注：#比较难的题目–12，19，22，24，25，42，46（基于oracle编写），其余题目均是基于mysql。
全部文档我放在了评论区，百度网盘提取
如下图所示。一共四个表student、score、course、teacher。其中student表中有字段s_id,s_name,s_birth,s_sex；score表中有字段s_id,c_id,s_score；course表中有字段c_id,c_name,t_id；teacher表中有字段t_id,t_name；各表两两相互关联。
需要的同学可直接复制下面这段
DROP DATABASE IF EXISTS my_test50; CREATE DATABASE my_test50; #建学生表 DROP TABLE IF EXISTS student; CREATE TABLE student( s_id VARCHAR(20) PRIMARY KEY, s_name VARCHAR(20) NOT NULL DEFAULT '', s_birth VARCHAR(20) NOT NULL DEFAULT '', s_sex VARCHAR(10) NOT NULL DEFAULT '' ); #建课程表 DROP TABLE IF EXISTS course; CREATE TABLE course( c_id VARCHAR(20) PRIMARY KEY, c_name VARCHAR(20) NOT NULL DEFAULT '', t_id VARCHAR(20) NOT NULL ); #建教师表 DROP TABLE IF EXISTS teacher; CREATE TABLE teacher( t_id VARCHAR(20) PRIMARY KEY, t_name VARCHAR(20) NOT NULL DEFAULT '' ); #建成绩表 DROP TABLE IF EXISTS score; CREATE TABLE score( s_id VARCHAR(20), c_id VARCHAR(20), s_score INT(3), PRIMARY KEY(s_id,c_id) ); #插入学生表数据 INSERT INTO student VALUES('01','赵雷','1990-01-01','男'), ('02','钱电','1990-12-21','男'), ('03','孙风','1990-05-20','男'), ('04','李云','1990-08-06','男'), ('05','周梅','1991-12-01','女'), ('06','吴兰','1992-03-01','女'), ('07','郑竹','1989-07-01','女'), ('08','王菊','1990-01-20','女'); #插入课程表数据 INSERT INTO course VALUES('01','语文','02'), ('02','数学','01'), ('03','英语','03'); #插入教师表数据 INSERT INTO teacher VALUES('01','张三'), ('02','李四'), ('03','王五'); #插入成绩表数据 INSERT INTO score VALUES('01','01','80'), ('01','02','90'), ('01','03','99'), ('02','01','70'), ('02','02','60'), ('02','03','80'), ('03','01','80'), ('03','02','80'), ('03','03','80'), ('04','01','50'), ('04','02','30'), ('04','03','20'), ('05','01','76'), ('05','02','87'), ('06','01','31'), ('06','03','34'), ('07','02','89'), ('07','03','98');1.查询课程编号为01的课程比02的课程成绩高的所有学生的学号（重点*）
#step1.在score表中查询课程编号为01的学生编号和分数 SELECT s_id,s_score FROM score WHERE c_id='01'; #step2.在score表中查询课程编号为02的学生编号和分数 SELECT s_id,s_score FROM score WHERE c_id='02'; #step3.将student表和step1与step2得出的表连接 SELECT st.s_id,st.s_name,a.s_score,b.s_score FROM student st INNER JOIN (SELECT s_id,s_score FROM score WHERE c_id='01') a ON st.s_id=a.s_id INNER JOIN (SELECT s_id,s_score FROM score WHERE c_id='02') b ON st.s_id=b.s_id WHERE a.s_score>b.s_score;2.查询平均成绩大于60分的学生的学号和平均成绩（重点）
SELECT s_id,AVG(s_score) FROM score GROUP BY s_id HAVING AVG(s_score)>60;3.查询所有学生的学号、姓名、选课数、总成绩（不重要）
SELECT st.s_id,st.s_name,COUNT(sc.c_id),SUM(sc.s_score) FROM student st INNER JOIN score sc ON st.s_id=sc.s_id GROUP BY st.s_id;4.查询姓“猴”的老师的个数（不重要）
SELECT COUNT(*) FROM teacher WHERE t_name LIKE '猴%';5.查询没学过“张三”老师课程的学生的学号、姓名（重点）
