本文介绍: 本文整理了经典的 50 道 SQL 题目文本分为建库建表、插入数据以及 SQL 50 题这三个部分。

目录

本文整理了经典的 50 道 SQL 题目文本分为建库建表插入数据以及 SQL 50 题这三个部分。
② 这些题目许多博主整理过,但本人不太了解这些题目具体的出处。第一次了解这些题目是本科期间老师出的题目。如果有网友知道这些题目的最原始出处,可以评论评论区中告知。
本文使用的 MySQL 版本为 5.5,虽然版本有一点旧,但是对 SQL 知识点复习没有太大的影响(除了一些旧版没有函数)。
④ 由于本文旨在对 SQL 基础知识进行复习,并且所涉及的数据量也十分的小,所以在编写 SQL 语句时,并未过多考虑 SQL 优化的方面。如果读者有其它的解法或者发现错误之处,可在评论留言,笔者在看到后会及时更新

1.建库建表

(1)建库:创建一个名为 sqlpractice数据库
(2)建表:建立 studentcourseteacherscore 这 4 张表。它们的字段以及之间的关系如下图所示
在这里插入图片描述
(3)建库建表的完整 SQL 语句如下所示。

# 建库
create database sqlpractice;
use sqlpractice;

# 建立 Student 学生
CREATE TABLE Student(
s_id VARCHAR(20),
s_name VARCHAR(20) NOT NULL,
s_birth VARCHAR(20) NOT NULL, 
s_sex VARCHAR(10) NOT NULL,
PRIMARY KEY(s_id)	# 主键
);

# 建立 Course 课程表
CREATE TABLE Course(
c_id VARCHAR(20),
c_name VARCHAR(20) NOT NULL,
t_id VARCHAR(20) NOT NULL,
PRIMARY KEY(c_id)	# 主键
);

# 建立 Teacher 教师表
CREATE TABLE Teacher(
t_id VARCHAR(20),
t_name VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(t_id)	# 主键
);

# 建立 Score 分数表
CREATE TABLE Score(
s_id VARCHAR(20),
c_id VARCHAR(20),
s_score INT(3),
PRIMARY KEY(s_id, c_id)  # 联合主键
);

# 添加外键
# 语法:ALTER TABLE 从表 ADD FOREIGN KEY(外键字段) REFERENCES 主表(主键字段)
ALTER TABLE Course ADD FOREIGN KEY(t_id) REFERENCES Teacher(t_id)
ALTER TABLE Score ADD FOREIGN KEY(s_id) REFERENCES Student(s_id)
ALTER TABLE Score ADD FOREIGN KEY(c_id) REFERENCES Course(c_id)

2.插入数据

(1)向上面创建的 4 张表中插入测试数据的 SQL 语句如下所示(需要注意表之间的关系,以免插入数据失败)。

# 分别向四张表中插入数据
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');

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

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

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);

(2)检验插入数据是否成功

SELECT * FROM Student;
SELECT * FROM Course;
SELECT * FROM Teacher;
SELECT * FROM Score;

Student 表
在这里插入图片描述

Course 表

在这里插入图片描述

Teacher 表

在这里插入图片描述

Score 表

在这里插入图片描述

3.SQL 50 题

3.1.✨SQL 01——查询“01”课程比”02″课程成绩高的学生信息课程分数

# 本题需要比较"01"课程比"02"课程成绩,故在 where 中将 score 表中字段 s_score 使用 2 次(即分别对应"01"课程成绩和"02"课程的成绩
# 所以可以使用为 s_score 表取别名方式多次使用 score 表中字段
SELECT
	student.*,
	score1.s_score 
FROM
	student,
	score AS score1,
	score AS score2 
WHERE
	student.s_id = score1.s_id 
	AND score1.s_id = score2.s_id # student, score1, score2 表连接条件是它们的 s_id 均相等
	AND score1.c_id = '01' 
	AND score2.c_id = '02' 
	AND score1.s_score > score2.s_score;

在这里插入图片描述

3.2.SQL 02——查询“01”课程比”02″课程成绩低的学生信息及课程分数

SELECT
	student.*,
	score1.s_score 
FROM
	student,
	score AS score1,
	score AS score2 
WHERE
	student.s_id = score1.s_id 
	AND score1.s_id = score2.s_id # student, score1, score2 表连接条件是它们的 s_id 均相等
	AND score1.c_id = '01' 
	AND score2.c_id = '02' 
	AND score1.s_score < score2.s_score;

