SQL每日一题挑战计划

2021/4/1 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	
-- 确定学生的信息在Student表中，课程的分数在Course表中
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'
join score c on a.s_id=c.s_id and c.c_id='02'  
where b.s_score>c.s_score

2021/4/1 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 01_score , c.s_score 02_score FROM Student a 
JOIN Score b ON a.s_id = b.s_id and b.c_id = '01'
JOIN Score c ON a.s_id = c.s_id and c.c_id  ='02'
where b.s_score < c.s_score;

2021/4/1 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 a.s_id , a.s_name , AVG(b.s_score) avg_score FROM Student a
join Score b ON a.s_id = b.s_id
GROUP BY a.s_id HAVING avg_score >= 60;

2021/4/1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 a.s_id , a.s_name , AVG(b.s_score) avg_score FROM Student a
JOIN Score b ON a.s_id = b.s_id GROUP BY a.s_id 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
)

2021/4/2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(DISTINCT(b.c_id)) , SUM(b.s_score) FROM Student a JOIN
Score b ON a.s_id = b.s_id GROUP BY a.s_id;

2021/4/2 查询"李"姓老师的数量

SELECT COUNT(a.t_id) FROM Teacher a where a.t_name like '李%'

2021/4/2 查询学过"张三"老师授课的同学的信息

SELECT a.*,d.t_name FROM Student a JOIN Score b ON a.s_id = b.s_id
JOIN Course c ON b.c_id = c.c_id
JOIN Teacher d ON c.t_id = d.t_id and d.t_name = '张三';

2021/4/2 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
	JOIN Course c ON b.c_id = c.c_id
	JOIN Teacher d ON c.t_id = d.t_id and d.t_name = '张三'
);

2021/4/2 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 * FROM Student where s_id in
(
	SELECT s_id FROM Score where 
	s_id in (SELECT DISTINCT s_id FROM Score  where c_id = '01') 
	and c_id = '02'
);

2021/4/2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 
	s_id not in (SELECT DISTINCT s_id FROM Score  where c_id = '02') 
	and c_id = '01'
);

2021/4/3查询没有学全所有课程的同学的信息

SELECT a.* FROM Student a where a.s_id not in(
	SELECT a.s_id FROM Score a GROUP BY a.s_id HAVING COUNT(a.s_id) = (SELECT count(*) FROM Course)
)

SELECT a.* FROM student a LEFT JOIN score b ON b.s_id = a.s_id 
GROUP BY a.s_id HAVING count(b.s_id) < (SELECT count(*) FROM course)

2021/4/3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 DISTINCT a.* FROM student a JOIN score b ON a.s_id = b.s_id
where b.c_id in (SELECT c_id FROM score where s_id = '01')

2021/4/3（这个题太难了）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 s_id FROM score where s_id in(	
		SELECT DISTINCT s_id FROM score where s_id in(		
			SELECT DISTINCT s_id FROM score where c_id = (SELECT c_id FROM score where s_id='01' LIMIT 0,1) and s_id != '01'
		) and c_id = (SELECT c_id FROM score where s_id='01' LIMIT 1,1)
	)and c_id = (SELECT c_id FROM score where s_id='01' LIMIT 2,1)
)

2021/4/4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(
-- 学过张三老师的课学生的id
	SELECT DISTINCT s_id FROM score where c_id = (
		SELECT a.c_id FROM course a JOIN teacher b ON a.t_id = b.t_id where b.t_name = '张三'
	)
)

2021/4/4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 JOIN score b
ON a.s_id = b.s_id where b.s_id in (
	-- 查询两门及其以上不及格课程的同学的学号
	SELECT s_id FROM score where s_score < 60  GROUP BY s_id HAVING count(s_id) >= 2
) GROUP BY b.s_id

2021/4/7检索"01"课程分数小于60，按分数降序排列的学生信息

SELECT a.s_id , a.s_name , b.s_score FROM student a JOIN score b ON a.s_id=b.s_id
and b.c_id = '01' where b.s_score < 60 ORDER BY b.s_score desc

2021/4/7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;

2021/4/7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(a.s_score) as 最高分, min(a.s_score) as 最低分 , avg(a.s_score) as 平均分 , sum(case when a.s_score >= 60 then 1 else 0 end)/sum(case when a.s_score then 1 end) as 及格率 , 
						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 end) as 中等率,
						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 end) as 优良率,
						sum(case when a.s_score >= 90  then 1 else 0 end)/sum(case when a.s_score then 1 end) as 优秀率
FROM score a JOIN course b ON a.c_id = b.c_id GROUP BY a.c_id

2021/4/8按各科成绩进行排序，并显示排名

1MySql模拟生成rownum的方法

Oracle中有一个伪列rownum，可以在生成查询结果表的时候生成一组递增的序列号。MySQL中没有这个伪列，但是有时候要用，可以用如下方法模拟生成一列自增序号。

（1）sql示例:

SELECT @i:=@i+1 as rownum ,s_id,s_name FROM student , (SELECT @i:=0) as init;

(2) 如果是多表联查，跟上述sql类似，连查完后定义一个初始化序列号即可：

SELECT @i:=@i+1  as rownum ,a.s_id , a.s_name , b.c_id , b.s_score FROM student a JOIN score b ON
a.s_id = b.s_id , (SELECT @i:=0) as init;

select s_id,c_id,s_score, @i:=@i+1 as rank from score,(SELECT @i:=0) as init GROUP BY s_id,c_id,s_score ORDER BY s_score DESC

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 score a, (select @k:=0,@i:=0,@score:=0) as init GROUP BY a.s_id ,a.c_id,a.s_score  ORDER BY a.s_score desc
				

2021/4/8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 as 老师ID , a.c_id as 课程号 , AVG(b.s_score) as 平均成绩 FROM course a JOIN 
score b ON a.c_id = b.c_id GROUP BY a.t_id , b.c_id ORDER BY 平均成绩 desc

***2021/4/9查询各科成绩前三名的记录

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