *  学生表a join 成绩表b 得到学生的语文成绩, 再left join on 成绩表c 得到语文, 数学成绩完整的表
 *  再得到一张大表的基础上 语文成绩大于数学成绩

### 1.2 什么时候用left join , 什么时候不用 ###

*  成绩表score只是有单一的成绩, 姓名&学生编号会重复, student表–>score表是一对多
 *  如果需求是多科成绩之间的比较, 就需要student left join score
 *  如果不涉及多科, 就不需要left join了,
 *  涉及几科, student就left join几次score, 每次关联就取一科成绩

## 2. 查询"01"课程比"02"课程成绩低的学生的信息及课程分数 ##

-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
    
    select a.* ,b.s_score as 01_score,c.s_score as 02_score from 
        student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL 
         join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score

## 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 ##

-- 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
    select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
        student b 
        join score a on b.s_id = a.s_id
        GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)>=60;

*  hql查询语句结构  
    ![在这里插入图片描述][20190323101941387.png]
 *  hql语句执行顺序:
    
     *  无作为连接 先from–>where–>group by–>having–>avg , sum --> select–>order by
        
        有连接查询 form–>on–>join–>where–>group by–>avg,sum–>having–>select–>distinct–>order by–>limit  
        ![在这里插入图片描述][20190323101956493.png]
 *  having 后面跟的条件, 不要写select 中起的别名, 要写起别名之前的原名
    
     *  因为先group by xxx having xxx, 再select xxx起别名
 *  原表的别名, 后面都可以使用, 因为是先from 表名 别名
 *  根据平均成绩排序: order by 的条件,
    
     *  一定要写别名, 因为最后才排序的, 最后已经将查询结果揉成一张大表了,
     *  大表的字段就是select 后面的字段 , 所以就不认识原名了, 只认识别名了
 *  模糊查询
    
     *  \_ 代表一个
     *  % 代表多个
     *  rlike ‘\[9\]’ 含有9的
     *  like关键字
 *  排序
    
     *  order by xxx asc/desc : 全局排序, 只能有一个reduce
     *  sort by xxx asc/desc: map端排序, 局部排序
     *  distribute by 分区字段 + sort by 排序字段 asc/desc : 分区排序
     *  cluster by 字段: 分区字段&排序字段相同时使用 --> 只能倒序,也就是降序

## 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 ##

-- 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)
    select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
        student b 
        left join score a on b.s_id = a.s_id
        GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)<60
        union all
    select a.s_id,a.s_name,0 as avg_score from 
        student a 
        where a.s_id not in (
                    select distinct s_id from score);

*  hql 联合查询是union all
 *  hql中不能省略all, sql中可以省略all
 *  sql语句别名
    
     *  想添加一列并且给列赋值 select 0 as avg\_score, 查出的数据后面添加一列avg\_score,列值均为0
     *  想给原表中的一列起别名 select round(avg(a.s\_score),2) as avg\_score–> 平均值赋值给了avg\_score(新列值)

**以上原理是一样的, 就是可以在原有查询数据基础上,添加列明&列值**

## 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、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 sum_course,sum(b.s_score) as sum_score --4
    	from student a ---1
        left join score b on a.s_id=b.s_id--2
        GROUP BY a.s_id,a.s_name;---3

*  收获**一定要注意思路和查询顺序**
    
     *  先把两张表结合起来
     *  再分组,
     *  再求b.c\_id的count 和b.s\_score的sum
     *  sum()是每个组中的s\_score的和, count()是求每个组中c\_id的个数

## 6. 查询"李"姓老师的数量 ##

-- 6、查询"李"姓老师的数量
    select count(t.t_id) from teacher t where t.t_name like '%李%' group by t.t_id;

## 7. 查询学过"张三"老师授课的同学的信息 ##

-- 7、查询学过"张三"老师授课的同学的信息
    SELECT stu.* FROM student stu LEFT JOIN score sco ON stu.s_id = sco.s_id WHERE sco.c_id = (SELECT c.c_id FROM course c WHERE c.t_id=(SELECT t.t_id FROM teacher t WHERE t.t_name='张三'));

## 8. 查询没学过"张三"老师授课的同学的信息 ##

-- 8、查询没学过"张三"老师授课的同学的信息
    SELECT s.* FROM student s WHERE s.s_id NOT IN 
    (
    	SELECT stu.s_id FROM student stu LEFT JOIN score sco ON stu.s_id = sco.s_id 
    	WHERE sco.c_id =(SELECT c.c_id FROM course c LEFT JOIN teacher t ON c.t_id=t.t_id WHERE 	t.t_name='张三') 
    )

