SQL语句练习（mysql）-1

本sql语句练习题目取自：https://blog.csdn.net/fashion2014/article/details/78826299
SQL语句均为自写，今后会持续发布SQL练习，用于自己记小笔记，练习时建议别看答案。
数据表：

-- 建表
-- 学生表
CREATE TABLE `Student` (
`s_id` VARCHAR ( 20 ) COMMENT '学生编号',
`s_name` VARCHAR ( 20 ) NOT NULL DEFAULT '' COMMENT '学生姓名',
`s_birth` VARCHAR ( 20 ) NOT NULL DEFAULT '' COMMENT '出生年月',
`s_sex` VARCHAR ( 10 ) NOT NULL DEFAULT '' COMMENT '学生性别',
PRIMARY KEY ( `s_id` ) 
) COMMENT = '学生表';
-- 课程表
CREATE TABLE `Course` (
`c_id` VARCHAR ( 20 ) COMMENT '课程编号',
`c_name` VARCHAR ( 20 ) NOT NULL DEFAULT '' COMMENT '课程名称',
`t_id` VARCHAR ( 20 ) NOT NULL COMMENT '教师编号',
PRIMARY KEY ( `c_id` ) 
) COMMENT = '课程表';
-- 教师表
CREATE TABLE `Teacher` ( `t_id` VARCHAR ( 20 ) COMMENT '教师编号', `t_name` VARCHAR ( 20 ) NOT NULL DEFAULT '' COMMENT '教师姓名', PRIMARY KEY ( `t_id` ) ) COMMENT = '教师表';
-- 成绩表
CREATE TABLE `Score` (
`s_id` VARCHAR ( 20 ) COMMENT '学生编号',
`c_id` VARCHAR ( 20 ) COMMENT '课程编号',
`s_score` INT ( 3 ) COMMENT '分数',
PRIMARY KEY ( `s_id`, `c_id` ) 
) COMMENT = '成绩表';

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 );

题目如下：

-- 1、查询"01"课程比"02"课程成绩高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
-- 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
-- 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)
-- 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
-- 6、查询"李"姓老师的数量 
-- 7、查询学过"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
-- 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
-- 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
-- 11、查询没有学全所有课程的同学的信息
-- 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 
-- 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
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名 
-- 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 
-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息
-- 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
-- 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
-- 19、按各科成绩进行排序，并显示排名
-- 20、查询学生的总成绩并进行排名
-- 21、查询不同老师所教不同课程平均分从高到低显示 
-- 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
-- 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
-- 24、查询学生平均成绩及其名次
-- 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
-- 26、查询每门课程被选修的学生数 
-- 27、查询出只有两门课程的全部学生的学号和姓名 
-- 28、查询男生、女生人数 
-- 29、查询名字中含有"风"字的学生信息
-- 30、查询同名同性学生名单，并统计同名人数 
-- 31、查询1990年出生的学生名单
-- 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
-- 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 
-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数 
-- 35、查询所有学生的课程及分数情况；
-- 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； 
-- 37、查询不及格的课程
-- 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；
-- 39、求每门课程的学生人数 
-- 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
-- 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 
-- 42、查询每门功成绩最好的前两名 
-- 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
-- 44、检索至少选修两门课程的学生学号 
-- 45、查询选修了全部课程的学生信息 
-- 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
-- 47、查询本周过生日的学生
-- 48、查询下周过生日的学生
-- 49、查询本月过生日的学生
-- 50、查询下月过生日的学生

题解如下：

-- 1-13题（后面题暂时没写，后面慢慢写）
-- 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
	Student.s_id AS '学生编号',
	Student.s_name AS '学生姓名',
	Student.s_birth AS '出生日期',
	Student.s_sex AS '性别',
	Score_01.c_id AS '课程编号',
	Score_01.s_score AS '01课程成绩',
	Score_02.c_id AS '课程编号',
	Score_02.s_score AS '02课程成绩' 
FROM
	Student
	LEFT JOIN ( SELECT * FROM Score WHERE c_id = '01' ) Score_01 ON Student.s_id = Score_01.s_id
	LEFT JOIN ( SELECT * FROM Score WHERE c_id = '02' ) Score_02 ON Score_01.s_id = Score_02.s_id 
WHERE
	Score_01.s_score > Score_02.s_score 
	*/

-- 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
	Student.s_id AS '学生编号',
	Student.s_name AS '学生姓名',
	Student.s_birth AS '出生日期',
	Student.s_sex AS '性别',
	Score_01.c_id AS '课程编号',
	Score_01.s_score AS '01课程成绩',
	Score_02.c_id AS '课程编号',
	Score_02.s_score AS '02课程成绩' 
FROM
	Student
	LEFT JOIN ( SELECT * FROM Score WHERE c_id = '01' ) Score_01 ON Student.s_id = Score_01.s_id
	LEFT JOIN ( SELECT * FROM Score WHERE c_id = '02' ) Score_02 ON Score_01.s_id = Score_02.s_id 
WHERE
	Score_01.s_score < Score_02.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
/*
SELECT
	* 
FROM
	(
SELECT
	Student.s_id AS '学生编号',
	Student.s_name AS '学生姓名',
	AVG( Score.s_score ) AS '平均成绩' 
FROM
	Student
	JOIN Score ON Student.s_id = Score.s_id 
GROUP BY
	Student.s_id 
	) avgScore 
WHERE
	平均成绩 > 60
*/

