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sql语句练习50题(Mysql版-详加注释)

表名和字段

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
2.课程表
Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号
3.教师表
Teacher(t_id,t_name) --教师编号,教师姓名
4.成绩表
Score(s_id,c_id,s_score) --学生编号,课程编号,分数

测试数据

--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`)
);

--插入测试数据

--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);

练习题和sql语句

-- 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
	
--也可以这样写，个人觉得第二种写法容易理解些。对于不懂join的同学来说。
	select a.*,b.s_score as 01_score,c.s_score as 02_score from student a,score b,score c 
			where a.s_id=b.s_id 
			and a.s_id=c.s_id 
			and b.c_id='01' 
			and c.c_id='02' 
			and b.s_score>c.s_score

关于"join"请查看：图解 SQL 里的各种 JOIN

-- 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

-- 或者
SELECT 
    st.*,
    s1.`s_score` AS 01_score, 
    s2.`s_score` AS 02_score
FROM
    score s1 JOIN score s2 JOIN student st 
    ON s1.`s_id` = s2.`s_id` 
    AND s2.`s_id` = st.`s_id`    
WHERE
    s1.`s_score` > s2.`s_score`
    AND
        s1.`c_id` = '01' 
    AND
        s2.`c_id` = '02'

-- 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 avg_score >=60;

ROUND 函数用于把数值字段舍入为指定的小数位数。

SELECT ROUND(column_name,decimals) FROM table_name

column_name，必需，要处理的字段；decimals，必需，需要返回的小数位数；

在 SQL 中增加 HAVING 子句原因是，WHERE 关键字无法与合计函数一起使用。如AVG, Group By。

SELECT column_name, aggregate_function(column_name)
FROM table_name
WHERE column_name operator value
GROUP BY column_name
HAVING aggregate_function(column_name) operator value

-- 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 avg_score <60
	union
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);
-- 无成绩：没有记录进入score，故选择student表中有的而score表中没有的。

在表中，可能会包含重复值。这并不成问题，不过，有时您也许希望仅仅列出不同（distinct）的值。

关键词 DISTINCT 用于返回唯一不同的值。SELECT DISTINCT 列名称 FROM 表名称

-- 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 from 
	student a 
	left join score b on a.s_id=b.s_id
	GROUP BY a.s_id,a.s_name;

-- 6、查询"李"姓老师的数量

-- 6、查询"李"姓老师的数量 
select count(t_id) from teacher where t_name like '李%';

-- 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 a.* from 
	student a 
	join score b on a.s_id=b.s_id where b.c_id in(
		select c_id from course where t_id =(
			select t_id from teacher where t_name = '张三'));

--另一版本，估计是错的
select a.* from
    student a where a.s_id in  
    select b.s_id from score b, teacher c
    where b.c_id=c.c_id and c.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 * from 
    student  
    where s_id not in(
        select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in(
        select c.c_id from course c join teacher b on c.t_id = b.t_id where b.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,score b,score c 
	where a.s_id = b.s_id  and a.s_id = c.s_id and 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 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')

-- 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 
--@wendiepei的写法
select s.* from student s 
left join Score s1 on s1.s_id=s.s_id
group by s.s_id having count(s1.c_id)<(select count(*) from course)

COUNT() 函数返回匹配指定条件的行数。

SQL COUNT(column_name) 语法

COUNT(column_name) 函数返回指定列的值的数目（NULL 不计入）：

SELECT COUNT(column_name) FROM table_name

SQL COUNT(*) 语法

COUNT(*) 函数返回表中的记录数：

SELECT COUNT(*) FROM table_name

SQL COUNT(DISTINCT column_name) 语法

COUNT(DISTINCT column_name) 函数返回指定列的不同值的数目：

SELECT COUNT(DISTINCT column_name) FROM table_name

注释：COUNT(DISTINCT) 适用于 ORACLE 和 Microsoft SQL Server，但是无法用于 Microsoft Access。

-- 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 distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01')
	);

-- 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 
--@ouyang_1993的写法
SELECT Student.*
FROM  Student
WHERE s_id IN 
(SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
    --#下面的语句是找到'01'同学学习的课程数
    SELECT COUNT(c_id) FROM Score WHERE s_id = '01')
 )
AND s_id NOT IN (
 --#下面的语句是找到学过‘01’同学没学过的课程，有哪些同学。并排除他们
 SELECT s_id FROM Score
 WHERE c_id IN(
   --#下面的语句是找到‘01’同学没学过的课程
   SELECT DISTINCT c_id FROM Score
   WHERE c_id NOT IN (
     --#下面的语句是找出‘01’同学学习的课程
     SELECT c_id FROM Score WHERE s_id = '01')
  ) GROUP BY s_id
) --#下面的条件是排除01同学
AND s_id NOT IN ('01')

--@k1051785839的写法
SELECT    t3.*    FROM
  (
  SELECT s_id, group_concat(c_id ORDER BY c_id) group1
  FROM score
  WHERE s_id > '01'
  GROUP BY s_id
  ) t1
INNER JOIN 
(
 SELECT
  group_concat(c_id ORDER BY c_id) group2
 FROM score
 WHERE s_id = '01'
 GROUP BY s_id
) t2 ON t1.group1 = t2.group2
INNER JOIN 
student t3 ON t1.s_id = t3.s_id

MySQL教程之concat以及group_concat的用法

-- 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 a.s_name from student a where a.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,ROUND(AVG(b.s_score)) from 
	student a 
	left join score b on a.s_id = b.s_id
	where a.s_id in(
			select s_id from score where s_score<60 GROUP BY  s_id having count(1)>=2)
	GROUP BY a.s_id,a.s_name