## 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 ##

-- 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
    SELECT a.* FROM student a 
    	LEFT JOIN score b ON a.s_id = b.s_id 
    		LEFT JOIN score c ON a.s_id = c.s_id 
    			WHERE b.c_id = '01' AND c.c_id ='02';

## 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 ##

-- 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 
    SELECT * FROM student WHERE s_id IN
    	(SELECT s_id FROM score WHERE c_id = '01') AND
    		s_id NOT IN (SELECT s_id FROM score WHERE c_id = '02')

### 10.1 判断为null ###

where xxx is null 或者where xxx is not null

## 11. 查询没有学全所有课程的同学的信息 ##

-- 11、查询没有学全所有课程的同学的信息
    SELECT a.* FROM student a LEFT JOIN score b ON a.s_id = b.s_id GROUP BY a.s_id
    	HAVING a.s_id NOT IN 
    		(SELECT s_id AS cts FROM score GROUP BY s_id HAVING COUNT(*) = 3);

## 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 ##

-- 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
    SELECT * FROM student WHERE s_id IN
    (SELECT s_id FROM score WHERE c_id IN
    	(SELECT c_id FROM score WHERE s_id = '01') GROUP BY s_id);

## 13. 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 ##

-- 13、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 
    SELECT * FROM student WHERE s_id IN
    (SELECT a.s_id FROM 
    	(SELECT s_id, COUNT(*) AS cts FROM score WHERE c_id IN
    		(SELECT c_id FROM score WHERE s_id = '01') GROUP BY s_id
    			HAVING cts = (SELECT COUNT(*) FROM score WHERE s_id = '01')) a WHERE a.s_id !='01')

## 14. 查询没学过"张三"老师讲授的任一门课程的学生姓名 ##

-- 14、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 =
    	(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 ##

-- 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
    SELECT a.s_id, a.s_name, AVG(b.s_score) FROM student a 
    	LEFT JOIN score b ON a.s_id = b.s_id WHERE b.s_score < 60 OR b.s_score IS NULL
    		GROUP BY a.s_id;

## 16. 检索"01"课程分数小于60，按分数降序排列的学生信息 ##

-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息 
    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;

## 17. 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 ##

-- 17、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 
    SELECT tempA.*, tempB.avgs FROM 
    	(SELECT a.*, b.s_score AS '01', c.s_score AS '02', d.s_score AS '03' FROM student a
    			LEFT JOIN score b ON a.s_id = b.s_id AND b.c_id = '01'
    				LEFT JOIN score c ON a.s_id = c.s_id AND c.c_id = '02'
    					LEFT JOIN score d ON a.s_id = d.s_id AND d.c_id = '03') tempA LEFT JOIN	
        (SELECT a.s_id, ROUND(AVG(b.s_score),2) AS avgs FROM student a 
    		LEFT JOIN score b ON a.s_id = b.s_id 
    			GROUP BY a.s_id 
    				ORDER BY avgs DESC) tempB ON tempA.s_id = tempB.s_id;

## 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 ##

-- 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 
    //及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90
    SELECT s.c_id AS 课程ID, c.c_name AS 课程NAME, ROUND(MAX(s.s_score),2) AS 最高分, ROUND(MIN(s.s_score),2) AS 最低分, ROUND(AVG(s.s_score),2) AS 平均分,
    	ROUND((SELECT SUM(CASE WHEN b.s_score >= 60 THEN 1 ELSE 0 END) FROM score b WHERE b.c_id = s.c_id)/
    	(SELECT COUNT(*) FROM score a WHERE a.c_id = s.c_id),2) AS 及格率,
    	ROUND((SELECT SUM(CASE WHEN b.s_score >= 70 AND b.s_score < 80 THEN 1 ELSE 0 END) FROM score b WHERE b.c_id = s.c_id)/
    	(SELECT COUNT(*) FROM score a WHERE a.c_id = s.c_id),2) AS 中等率,
    	ROUND((SELECT SUM(CASE WHEN b.s_score >= 80 AND b.s_score < 90 THEN 1 ELSE 0 END) FROM score b WHERE b.c_id = s.c_id)/
    	(SELECT COUNT(*) FROM score a WHERE a.c_id = s.c_id),2) AS 优良率,
    	ROUND((SELECT SUM(CASE WHEN b.s_score >= 90 THEN 1 ELSE 0 END) FROM score b WHERE b.c_id = s.c_id)/
    	(SELECT COUNT(*) FROM score a WHERE a.c_id = s.c_id),2) AS 优秀率
    	FROM score s LEFT JOIN course c ON s.c_id = c.c_id
    	GROUP BY s.c_id;