在这里插入图片描述

3.3.SQL 03——查询平均成绩大于等于 60 分的同学学生编号学生姓名和平均成绩

# 1.创建临时表 ss
EXPLAIN SELECT
	student.s_id,
	student.s_name,
	ss.avg_score 
FROM
	student,
	(SELECT s_id, AVG(s_score) AS avg_score FROM score GROUP BY s_id) AS ss 
WHERE
	student.s_id = ss.s_id 
	AND ss.avg_score &gt;= 60;

# 2.先进行内连接然后分组
SELECT
	student.s_id,
	s_name,
	round(AVG(score.s_score), 2) as avg_score
FROM
	student
	INNER JOIN score ON student.s_id = score.s_id 
GROUP BY
	student.s_id,
	s_name 
HAVING
	AVG(score.s_score) &gt;= 60

在这里插入图片描述

3.4.✨SQL 04——查询平均成绩小于 60 分的同学学生编号学生姓名和平均成绩(包括有成绩的和无成绩的)

# isnull(exper) 判断 exper 是否为空,是则返回 1,否则返回 0
# ifnull(exper1, exper2) 判断 exper1 是否为空,是则用 exper2 代替
# nullif(exper1, exper2) 如果 expr1 = expr2 成立,那么返回值为 NULL,否则返回值为 expr1。
SELECT
	student.s_id,
	s_name,
	round(AVG(score.s_score), 2) as avg_score
FROM
	student
	LEFT OUTER JOIN score ON student.s_id = score.s_id 
GROUP BY
	student.s_id,
	s_name 
HAVING
	AVG(IFNULL(score.s_score,0)) < 60

在这里插入图片描述

3.5.SQL 05——查询所有同学学生编号学生姓名、选课总数、所有课程的总成绩

SELECT
	student.s_id,
	student.s_name,
	COUNT(DISTINCT c_id) AS totalCourses,
	SUM(s_score) AS totalScores 
FROM
	student
	# 由于要查询所有的学生,故无论其是否有课程信息都要查询,所以使用 LEFT OUTER JOIN
	LEFT OUTER JOIN score ON student.s_id = score.s_id 
GROUP BY
	student.s_id,
	student.s_name;

在这里插入图片描述

3.6.SQL 06——查询”李”姓老师的数量

# 1.模糊查询
SELECT
	COUNT(*) 
FROM
	teacher 
WHERE
	t_name LIKE '李%'

# 2.正则表达式查询,字符 '^' 匹配以特定字符或者字符串开头的文本
SELECT
	count(*) 
FROM
	teacher 
WHERE
	t_name REGEXP '^李'

在这里插入图片描述

3.7.✨SQL 07——查询学过”张三”老师授课的同学的信息

# 1.使用多表连接(score, course, teacher)找到上张三老师课的同学的 s_id,然后再根据 s_id 从 student 表中查询同学信息
SELECT
	student.* 
FROM
	student 
WHERE
	s_id IN (
		SELECT
			s_id 
		FROM
			score,
			course,
			teacher 
		WHERE
			teacher.t_name = '张三' 
			AND teacher.t_id = course.t_id 
			AND course.c_id = score.c_id 
	)

# 2.多层嵌套子查询(当数据量较大时,一般不推荐使用子查询)
# 在 student 表中根据上过张三老师教的课的学生 s_id 来查询他们的信息
SELECT
	student.* 
FROM
	student 
WHERE
	student.s_id IN (
			# 在 score 表中根据张三老师教的课程 c_id 来查找上这些课的学生 s_id
			SELECT DISTINCT
				s_id 
			FROM
				score 
			WHERE
				score.c_id IN (
					# 在 course 表中根据张三老师的 t_id 查询他所教的课程 c_id
					SELECT 
						c_id 
					FROM 
						course 
					WHERE 
						course.t_id = (
							# 在 teacher 表中查询张三老师的 t_id
							SELECT 
								t_id 
							FROM 
								teacher 
							WHERE 
								t_name = '张三'
						)
				)
	)

