① 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 本文所使用的 MySQL 版本为 5.5,虽然版本有一点旧,但是对 SQL 知识点的复习没有太大的影响(除了一些旧版没有的函数)。
③ 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,可在评论区留言,笔者在看到后会及时更新!
(1)建库:创建一个名为 sqlpractice 的数据库。
(2)建表:建立 student、course、teacher 和 score 这 4 张表。它们的字段以及之间的关系如下图所示。
(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 学生表
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)
(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 Course VALUES('01', '语文', '02');
INSERT INTO Course VALUES('02', '数学', '01');
INSERT INTO Course VALUES('03', '英语', '03');INSERT INTO Teacher VALUES('01', '张三');
INSERT INTO Teacher VALUES('02', '李四');
INSERT INTO Teacher VALUES('03', '王五');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
Course 表
Teacher 表
Score 表
# 本题需要比较"01"课程比"02"课程的成绩,故在 where 中将 score 表中的字段 s_score 使用 2 次(即分别对应"01"课程的成绩和"02"课程的成绩)
# 所以可以使用为 s_score 表取别名的方式来多次使用 score 表中的字段
SELECTstudent.*,score1.s_score
FROMstudent,score AS score1,score AS score2
WHEREstudent.s_id = score1.s_id AND score1.s_id = score2.s_id # student, score1, score2 表连接的条件是它们的 s_id 均相等AND score1.c_id = '01' AND score2.c_id = '02' AND score1.s_score > score2.s_score;
SELECTstudent.*,score1.s_score
FROMstudent,score AS score1,score AS score2
WHEREstudent.s_id = score1.s_id AND score1.s_id = score2.s_id # student, score1, score2 表连接的条件是它们的 s_id 均相等AND score1.c_id = '01' AND score2.c_id = '02' AND score1.s_score < score2.s_score;
# 1.创建临时表 ss
EXPLAIN SELECTstudent.s_id,student.s_name,ss.avg_score
FROMstudent,(SELECT s_id, AVG( s_score ) AS avg_score FROM score GROUP BY s_id) AS ss
WHEREstudent.s_id = ss.s_id AND ss.avg_score >= 60;# 2.先进行内连接,然后再分组
SELECTstudent.s_id,s_name,AVG(score.s_score) avg_score
FROMstudentINNER JOIN score ON student.s_id = score.s_id
GROUP BYstudent.s_id,s_name
HAVINGAVG(score.s_score) >= 60
# 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。
SELECTstudent.s_id,s_name,AVG(score.s_score) avg_score
FROMstudentLEFT OUTER JOIN score ON student.s_id = score.s_id
GROUP BYstudent.s_id,s_name
HAVINGAVG(IFNULL(score.s_score,0)) < 60
SELECTstudent.s_id,student.s_name,COUNT(DISTINCT c_id) AS totalCourses,SUM(s_score) AS totalScores
FROMstudent# 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 JOINLEFT OUTER JOIN score ON student.s_id = score.s_id
GROUP BYstudent.s_id,student.s_name;
SELECTCOUNT(*)
FROMteacher
WHEREt_name LIKE '李%'
# 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
SELECTstudent.*
FROMstudent
WHEREs_id IN (SELECTs_id FROMscore,course,teacher WHEREteacher.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
SELECTstudent.*
FROMstudent
WHEREstudent.s_id = ANY (# 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_idSELECT DISTINCTs_id FROMscore WHEREscore.c_id = (# 在 course 表中根据张三老师的 t_id 查询他所教的课程 c_idSELECT c_id FROM course WHERE course.t_id = (# 在 teacher 表中查询张三老师的 t_idSELECT t_id FROM teacher WHERE t_name = '张三')))
SELECTstudent.*
FROMstudent
WHEREstudent.s_id != ALL (SELECT DISTINCTs_id FROMscore WHEREscore.c_id = (SELECT c_id FROM course WHERE course.t_id = (SELECT t_id FROM teacher WHERE t_name = '张三')))
SELECTstudent.*
FROMstudent
WHEREstudent.s_id = ANY (SELECTs1.s_id FROMscore AS s1,score AS s2 WHEREs1.s_id = s2.s_id AND s1.c_id = '01' AND s2.c_id = '02' )
SELECTstudent.*
FROMstudent
WHEREstudent.s_id NOT IN (SELECTs1.s_id FROMscore AS s1,score AS s2 WHEREs1.s_id = s2.s_id AND s1.c_id = '01' AND s2.c_id = '02' )