### 18.1 myql流程控制函数之case ###

*  基本语法
    
        select xxx, xxx, -- 一定要有逗号
        	case 
        		when 条件1 then 条件1查询结果对应的列值,
        		when 条件2 then 条件2查询结果对应的列值,
                 else 不符合以上条件对应的列值,
                 end 查询结果对应的列名 -- 一定不可以有逗号
            from 表名;
 *  case when … else… end… 就是把查询结果单独赋值,单独赋列名

## 19. 按各科成绩进行排序，并显示排名 ##

-- 19、按各科成绩进行排序，并显示排名
     -- 成绩相同不并列
    SELECT ss01.*, @rank1 := @rank1 + 1 AS rank
    	FROM (SELECT c_id, s_score, s_id FROM score WHERE c_id = '01' ORDER BY s_score DESC) ss01,
    	(SELECT @rank1 := 0) r
    UNION ALL
    SELECT ss02.*, @rank2 := @rank2 + 1 AS rank
    	FROM (SELECT c_id, s_score, s_id FROM score WHERE c_id = '02' ORDER BY s_score DESC) ss02,
    	(SELECT @rank2 := 0) r
    UNION ALL
    SELECT ss03.*, @rank3 := @rank3 + 1 AS rank
    	FROM (SELECT c_id, s_score, s_id FROM score WHERE c_id = '03' ORDER BY s_score DESC) ss03,
    	(SELECT @rank3 := 0) r;

### 19.1 rank函数 ###

*  语法
    
        -- 1,2,3 顺序排名
        select @rank := @rank + 1 as rank from (select * from 表名 where xxx=xxx order by xxx desc) 表名, (select @rank := 0) r
        -- 1,1,2 并列排名
        select t1.*,
        		(case when @avgScore = t1.s_score then @rank 
        		      when @avgScore := t1.s_score THEN @rank := @rank+1 end) as rank
        		from (SELECT * FROM score WHERE c_id= '01' ORDER BY s_score DESC) t1,
        		(SELECT @rank := 0, @avgScore := 0)
 *  union all使用
 *  order by & union all 使用
 *  rank使用

### 19.2 :=和=区别 ###

*  := 代表赋值,
 *  = 代表判断

## 20. 查询学生的总成绩并进行排名 ?? 并列排名 ##

-- 20、查询学生的总成绩并进行排名 ?? 并列排名
    SELECT s.s_id, s.sums, @rank := @rank + 1 AS 总成绩排名
    	FROM (SELECT s_id, SUM(s_score) AS sums FROM score GROUP BY s_id ORDER BY sums DESC) s,
    	(SELECT @rank := 0) r

## 21. 查询不同老师所教不同课程平均分从高到低显示 ##

-- 21、查询不同老师所教不同课程平均分从高到低显示
    SELECT c.t_id, c.c_id, c.c_name,t.t_name, AVG(s.s_score) AS avgs FROM score s 
    	LEFT JOIN course c ON s.c_id = c.c_id
    	LEFT JOIN teacher t ON c.t_id = t.t_id
    	GROUP BY t.t_id
    	ORDER BY avgs 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(采取的是1,1,2的方式进行排名) ##