在这里插入图片描述

3.8.SQL 08——查询没学过”张三”老师授课的同学的信息

SELECT
	student.* 
FROM
	student 
WHERE
	student.s_id NOT IN (
			SELECT DISTINCT
				s_id 
			FROM
				score 
			WHERE
				score.c_id IN (
					SELECT 
						c_id 
					FROM 
						course 
					WHERE 
						course.t_id = (
							SELECT 
								t_id 
							FROM 
								teacher 
							WHERE 
								t_name = '张三'
						)
				)
	)

在这里插入图片描述

3.9.SQL 09——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
	student.* 
FROM
	student 
WHERE
	student.s_id IN (
		SELECT
			s1.s_id 
		FROM
			score AS s1,
			score AS s2 
		WHERE
			s1.s_id = s2.s_id 
			AND s1.c_id = '01' 
			AND s2.c_id = '02' 
	)

在这里插入图片描述

3.10.✨SQL 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
	stu.s_id,
	stu.s_name,
	stu.s_birth,
	stu.s_sex 
FROM
	student AS stu
	JOIN score AS sc ON stu.s_id = sc.s_id
	JOIN course AS co ON co.c_id = sc.c_id 
WHERE
	co.c_id = '01' 
	AND stu.s_id NOT IN (
		# 查询学过编号为 "02" 的课程的同学 id
		SELECT
			stu.s_id
		FROM
			student AS stu
			JOIN score AS sc ON stu.s_id = sc.s_id
			JOIN course AS co ON co.c_id = sc.c_id 
		WHERE
			co.c_id = '02' 
	)

在这里插入图片描述

3.11.SQL 11——查询没有学全所有课程的同学的信息

# 下面的课程数量 3 也可以用 (SELECT count(*) FROM course) 来代替
SELECT
	* 
FROM
	student 
WHERE
	s_id IN (SELECT s_id FROM score GROUP BY s_id HAVING count(c_id) < 3)

在这里插入图片描述

3.12.SQL 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

# 不包括学号为 '01' 学生自己
SELECT
	* 
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') 
			  AND s_id != '01'
	)

在这里插入图片描述

3.13.✨SQL 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
			s_id 
		FROM
			score 
		WHERE
			# 保证学习的课程相同
			c_id IN (SELECT DISTINCT c_id FROM score WHERE s_id = '01') 
			AND s_id != '01' 
		GROUP BY
			s_id 
		HAVING
			# 保证学习的课程数量相同
			count(c_id) = (select count(*) from score where s_id = '01')
	)

在这里插入图片描述

3.14.SQL 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 (
		# 查询学习过"张三"老师讲授的任一门课程的学生 id
		SELECT
			s_id 
		FROM
			score 
		WHERE
			c_id IN ( 
				# 查询由姓名为张三的老师所讲授的课程 id
				SELECT
					c_id 
				FROM
					course 
				WHERE
					t_id IN (SELECT t_id FROM teacher WHERE t_name = '张三')
			)
	)

在这里插入图片描述

3.15.✨SQL 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
	stu.s_id,
	stu.s_name,
	tmp_t.avg_score 
FROM
	student AS stu
	INNER JOIN 
		(
			SELECT 
				s_id, 
				round(avg(s_score), 2) AS avg_score 
			FROM 
				score 
			WHERE 
				s_score < 60 
			GROUP BY 
				s_id 
			HAVING 
				count(s_score) &gt;= 2
		) AS tmp_t 
	ON 
		stu.s_id = tmp_t.s_id

在这里插入图片描述

3.16.SQL 16——检索“01”课程分数小于60,按分数降序排列的学生信息

SELECT
	stu.* 
FROM
	student AS stu
	INNER JOIN score ON stu.s_id = score.s_id 
WHERE
	c_id = '01' 
	AND 
	s_score < 60 
ORDER BY
	s_score DESC

在这里插入图片描述

3.17.SQL 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
	s_id,
	max(CASE c_id WHEN '01' THEN s_score ELSE 0 END) AS '01',
	max(CASE c_id WHEN '02' THEN s_score ELSE 0 END) AS '02',
	max(CASE c_id WHEN '03' THEN s_score ELSE 0 END) AS '03',
	avg(s_score) AS avg_score 
FROM
	score 
GROUP BY
	s_id 
ORDER BY
	avg_score DESC

在这里插入图片描述

3.18.✨SQL 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,课程名称,最高分,最低分,平均分,及格率,中等率,优良率,优秀率。及格为 60-70,中等为 71-80,优良为 81-90,优秀为 &gt;= 91。