# 下面的课程数量 3 也可以用 (SELECT count(*) FROM course) 来代替
SELECT*
FROMstudent
WHEREs_id IN (SELECT s_id FROM score GROUP BY s_id HAVING count(c_id) < 3)
# 不包括学号为 '01' 学生自己
SELECT*
FROMstudent
WHEREs_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')
SELECT*
FROMstudent
WHEREs_id IN (SELECTs_id FROMscore WHEREc_id IN (SELECT DISTINCT c_id FROM score WHERE s_id = '01') AND s_id != '01' GROUP BYs_id HAVING# 下面的课程数量 3 也可以用 (SELECT count(*) FROM course) 来代替count( c_id ) >= 3 )
SELECTs_name
FROMstudent
WHEREs_id NOT IN (SELECTs_id FROMscore WHEREc_id IN ( # 查询由姓名为张三的老师所讲授的课程 idSELECTc_id FROMcourse WHERE# 姓名为张三的老师可能有多个,所以此处不能直接使用 =t_id IN (SELECT t_id FROM teacher WHERE t_name = '张三')))
SELECTstu.s_id,stu.s_name,tmp_t.avg_score
FROMstudent AS stuRIGHT JOIN (SELECT s_id, avg(s_score) AS avg_score FROM score WHERE s_score < 60 GROUP BY s_id HAVING count(s_score) >= 2) AS tmp_t ON stu.s_id = tmp_t.s_id
SELECTstu.*
FROMstudent AS stuINNER JOIN score ON stu.s_id = score.s_id
WHEREc_id = '01' AND s_score < 60
ORDER BYs_score DESC
SELECTs_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
FROMscore
GROUP BYs_id
ORDER BYavg_score DESC
SELECTsc.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 70 AND 80, 1, 0)) / COUNT(*) as '中等率',SUM(IF (sc.s_score BETWEEN 80 AND 90, 1, 0)) / COUNT(*) as '优良率',SUM(IF (sc.s_score >= 90, 1, 0)) / COUNT(*) as '优秀率'
FROMscore AS scJOIN course AS c ON sc.c_id = c.c_id
GROUP BYsc.c_id
SELECTsc1.c_id,sc1.s_id,sc1.s_score,count(sc2.s_score) + 1 AS rank
FROMscore AS sc1LEFT JOIN score AS sc2 ON sc1.s_score < sc2.s_score AND sc1.c_id = sc2.c_id
GROUP BYsc1.c_id,sc1.s_id,sc1.s_score
ORDER BYsc1.c_id,rank
SELECTstudent.s_id,s_name,sum(s_score) AS '总成绩'
FROMstudentLEFT JOIN score ON student.s_id = score.s_id
GROUP BYstudent.s_id,s_name
ORDER BYsum(s_score) DESC
SELECTteacher.t_id,t_name,avg(s_score * 1.0) AS avg_score
FROMteacher,course,score
WHEREteacher.t_id = course.t_id AND course.c_id = score.c_id
GROUP BYteacher.t_id,t_name,score.c_id
ORDER BYavg(s_score * 1.0) DESC
# 1.分别对每门课程进行查询,然后再合并查询结果,但是如果课程太多,该方法就不太合适
SELECTt1.*
FROM(SELECTst.*,c.c_id,c.c_name,sc.s_score FROMstudent stLEFT JOIN score sc ON sc.s_id = st.s_idINNER JOIN course c ON c.c_id = sc.c_id AND c.c_id = "01" ORDER BYsc.s_score DESC LIMIT 1,2 ) as t1UNION ALLSELECTt2.*
FROM(SELECTst.*,c.c_id,c.c_name,sc.s_score FROMstudent stLEFT JOIN score sc ON sc.s_id = st.s_idINNER JOIN course c ON c.c_id = sc.c_id AND c.c_id = "02" ORDER BYsc.s_score DESC LIMIT 1,2 ) as t2UNION ALLSELECTt3.*
FROM(SELECTst.*,c.c_id,c.c_name,sc.s_score FROMstudent stLEFT JOIN score sc ON sc.s_id = st.s_idINNER JOIN course c ON c.c_id = sc.c_id AND c.c_id = "03" ORDER BYsc.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
SELECTc_id,student.*,s_score
FROMstudentINNER 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
WHEREtmp_t.rank IN (2, 3)
SELECTscore.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',sum(CASE WHEN s_score BETWEEN 0 AND 60 THEN 1 ELSE 0 END) / count( * ) AS '[0-60]人数所占百分比',sum(CASE WHEN s_score BETWEEN 61 AND 70 THEN 1 ELSE 0 END) / count( * ) AS '[61-70]人数所占百分比',sum(CASE WHEN s_score BETWEEN 71 AND 85 THEN 1 ELSE 0 END) / count( * ) AS '[71-85]人数所占百分比',sum(CASE WHEN s_score BETWEEN 86 AND 100 THEN 1 ELSE 0 END) / count( * ) AS '[86-100]人数所占百分比'
FROMscoreLEFT JOIN course ON score.c_id = course.c_id
GROUP BYscore.c_id,course.c_name
SELECTstudent.s_id,avg(s_score) AS '平均成绩'
FROMstudent,score