-- 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(采取的是1,1,2的方式进行排名)
    SELECT student.*,tt.c_id,tt.s_score,tt.rank FROM 
    	(	
    		SELECT t1.*,
    		(CASE WHEN @avgScore = t1.s_score THEN @rank1 
    		      WHEN @avgScore := t1.s_score THEN @rank1 := @rank1+1 END) AS rank 
    		FROM (SELECT * FROM score WHERE c_id= '01' ORDER BY s_score DESC) t1,(SELECT @rank1:=0,@avgScore:=0)r
    	UNION ALL
    		SELECT t1.*,
    		(CASE WHEN @avgScore = t1.s_score THEN @rank2 
    		      WHEN @avgScore := t1.s_score THEN @rank2 := @rank2+1 END) AS rank 
    		FROM (SELECT * FROM score WHERE c_id= '02' ORDER BY s_score DESC) t1,(SELECT @rank2:=0,@avgScore:=0)r
    	UNION ALL
    		SELECT t1.*, 
    		(CASE WHEN @avgScore = t1.s_score THEN @rank3
    		      WHEN @avgScore := t1.s_score THEN @rank3 := @rank3+1 END) AS rank 
    		FROM (SELECT * FROM score WHERE c_id= '03' ORDER BY s_score DESC) t1,(SELECT @rank3:=0,@avgScore:=0)r	
    	) tt LEFT JOIN student ON tt.s_id = student.s_id 
    	WHERE tt.rank = 2 OR tt.rank = 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 ##

-- 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
    SELECT c.*,
    	(SELECT COUNT(*) FROM score s WHERE s.s_score >= 85 AND s.s_score <= 100 AND s.c_id = c.c_id) AS `[100-85]`,
    	CONCAT(
    		ROUND(100 * (SELECT COUNT(*) FROM score s WHERE s.s_score >= 85 AND s.s_score <= 100 AND s.c_id = c.c_id)/
    		(SELECT COUNT(*) FROM score s WHERE s.c_id = c.c_id),2),
    	'%') AS '[100-85]占比',
    	(SELECT COUNT(*) FROM score s WHERE s.s_score >= 70 AND s.s_score < 85 AND s.c_id = c.c_id) AS `[85-70]`,
    	CONCAT(
    		ROUND(100 * (SELECT COUNT(*) FROM score s WHERE s.s_score >= 70 AND s.s_score < 85 AND s.c_id = c.c_id)/
    		(SELECT COUNT(*) FROM score s WHERE s.c_id = c.c_id),2),
    	'%') AS '[85-70]占比',
    	(SELECT COUNT(*) FROM score s WHERE s.s_score >= 60 AND s.s_score < 70 AND s.c_id = c.c_id) AS '[70-60]',	
    	CONCAT(
    		ROUND(100 * (SELECT COUNT(*) FROM score s WHERE s.s_score >= 60 AND s.s_score < 70 AND s.c_id = c.c_id)/
    		(SELECT COUNT(*) FROM score s WHERE s.c_id = c.c_id),2),
    	'%') AS '[70-60]占比',
    	(SELECT COUNT(*) FROM score s WHERE s.s_score >= 0 AND s.s_score < 60 AND s.c_id = c.c_id) AS '[60-0]',	
    	CONCAT(
    		ROUND(100 * (SELECT COUNT(*) FROM score s WHERE s.s_score >= 0 AND s.s_score < 60 AND s.c_id = c.c_id)/
    		(SELECT COUNT(*) FROM score s WHERE s.c_id = c.c_id),2),
    	'%') AS '[60-0]占比'
    	FROM course c;

### 23.1 concat()函数 ###

*  字符串拼接函数
 *  SELECT CONCAT (s\_score,‘分’) as 成绩 FROM score;
    
    字符串拼接, 查出来的结果拼接一个字符串

## 24. 查询学生平均成绩及其名次 采取的是（1,1,2）的方式排序 ##

-- 24、查询学生平均成绩及其名次 采取的是（1,1,2）的方式排序
    SELECT temp.*, 
    	(CASE WHEN @avgs = temp.avgs THEN @rank
    	      WHEN @avgs := temp.avgs THEN @rank := @rank + 1 END) AS rank FROM 
    	(SELECT s.s_id, ROUND(AVG(s.s_score),2) AS avgs FROM score s 
    		LEFT JOIN student stu ON s.s_id = stu.s_id 
    		GROUP BY s.s_id ORDER BY avgs DESC) temp,
    	(SELECT @avgs := NULL, @rank := 0) r;

*  rank 112排名

## 25. 查询各科成绩前三名的记录 ##

-- 25、查询各科成绩前三名的记录 
    select * from score a where 3>(select count(*) from score where a.c_id = c_id and s_score > a.s_score ) order by a.c_id,s_score desc

## 26. 查询每门课程被选修的学生数 ##

-- 26、查询每门课程被选修的学生数
    SELECT s.c_id, COUNT(*) AS 选修人数 FROM score s GROUP BY s.c_id;