SELECT
	sc.c_id AS "课程ID",
	c.c_name AS '课程名称',
	MAX(sc.s_score) AS "最高分",
	MIN(sc.s_score) AS '最低分',
	AVG(sc.s_score) AS '平均分',
	SUM(IF (sc.s_score BETWEEN 60 AND 70, 1, 0)) / COUNT(*) as '及格率',
	SUM(IF (sc.s_score BETWEEN 71 AND 80, 1, 0)) / COUNT(*) as '中等率',
	SUM(IF (sc.s_score BETWEEN 81 AND 90, 1, 0)) / COUNT(*) as '优良率',
	SUM(IF (sc.s_score &gt;= 91, 1, 0)) / COUNT(*) as '优秀率' 
FROM
	score AS sc
	JOIN course AS c ON sc.c_id = c.c_id 
GROUP BY
	sc.c_id

在这里插入图片描述

3.19.✨SQL 19——按各科成绩进行排序,并显示排名,成绩重复合并名次

SELECT
	sc1.c_id,
	sc1.s_id,
	sc1.s_score,
	count(sc2.s_score) + 1 AS rank 
FROM
	score AS sc1 LEFT JOIN score AS sc2 
	ON sc1.s_score < sc2.s_score
	   AND sc1.c_id = sc2.c_id 
GROUP BY
	sc1.c_id,
	sc1.s_id,
	sc1.s_score 
ORDER BY
	sc1.c_id,
	rank

在这里插入图片描述

3.20.✨SQL 20——查询学生的总成绩并进行排名,总分重复时不保留名次空缺

SELECT
    stu.s_id,
    stu.s_name,
    total_score,
    (
			SELECT COUNT(DISTINCT total_score) 
			FROM (SELECT SUM(s_score) AS total_score FROM score GROUP BY s_id) AS sub 
			WHERE total_score &gt;= tmp.total_score
		) AS rank
FROM
    student as stu
    INNER JOIN (
        SELECT
            s_id,
            SUM(s_score) AS total_score
        FROM
            score
        GROUP BY
            s_id
    ) AS tmp ON stu.s_id = tmp.s_id
ORDER BY
    total_score DESC;

在这里插入图片描述

3.21.SQL 21——查询不同老师所教不同课程平均分,并从高到低显示

SELECT
	teacher.t_id,
	t_name,
	round(avg(s_score), 2) AS avg_score 
FROM
	teacher,
	course,
	score 
WHERE
	teacher.t_id = course.t_id 
	AND course.c_id = score.c_id 
GROUP BY
	teacher.t_id,
	t_name,
	score.c_id 
ORDER BY
	avg(score.s_score) DESC

在这里插入图片描述

3.22.SQL 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

# 1.分别对每门课程进行查询,然后合并查询结果,但是如果课程太多,该方法就不太合适
SELECT
	t1.* 
FROM
	(
		SELECT
			st.*,
			c.c_id,
			c.c_name,
			sc.s_score 
		FROM
			student st
			LEFT JOIN score sc ON sc.s_id = st.s_id
			INNER JOIN course c ON c.c_id = sc.c_id 
			AND c.c_id = "01" 
		ORDER BY
			sc.s_score DESC 
			LIMIT 1,
			2 
	) as t1

UNION ALL

SELECT
	t2.* 
FROM
	(
		SELECT
			st.*,
			c.c_id,
			c.c_name,
			sc.s_score 
		FROM
			student st
			LEFT JOIN score sc ON sc.s_id = st.s_id
			INNER JOIN course c ON c.c_id = sc.c_id 
			AND c.c_id = "02" 
		ORDER BY
			sc.s_score DESC 
			LIMIT 1,
			2 
	) as t2

UNION ALL

SELECT
	t3.* 
FROM
	(
		SELECT
			st.*,
			c.c_id,
			c.c_name,
			sc.s_score 
		FROM
			student st
			LEFT JOIN score sc ON sc.s_id = st.s_id
			INNER JOIN course c ON c.c_id = sc.c_id 
			AND c.c_id = "03" 
		ORDER BY
			sc.s_score DESC 
			LIMIT 1,
			2 
	) as t3