-- 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
	* 
FROM
	(
SELECT
	Student.s_id AS '学生编号',
	Student.s_name AS '学生姓名',
	AVG( Score.s_score ) AS '平均成绩' 
FROM
	Student
	LEFT JOIN Score ON Student.s_id = Score.s_id 
GROUP BY
	Student.s_id 
	) avgScore 
WHERE
	平均成绩 < 60 UNION
SELECT
	Student.s_id,
	Student.s_name,
	0 AS s_score 
FROM
	Student 
WHERE
	Student.s_id NOT IN ( SELECT DISTINCT s_id FROM Score )
*/

-- 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
	Student.s_id AS '学生编号',
	Student.s_name AS '学生姓名',
	count( Score.c_id ) AS '选课总数',
	sum( Score.s_score ) 
FROM
	Student
	LEFT JOIN Score ON Student.s_id = Score.s_id 
GROUP BY
	student.s_id,
	student.s_name
	*/

-- 6、查询"李"姓老师的数量 
/*
SELECT
	COUNT( * ) 
FROM
	Teacher 
WHERE
	Teacher.t_name LIKE '李%'
	*/

-- 7、查询学过"张三"老师授课的同学的信息
/*第一种解决方法
SELECT
	Student.s_id as '学生编号',Student.s_name as '学生姓名',Student.s_birth as '学生生日',Student.s_sex as '学生性别' 
FROM
	Student
	JOIN (
SELECT
	Score.s_id 
FROM
	Score
	JOIN ( SELECT Course.c_id FROM Course JOIN Teacher ON Course.t_id = Teacher.t_id WHERE Teacher.t_name = '张三' ) Cid ON Score.c_id = Cid.c_id 
WHERE
	Score.c_id = Cid.c_id 
	) Sid ON Student.s_id = Sid.s_id
	*/
	
	/*第二种解决方法
SELECT
	stu.s_id as '学生编号',stu.s_name as '学生姓名',stu.s_birth as '学生生日',stu.s_sex as '学生性别' 
FROM
	student stu
	JOIN score sco ON stu.s_id = sco.s_id 
WHERE
	sco.c_id IN ( SELECT c_id FROM course WHERE t_id = ( SELECT t_id FROM teacher WHERE t_name = '张三' ) );
	*/

-- 8、查询没学过"张三"老师授课的同学的信息 
/*
SELECT
	s_id AS '学生编号',
	s_name AS '学生姓名',
	s_birth AS '学生生日',
	s_sex AS '学生性别' 
FROM
	Student 
WHERE
	s_id NOT IN (
SELECT
 s_id 
FROM
	Score 
WHERE
	Score.c_id = ( SELECT c_id FROM Course WHERE t_id = ( SELECT t_id FROM Teacher WHERE 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
/*
SELECT
	s_id AS '学生编号',
	s_name AS '学生姓名',
	s_birth AS '出生日期',
	s_sex AS '性别' 
	FROM
		student 
WHERE
		student.s_id IN (
	SELECT
		sc1.s_id 
	FROM
		( SELECT * FROM Score WHERE c_id = 01 ) sc1
		JOIN ( SELECT * FROM Score WHERE c_id = 02 ) sc2 ON sc1.s_id = sc2.s_id 
	)
	*/
	
-- 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
	s_id AS '学生编号',
	s_name AS '学生姓名',
	s_birth AS '出生日期',
	s_sex AS '性别' 
FROM
	Student 
WHERE
	s_id IN ( SELECT s_id FROM Score WHERE s_id NOT IN ( SELECT s_id FROM Score WHERE c_id = 02 ) AND c_id = 01 )
*/

-- 11、查询没有学全所有课程的同学的信息
/*解决方法1:找出有三门课程的同学然后排除掉
SELECT
	s_id AS '学生编号',
	s_name AS '学生姓名',
	s_birth AS '出生日期',
	s_sex AS '性别' 
FROM
	student 
WHERE
	s_id NOT IN (
SELECT
	sc1.s_id 
FROM
	( SELECT * FROM score WHERE c_id = 01 ) sc1
	JOIN ( SELECT * FROM score WHERE c_id = 02 ) sc2 ON sc2.s_id = sc1.s_id
	JOIN ( SELECT * FROM score WHERE c_id = 03 ) sc3 ON sc2.s_id = sc3.s_id 
	)
*/

/*解决方法2：首先查出课程总数，然后将课程数为总课程数的筛选出来排除掉
SELECT
	* 
FROM
	student 
WHERE
	s_id NOT IN ( SELECT s_id FROM score t1 GROUP BY s_id HAVING count( * ) = ( SELECT count( DISTINCT c_id ) 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 
/*
SELECT
	s_id AS '学生编号',
	s_name AS '学生姓名',
	s_birth AS '出生日期',
	s_sex AS '性别' 
FROM
	Student 
WHERE
	s_id IN ( SELECT DISTINCT s_id FROM Score WHERE c_id IN ( SELECT c_id FROM Score WHERE 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
/* 解决方案:首先找出学过的课程数跟01号学生学过的课程数相同的学生，然后第二大步（首先查找出01学生学过的课程，再过滤01同学学过的课程得到01同学没学过的课程，
然后将学过01同学没学过的课程的同学剔除掉）
SELECT
	* 
FROM
	Student 
WHERE
	s_id IN (
SELECT
	s_id 
FROM
	score 
GROUP BY
	s_id 
HAVING
	count( s_id ) = ( SELECT count( c_id ) FROM score WHERE s_id = '01' ) 
	) 
	AND s_id NOT IN (
SELECT DISTINCT
	s_id 
FROM
	score 
WHERE
	c_id IN ( SELECT c_id FROM score WHERE c_id NOT IN ( SELECT c_id FROM Score WHERE s_id = '01' ) GROUP BY s_id ) 
	)
*/
*/

本节仅1-13题题解