## 27、查询出只有两门课程的全部学生的学号和姓名 ##

-- 27、查询出只有两门课程的全部学生的学号和姓名 
    SELECT s.s_id, s.s_name FROM student s LEFT JOIN score sco ON s.s_id = sco.s_id
    	GROUP BY s.s_id HAVING COUNT(*) = 2;

## 28、查询男生、女生人数 ##

-- 28、查询男生、女生人数
    SELECT s.s_sex, COUNT(*) AS nums FROM student s GROUP BY s.s_sex

## 29、查询名字中含有"风"字的学生信息 ##

-- 29、查询名字中含有"风"字的学生信息
    SELECT s.* FROM student s WHERE s.s_name LIKE '%风%'

## 30、查询同名同性学生名单，并统计同名人数 ##

-- 30、查询同名同性学生名单，并统计同名人数
    SELECT s.s_name, COUNT(*) AS nums FROM student s GROUP BY s.s_name HAVING nums > 1

### 30.1 统计同名同姓学生名单: ###

*  按照姓名分组,
 *  求count()
 *  count() 大于1的就是同名同姓的人数

### 30.2 substr(列名 from n \[for nums\]) ###

*  截取字符串的函数
 *  substr(列名 from 第几个字符 \[for 要截取几个字符\])

## 31、查询1990年出生的学生名单 ##

-- 31、查询1990年出生的学生名单
    SELECT * FROM student WHERE SUBSTR(s_birth FROM 1 FOR 4) = '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，按课程编号升序排列 ##

-- 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，按课程编号升序排列
    select s.c_id, round(avg(s.s_score),2) as avgs from score s group by s.c_id order by avgs desc, s.c_id asc;

*  32题: 注意, order by avgs 是正确的, order by s.avgs是错误的, group by s.c\_id后就不能s.xxx

## 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 ##

-- 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
    SELECT stu.s_id, stu.s_name, ROUND(AVG(s.s_score),2) AS avgs FROM score s LEFT JOIN student stu
    	ON s.s_id = stu.s_id
    	GROUP BY s.s_id
    	HAVING avgs >= 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 ##

-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数 
    SELECT stu.s_name, s.s_score FROM score s LEFT JOIN student stu 
    	ON s.s_id = stu.s_id 
    	WHERE s.c_id = (SELECT c_id FROM course WHERE c_name = '数学')
    	AND s.s_score < 60

## 35. 查询所有学生的课程及分数情况 ##

-- 35、查询所有学生的课程及分数情况；
    SELECT stu.*, c.c_name, s.s_score FROM score s
    	LEFT JOIN course c ON s.c_id = c.c_id
    	LEFT JOIN student stu ON s.s_id = stu.s_id

## 36. 查询任何一门课程成绩在70分以上的学生的姓名、课程名称和分数 ##

-- 36、查询任何一门课程成绩在70分以上的学生的姓名、课程名称和分数； 
    SELECT stu.*, c.c_name, s.s_score FROM score s 
    	LEFT JOIN course c ON s.c_id = c.c_id 
    	LEFT JOIN student stu ON s.s_id = stu.s_id 
    	WHERE s.s_score > 70

## 37. 查询不及格的课程 ##

-- 37、查询不及格的课程 
    SELECT s.s_id, s.c_id, c.c_name,s.s_score FROM score s 
    	LEFT JOIN course c ON s.c_id = c.c_id 
    	WHERE s.s_score < 60

## 38. 查询课程编号为01且课程成绩大于等于80分的学生的学号和姓名 ##

-- 38、查询课程编号为01且课程成绩大于等于80分的学生的学号和姓名；
    SELECT stu.s_id, stu.s_name, s.c_id, s.s_score FROM score s 
    	LEFT JOIN student stu ON s.s_id = stu.s_id
    	WHERE s.c_id = '01' AND s.s_score >= 80

## 39. 求每门课程的学生人数 ##

-- 39、求每门课程的学生人数 
    SELECT c_id, COUNT(s_id) AS cts FROM score GROUP BY c_id

## 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(可能包含相同成绩) ##

-- 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(可能包含相同成绩)
    SELECT stu.*, s.c_id, MAX(s.s_score) AS max_score FROM score s 
    	LEFT JOIN student stu ON s.s_id = stu.s_id
    	WHERE c_id = (SELECT c_id FROM course WHERE t_id = (SELECT t_id FROM teacher WHERE t_name = '张三'))_id