-- 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,score b 
	where a.s_id = b.s_id and 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 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 a  GROUP BY a.s_id ORDER BY 平均分 DESC;
--@喝完这杯还有一箱的写法
SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文, 
MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学, 
MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语, 
avg(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 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 --及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

-- 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 a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
	ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
	ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
	ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
	ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
	from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name

-- 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
-- mysql没有rank函数
	select a.s_id,a.c_id,
        @i:=@i +1 as i保留排名,
        @k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
        @score:=a.s_score as score
    from (
        select s_id,c_id,s_score from score GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s

--@k1051785839的写法
select * from 
    (select t1.c_id, t1.s_score, (select count(distinct t2.s_score) 
    from score t2 where t2.s_score>=t1.s_score and t2.c_id='01') 
    rank FROM score t1 where t1.c_id='01'
    order by t1.s_score desc) t1)
    union
    (select * from (select t1.c_id, t1.s_score, (select count(distinct t2.s_score) 
    from score t2 where t2.s_score>=t1.s_score and t2.c_id='02') 
    rank FROM score t1 where t1.c_id='02' order by t1.s_score desc) t2)
    union
    (select * from (select t1.c_id, t1.s_score,(select count(distinct t2.s_score) 
    from score t2 where t2.s_score>=t1.s_score and t2.c_id='03') 
    rank FROM score t1 where t1.c_id='03'
    order by t1.s_score desc) t3

Mysql中（@i:=@i+1）的作用

mysql @value := 用法

-- 20、查询学生的总成绩并进行排名

-- 20、查询学生的总成绩并进行排名
select a.s_id,
	@i:=@i+1 as i,
	@k:=(case when @score=a.sum_score then @k else @i end) as rank,
	@score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
	(select @k:=0,@i:=0,@score:=0)s

-- 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 a.t_id,c.t_name,a.c_id,ROUND(avg(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 a.c_id,a.t_id,c.t_name ORDER BY avg_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

-- 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		
			select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'  
								ORDER BY a.s_score DESC  
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3
            UNION
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'  
								ORDER BY a.s_score DESC
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3
            UNION
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03' 
								ORDER BY a.s_score DESC
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 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

select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a
				left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
											ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
								from score GROUP BY c_id)b on a.c_id=b.c_id
				left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
											ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
								from score GROUP BY c_id)c on a.c_id=c.c_id
				left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
											ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
								from score GROUP BY c_id)d on a.c_id=d.c_id
				left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
											ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
								from score GROUP BY c_id)e on a.c_id=e.c_id
				left join course f on a.c_id = f.c_id

-- 24、查询学生平均成绩及其名次

-- 24、查询学生平均成绩及其名次 
		select a.s_id,
				@i:=@i+1 as '不保留空缺排名',
				@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
				@avg_score:=avg_s as '平均分'
		from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;

-- 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
			-- 1.选出b表比a表成绩大的所有组
			-- 2.选出比当前id成绩大的 小于三个的
		select a.s_id,a.c_id,a.s_score from score a 
			left join score b on a.c_id = b.c_id and a.s_score<b.s_score
			group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
			ORDER BY a.c_id,a.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 c_id,count(s_id) from score a GROUP BY 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_id,s_name from student where s_id in(
				select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);

-- 28、查询男生、女生人数

-- 28、查询男生、女生人数 
		select s_sex,COUNT(s_sex) as 人数  from student GROUP BY 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 * from student where 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 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

-- 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 s_name from student where s_birth like '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 c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,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

-- 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 a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
		left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_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

-- 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 a.s_name,b.s_score from score b join student a on a.s_id=b.s_id where b.c_id=(
					select c_id from course where c_name ='数学') and b.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 a.s_id,a.s_name,
					SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
					SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
					SUM(case c.c_name when '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 b.c_id = c.c_id 
		GROUP BY a.s_id,a.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；

 -- 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； 
			select a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id
				left join student a on a.s_id=c.s_id where c.s_score>=70

-- 37、查询不及格的课程

-- 37、查询不及格的课程
	select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id
			where a.s_score<60

--38、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；

--38、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,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id
			where a.c_id = '01'	and a.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 count(*) 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

-- 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
		
	-- 查询老师id	
	select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
	-- 查询最高分（可能有相同分数）
	select MAX(s_score) from score where c_id='02'
	-- 查询信息
	select a.*,b.s_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 b.c_id=c.c_id
		where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')
		and b.s_score in (select MAX(s_score) from score where c_id='02')

-- 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 DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score

-- 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 
		-- 牛逼的写法
	select a.s_id,a.c_id,a.s_score from score a
		where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id

-- 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

-- 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 c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,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_id,count(*) as sel from score GROUP BY s_id HAVING sel>=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_id from score GROUP BY s_id HAVING count(*)=(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，年龄减一

--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 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;

-- 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 WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
	select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))	
	select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

-- 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 WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)

-- 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(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)

-- 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(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

重要参考：

sql语句练习50题(Mysql版)：https://blog.csdn.net/fashion2014/article/details/78826299

W3School在线教程：https://www.w3school.com.cn/index.html

经典SQL练习题(MySQL版)（有运行结果）：https://blog.csdn.net/paul0127/article/details/82529216

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原文地址：https://www.cnblogs.com/JasonLGJnote/p/11876165.html

sql语句练习50题(Mysql版-详加注释)

表名和字段

测试数据

练习题和sql语句

​​​​​​​​​​​重要参考：

重要参考：