# 2.一次性查询,需要注意的是 row_number() 在 MySQL 8.0 中才支持
SELECT
	c_id,
	student.*,
	s_score 
FROM
	student
	INNER JOIN (
			SELECT 
				s_id, 
				s_score, 
				c_id, 
				row_number() over (PARTITION BY c_id ORDER BY s_score DESC) AS rank 
			FROM 
				score
	) AS tmp_t ON tmp_t.s_id = student.s_id 
WHERE
	tmp_t.rank IN (2, 3)

在这里插入图片描述

3.23.✨SQL 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
	score.c_id,
	course.c_name,
	sum(CASE WHEN s_score BETWEEN 0 AND 60 THEN 1 ELSE 0 END) AS '[0-60]人数',
	sum(CASE WHEN s_score BETWEEN 61 AND 70 THEN 1 ELSE 0 END) AS '[61-70]人数',
	sum(CASE WHEN s_score BETWEEN 71 AND 85 THEN 1 ELSE 0 END) AS '[71-85]人数',
	sum(CASE WHEN s_score BETWEEN 86 AND 100 THEN 1 ELSE 0 END) AS '[86-100]人数',
	round(sum(CASE WHEN s_score BETWEEN 0 AND 60 THEN 1 ELSE 0 END) / count(*), 2) AS '[0-60]人数所占百分比',
	round(sum(CASE WHEN s_score BETWEEN 61 AND 70 THEN 1 ELSE 0 END) / count(*), 2) AS '[61-70]人数所占百分比',
	round(sum(CASE WHEN s_score BETWEEN 71 AND 85 THEN 1 ELSE 0 END) / count(*), 2) AS '[71-85]人数所占百分比',
	round(sum(CASE WHEN s_score BETWEEN 86 AND 100 THEN 1 ELSE 0 END) / count(*), 2) AS '[86-100]人数所占百分比' 
FROM
	score LEFT JOIN course 
	ON score.c_id = course.c_id 
GROUP BY
	score.c_id,
	course.c_name

在这里插入图片描述

3.24.SQL 24——查询学生平均成绩及其名次

# 在 MySQL 8 中可以使用 rank 函数实现排名
SELECT
    stu.s_id,
    stu.s_name,
    round(avg(sc.s_score), 2) AS average_score,
    (
		SELECT COUNT(DISTINCT avg_score) 
		FROM (SELECT AVG(s_score) AS avg_score FROM score GROUP BY s_id) AS sub 
		WHERE avg_score >= AVG(sc.s_score)
	) AS rank
FROM
    student as stu
    INNER JOIN score as sc ON stu.s_id = sc.s_id
GROUP BY
    stu.s_id,
    stu.s_name
ORDER BY
    average_score DESC;

在这里插入图片描述

3.25.SQL 25——查询各科成绩前三名的记录

# 1.分别对每科进行查询,然后合并查询结果,但是如果课程太多,该方法就不太合适
(SELECT c_id, s_score FROM score WHERE c_id = '01' ORDER BY s_score DESC LIMIT 3) UNION ALL
(SELECT c_id, s_score FROM score WHERE c_id = '02' ORDER BY s_score DESC LIMIT 3) UNION ALL
(SELECT c_id, s_score FROM score WHERE c_id = '03' ORDER BY s_score DESC LIMIT 3)

# 2.一次性查询出结果
SELECT DISTINCT
	tmp_t.c_id,
	tmp_t.s_score 
FROM
	(
		SELECT DISTINCT
			student.*,
			sc1.c_id,
			sc1.s_score,
			count(DISTINCT sc2.s_score) + 1 AS rank 
		FROM
			score AS sc1
			LEFT JOIN score AS sc2 ON sc1.c_id = sc2.c_id 
			AND sc1.s_score < sc2.s_score
			LEFT JOIN student ON sc1.s_id = student.s_id 
		GROUP BY
			sc1.c_id,
			sc1.s_id 
		ORDER BY
			sc1.c_id,
			sc1.s_score DESC 
	) AS tmp_t 
WHERE
	tmp_t.rank BETWEEN 1 AND 3

在这里插入图片描述

3.26.SQL 26——查询每门课程被选修的学生数

SELECT
	c_id,
	count( s_id ) AS '选修该门课程的学生数' 
FROM
	score 
GROUP BY
	c_id

在这里插入图片描述

3.27.SQL 27——查询出只有两门课程的全部学生的学号和姓名

SELECT
	student.s_id,
	student.s_name 
FROM
	student,
	score 
WHERE
	student.s_id = score.s_id 
GROUP BY
	s_id 
HAVING
	count(c_id) = 2