WHEREstudent.s_id = score.s_id
GROUP BYstudent.s_id
ORDER BYavg(s_score) DESC
# 1.分别对每科进行查询,然后再合并查询结果,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 DISTINCTtmp_t.c_id,tmp_t.s_score
FROM(SELECT DISTINCTstudent.*,sc1.c_id,sc1.s_score,count(DISTINCT sc2.s_score) + 1 AS rank FROMscore AS sc1LEFT JOIN score AS sc2 ON sc1.c_id = sc2.c_id AND sc1.s_score < sc2.s_scoreLEFT JOIN student ON sc1.s_id = student.s_id GROUP BYsc1.c_id,sc1.s_id ORDER BYsc1.c_id,sc1.s_score DESC ) AS tmp_t
WHEREtmp_t.rank BETWEEN 1 AND 3
SELECTc_id,count( s_id ) AS '选修该门课程的学生数'
FROMscore
GROUP BYc_id
SELECTstudent.s_id,student.s_name
FROMstudent,score
WHEREstudent.s_id = score.s_id
GROUP BYs_id
HAVINGcount(c_id) = 2
SELECTsum(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'
FROMstudent
SELECT*
FROMstudent
WHEREs_name LIKE '%风%'
SELECTstu1.s_name,tmp_t.cnt AS '同名人数'
FROMstudent AS stu1LEFT 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
WHEREtmp_t.cnt > 1
SELECT*
FROMstudent
WHEREs_birth LIKE '1990%'
SELECTscore.c_id,avg( s_score ) AS avg_score
FROMscore
GROUP BYc_id
ORDER BYavg( s_score ) DESC,c_id ASC
SELECTstudent.s_id,student.s_name,avg(s_score) AS '平均成绩'
FROMstudentINNER JOIN score ON student.s_id = score.s_id
GROUP BYscore.s_id,student.s_id,student.s_name
HAVINGavg(score.s_score) > 85
SELECTs_name,s_score
FROMstudent,score
WHEREstudent.s_id = score.s_id AND c_id IN (SELECT c_id FROM course WHERE c_name = '数学') AND s_score < 60
SELECTstudent.s_id,student.s_name,course.c_name,score.s_score
FROMstudent,course,score
WHEREstudent.s_id = score.s_id AND score.c_id = course.c_id
ORDER BYs_id
SELECTstudent.s_name,course.c_name,score.s_score
FROMstudent,score,course
WHEREstudent.s_id = score.s_id AND score.c_id = course.c_id AND s_score > 70
SELECTstudent.s_name,course.c_name,score.s_score
FROMstudent,score,course
WHEREstudent.s_id = score.s_id AND score.c_id = course.c_id AND s_score < 60
SELECT student.s_id, s_name
FROM student, score
WHERE student.s_id = score.s_id AND c_id = '01' AND s_score >= 80
SELECTc_name,count( s_id ) AS '学生人数'
FROMscore,course
WHEREscore.c_id = course.c_id
GROUP BYscore.c_id,c_name
# 这里默认的是一门老师只教授一门课程
SELECTstudent.*,score.s_score
FROMstudent,score
WHEREstudent.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 BYs_score DESC LIMIT 1
SELECTsc1.s_id,sc1.c_id,sc2.c_id,sc1.s_score,sc2.s_score
FROMscore AS sc1,score AS sc2
WHEREsc1.s_id = sc2.s_id AND sc1.s_score = sc2.s_score AND sc1.c_id != sc2.c_id
SELECTsc1.c_id,sc1.s_id,count( sc2.s_score ) + 1 AS rank
FROMscore AS sc1LEFT JOIN score AS sc2 ON sc1.c_id = sc2.c_id AND sc1.s_score < sc2.s_score
GROUP BYsc1.c_id,sc1.s_score,sc1.s_id
HAVINGcount( sc2.s_score ) < 2
ORDER BYsc1.c_id,rank
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
SELECT s_id FROM score GROUP BY s_id HAVING count(c_id) >= 2
SELECT*
FROMstudent
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))
# 1.按照年份来计算
SELECTs_id,s_name,(YEAR(now()) - YEAR(s_birth)) AS age
FROMstudent/*
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。
*/
SELECTs_id,s_name,timestampdiff(YEAR, s_birth, now()) AS age
FROMstudent
# 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
SELECTs_id,s_name
FROMstudent
WHEREWEEK (s_birth) = WEEK (now(), 1)
SELECTs_id,s_name
FROMstudent
WHEREWEEK (s_birth) = WEEK (now(), 1) + 1
SELECTs_id,s_name
FROMstudent
WHEREMONTH (s_birth) = MONTH (now())
SELECTs_id,s_name
FROMstudent
WHEREMONTH (s_birth) = MONTH (now()) + 1