## 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 ##

-- 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 
    SELECT a.*
    FROM score a 
    LEFT JOIN score b
    ON a.s_id = b.s_id
    WHERE a.c_id <> b.c_id
    AND a.s_score = b.s_score
    GROUP BY a.c_id;

## 42. 查询每门功成绩最好的前两名 ##

-- 42、查询每门功成绩最好的前两名
    SELECT 
    temp.* 
    FROM 
    	(SELECT 
    	s1.*, 
    	@rank1 := @rank1 + 1 AS rank
    	FROM (
    		SELECT a.*
    		FROM score a
    		WHERE a.c_id = '01' ORDER BY a.s_score DESC) s1,
    	(SELECT @rank1 := 0) r	
    	UNION ALL
    	SELECT 
    	s2.*, 
    	@rank2 := @rank2 + 1 AS rank
    	FROM (
    		SELECT a.*
    		FROM score a
    		WHERE a.c_id = '02' ORDER BY a.s_score DESC) s2,
    	(SELECT @rank2 := 0) r
    	UNION ALL
    	SELECT 
    	s3.*, 
    	@rank3 := @rank3 + 1 AS rank
    	FROM (
    		SELECT a.*
    		FROM score a
    		WHERE a.c_id = '03' ORDER BY a.s_score DESC) s3,
    	(SELECT @rank3 := 0) r) temp 
    WHERE temp.rank = 1 OR temp.rank = 2;		
    	
    -- 简单方法, 不明白
    SELECT 
    * 
    FROM score s1 
    WHERE 2 > (SELECT COUNT(*) FROM score WHERE c_id = s1.c_id AND s_score > s1.s_score)
    ORDER BY c_id, s_score DESC

## 43. 统计每门课程的学生选修人数（超过5人的课程才统计） ##

-- 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 
    SELECT
    s.c_id,
    COUNT(*) AS cts
    FROM score s
    GROUP BY s.c_id
    HAVING cts > 5
    ORDER BY cts DESC, s.c_id ASC;

## 44. 检索至少选修两门课程的学生学号 ##

-- 44、检索至少选修两门课程的学生学号 
    SELECT 
    s.s_id,
    COUNT(*) AS cts
    FROM score s
    GROUP BY s.s_id
    HAVING cts >= 2;

## 45. 查询选修了全部课程的学生信息 ##

-- 45、查询选修了全部课程的学生信息 
    SELECT
    * 
    FROM student 
    WHERE s_id IN
    	(SELECT 
    	s.s_id
    	FROM score s 
    	GROUP BY s.s_id
    	HAVING COUNT(*) = 3)

*  month(日期类型的数据),可以取整–>比较大小&加减乘除
 *  weekofyear(xxx), 一年中的第几周

## 46. 查询各学生的年龄 ##

-- 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，年龄减一
    SELECT 
    *,
    (CASE WHEN MONTH(s_birth) > MONTH(NOW()) THEN YEAR(NOW()) - YEAR(s_birth) -1
    	WHEN MONTH(s_birth) < MONTH(NOW()) THEN YEAR(NOW()) - YEAR(s_birth)
    	WHEN MONTH(s_birth) = MONTH(NOW()) AND DAY(s_birth) > DAY(NOW()) THEN YEAR(NOW()) - YEAR(s_birth) - 1
    	WHEN MONTH(s_birth) = MONTH(NOW()) AND DAY(s_birth) < DAY(NOW()) THEN YEAR(NOW()) - YEAR(s_birth)
    	ELSE NULL END) AS age
    FROM student;

## 47. 查询本周过生日的学生 ##

-- 47、查询本周过生日的学生
    SELECT 
    *
    FROM student
    WHERE WEEKOFYEAR(s_birth) = WEEKOFYEAR(NOW())

## 48. 查询下周过生日的学生 ##

-- 48、查询下周过生日的学生
    SELECT 
    *
    FROM student
    WHERE (WEEKOFYEAR(s_birth) - WEEKOFYEAR(NOW()) = 1);

## 49. 查询本月过生日的学生 ##

-- 49、查询本月过生日的学生
    SELECT 
    *
    FROM student
    WHERE MONTH(s_birth) = MONTH(NOW());

## 50. 查询下月过生日的学生 ##

-- 50、查询下月过生日的学生
    SELECT 
    *
    FROM student
    WHERE (MONTH(s_birth) - MONTH(NOW()) = 1)

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