在这里插入图片描述

3.28.SQL 28——查询男生、女生人数

SELECT
	sum(CASE WHEN s_sex = '男' THEN 1 ELSE NULL END) AS '男生人数',
	sum(CASE WHEN s_sex = '女' THEN 1 ELSE NULL END) AS '女生人数' 
FROM
	student

在这里插入图片描述

3.29.SQL 29——查询名字中含有”风”字的学生信息

# 1.使用模糊匹配
SELECT
	* 
FROM
	student 
WHERE
	s_name LIKE '%风%'

# 2.使用正则表达式
SELECT
	* 
FROM
	student 
WHERE
	s_name REGEXP '风'

在这里插入图片描述

在 MySQL 中,LIKE 操作符用于文本字段搜索特定的模式如果需要文本字段匹配通配符本身,可以使用反斜杠字符转义通配符。例如,如果要在一个名为 ’mytable’ 的表中查找包含下划线字符字符串,可以使用以下查询:
SELECT * FROM mytable WHERE mycolumn LIKE '%_%' ESCAPE '';
在上面的查询中,ESCAPE 关键字指定转义字符为反斜杠,因此我们通配符前添加了一个反斜杠字符。这将告诉 MySQL 仅匹配下划线字符本身,而不是作为通配符进行匹配

3.30.✨SQL 30——查询同姓名同性别学生名单,并统计同名人数

SELECT
	stu1.s_name,
	tmp_t.cnt AS '同名人数' 
FROM
	student AS stu1
	LEFT JOIN (
		SELECT s_name, s_sex, count(*) AS cnt 
		FROM student 
		GROUP BY s_name, s_sex
	) AS tmp_t 
	ON stu1.s_name = tmp_t.s_name AND stu1.s_sex = tmp_t.s_sex 
WHERE
	tmp_t.cnt > 1

在这里插入图片描述

3.31.SQL 31——查询 1990 年出生的学生名单

# 1.使用模糊匹配
SELECT
	* 
FROM
	student 
WHERE
	s_birth LIKE '1990%'

# 2.使用正则表达式
SELECT
	* 
FROM
	student 
WHERE
	s_birth REGEXP '^1990'

在这里插入图片描述

3.32.✨SQL 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
	score.c_id,
	course.c_name,
	round(avg(s_score), 2) AS avg_score 
FROM
	score, course
where score.c_id = course.c_id
GROUP BY
	c_id
ORDER BY
	avg_score DESC,
	c_id ASC

在这里插入图片描述

3.33.SQL 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
	student.s_id,
	student.s_name,
	round(avg(s_score), 2) AS '平均成绩' 
FROM
	student
	INNER JOIN score ON student.s_id = score.s_id 
GROUP BY
	score.s_id,
	student.s_id,
	student.s_name 
HAVING
	avg(score.s_score) > 85

在这里插入图片描述

3.34.SQL 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
	s_name,
	s_score 
FROM
	student,
	score 
WHERE
	student.s_id = score.s_id 
	AND c_id IN (SELECT c_id FROM course WHERE c_name = '数学') 
	AND s_score < 60

在这里插入图片描述

3.35.SQL 35——查询所有学生的课程及分数情况

SELECT
	student.s_id,
	student.s_name,
	course.c_name,
	score.s_score 
FROM
	student,
	course,
	score 
WHERE
	student.s_id = score.s_id 
	AND score.c_id = course.c_id 
ORDER BY
	s_id

在这里插入图片描述

3.36.SQL 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
	student.s_name,
	course.c_name,
	score.s_score 
FROM
	student,
	score,
	course 
WHERE
	student.s_id = score.s_id 
	AND score.c_id = course.c_id 
	AND s_score > 70

在这里插入图片描述

3.37.SQL 37——查询不及格的课程的学生姓名、课程名称和分数

SELECT
	student.s_name,
	course.c_name,
	score.s_score 
FROM
	student,
	score,
	course 
WHERE
	student.s_id = score.s_id 
	AND score.c_id = course.c_id 
	AND s_score < 60

在这里插入图片描述

3.38.SQL 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 
	student.s_id, 
	s_name 
FROM 
	student, 
	score 
WHERE 
	student.s_id = score.s_id 
	AND c_id = '01' 
	AND s_score >= 80

在这里插入图片描述

3.39.SQL 39——求每门课程的学生人数

SELECT
	c_name,
	count(s_id) AS '学生人数' 
FROM
	score,
	course 
WHERE
	score.c_id = course.c_id 
GROUP BY
	score.c_id

在这里插入图片描述

3.40.✨SQL 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

# 这里默认的是一门老师只教授一门课程
SELECT
	student.*,
	score.s_score 
FROM
	student,
	score 
WHERE
	student.s_id = score.s_id 
	AND c_id IN (SELECT c_id FROM course WHERE t_id IN (SELECT t_id FROM teacher WHERE t_name = '张三')) 
ORDER BY
	s_score DESC 
	LIMIT 1

在这里插入图片描述

3.41.SQL 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

SELECT
	sc1.s_id,
	sc1.c_id,
	sc2.c_id,
	sc1.s_score,
	sc2.s_score 
FROM
	score AS sc1,
	score AS sc2 
WHERE
	sc1.s_id = sc2.s_id 
	AND sc1.s_score = sc2.s_score 
	AND sc1.c_id != sc2.c_id

在这里插入图片描述

3.42.✨SQL 42——查询每门课程成绩最好的前两名

SELECT
	sc1.c_id,
	sc1.s_id,
	count(sc2.s_score) + 1 AS rank 
FROM
	score AS sc1
	LEFT JOIN score AS sc2 ON sc1.c_id = sc2.c_id 
	AND sc1.s_score < sc2.s_score 
GROUP BY
	sc1.c_id,
	sc1.s_score,
	sc1.s_id 
HAVING
	count(sc2.s_score) < 2 
ORDER BY
	sc1.c_id,
	rank

在这里插入图片描述

3.43.SQL 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 '选修人数' 
FROM 
	score 
GROUP BY 
	c_id 
HAVING 
	count(*) > 5 
ORDER BY
	'选修人数' DESC,
	c_id ASC

在这里插入图片描述

3.44.SQL 44——查询至少选修两门课程的学生学号

SELECT s_id FROM score GROUP BY s_id HAVING count(c_id) >= 2

在这里插入图片描述

3.45.SQL 45——查询选修了全部课程的学生信息

SELECT
	* 
FROM
	student 
WHERE
	# SELECT count(*) FROM course) 查询的是总课程的数量
	s_id IN (SELECT s_id FROM score GROUP BY s_id HAVING count(c_id) = (SELECT count(*) FROM course))

在这里插入图片描述

3.46.✨SQL 46——查询各学生的年龄

# 1.按照年份计算
SELECT
	s_id,
	s_name,
	(YEAR(now()) - YEAR(s_birth)) AS age 
FROM
	student

/*
2.使用 timestampdiff()
(1) TIMESTAMPDIFF(): 第一个参数设置时间单位,可以精确到年(YEAR)、天(DAY)、小时(HOUR),分钟(MINUTE)和秒(SECOND)。对于比较
	的两个时间,时间小的放在前面,时间大的放在后面。
(3) datediff(): 返回值是相差的天数,无法定位到小时、分钟和秒。
*/
SELECT
	s_id,
	s_name,
	timestampdiff(YEAR, s_birth, now()) AS age 
FROM
	student

在这里插入图片描述

3.47.SQL 47——查询本周过生日的学生

# week(时间): 默认从 0 开始,表示星期天为一个星期的第一天,国外算法
# week(时间, 1): 从 1 开始,表示星期一为一个星期的第一天,国内算法
SELECT
	s_id,
	s_name 
FROM
	student 
WHERE
	WEEK (s_birth) = WEEK (now(), 1)

在这里插入图片描述

3.48.SQL 48——查询下周过生日的学生

SELECT
	s_id,
	s_name 
FROM
	student 
WHERE
	WEEK (s_birth) = WEEK (now(), 1) + 1

在这里插入图片描述

3.49.SQL 49——查询本月过生日的学生

SELECT
	s_id,
	s_name 
FROM
	student 
WHERE
	MONTH (s_birth) = MONTH (now())

在这里插入图片描述

3.50.SQL 50——查询下月过生日的学生

SELECT 
    s_id, 
    s_name 
FROM 
    student 
WHERE 
    (MONTH(s_birth) = (((MONTH(NOW()) + 12) % 12) + 1))

在这里插入图片描述

原文地址:https://blog.csdn.net/weixin_43004044/article/details/126557968

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