SELECT st.s_id,st.s_name FROM student st WHERE st.s_id NOT IN ( SELECT sc.s_id FROM score sc INNER JOIN course co ON sc.c_id=co.`c_id` INNER JOIN teacher te ON co.t_id =te.t_id WHERE te.t_name='张三' );6.查询学过“张三”老师课程的学生的学号、姓名（重点）
SELECT st.s_id,st.s_name FROM student st WHERE st.s_id IN ( SELECT sc.s_id FROM score sc INNER JOIN course co ON sc.c_id=co.`c_id` INNER JOIN teacher te ON co.t_id =te.t_id WHERE te.t_name='张三' );7.查询学过编号为’01’课程并且也学过’02’课程的学生的学号、姓名（重点）
#step1.查询学过编号为'01'课程的学生编号 SELECT sc.s_id FROM score sc WHERE sc.c_id='01'; #step2.查询学过编号为'02'课程的学生编号 SELECT sc.s_id FROM score sc WHERE sc.c_id='02'; #step3.查询出step1与step2的交集 SELECT s_id1.s_id FROM ( SELECT sc.s_id FROM score sc WHERE sc.c_id='01' ) s_id1 INNER JOIN ( SELECT sc.s_id FROM score sc WHERE sc.c_id='02' ) s_id2 ON s_id1.s_id=s_id2.s_id; #step4.查询出step3结果下的学生学号和姓名 SELECT st.s_id,st.s_name FROM student st WHERE st.s_id IN ( SELECT s_id1.s_id FROM ( SELECT sc.s_id FROM score sc WHERE sc.c_id='01' ) s_id1 INNER JOIN ( SELECT sc.s_id FROM score sc WHERE sc.c_id='02' ) s_id2 ON s_id1.s_id=s_id2.s_id );7.1.查询学过编号为’01’课程但没学过’02’课程的学生的学号、姓名（重点）
#step1. SELECT sc.s_id FROM score sc WHERE sc.c_id='01' #step2. SELECT sc.s_id FROM score sc WHERE sc.c_id='02' #step3. SELECT st.s_id,st.s_name FROM student st WHERE st.s_id IN( SELECT sc.s_id FROM score sc WHERE sc.c_id='01' ) AND st.s_id NOT IN ( SELECT sc.s_id FROM score sc WHERE sc.c_id='02' );8.查询课程编号为’02’的总成绩（不重点）
SELECT SUM(sc.s_score) FROM score sc WHERE sc.c_id='02';9.查询所有课程成绩小于60分的学生的学号、姓名（重点）
SELECT st.s_id,st.s_name FROM score sc INNER JOIN student st ON st.s_id=sc .s_id GROUP BY sc.s_id HAVING SUM(sc.s_score)<60*COUNT(sc.c_id);10.查询没有学完所有课的学生的学号、姓名（重点）
#step1.计算出课程总数目 SELECT COUNT(co.c_id) FROM course co; #step2.在表score查询出课程数目小于step1的学生编号 SELECT sc.s_id FROM score sc GROUP BY sc.s_id HAVING COUNT(sc.c_id)<( SELECT COUNT(co.c_id) FROM course co ); #step3.在表student中找出学号符合step2的学号和姓名 SELECT st.s_id,st.s_name FROM student st WHERE st.s_id IN ( SELECT sc.s_id FROM score sc GROUP BY sc.s_id HAVING COUNT(sc.c_id)<( SELECT COUNT(co.c_id) FROM course co ) );11.查询至少有一门课与学号为’01’的学生所学课程相同的学生的学号和姓名（重点）
#step1.在score表中查询出学号为'01'的学生所选的所有课程编号 SELECT sc.c_id FROM score sc WHERE sc.s_id='01'; #step2.将表student与表score连接，选出其中除了课程'01'，任意有一门课程编号在step1中的学号和姓名 SELECT DISTINCT st.s_id,st.s_name FROM student st INNER JOIN score sc ON st.s_id=sc.s_id WHERE sc.c_id = ANY( #此处的=any换为 in 效果一样 SELECT sc.c_id FROM score sc WHERE sc.s_id='01' ) AND sc.s_id != '01';12.查询和’01’号同学所学课程完全相同的其他同学的学号（重点,超级难）
SELECT s_id FROM score WHERE s_id!='01' AND s_id NOT IN ( SELECT s_id FROM score WHERE c_id NOT IN ( SELECT c_id FROM score WHERE s_id='01' ) ) GROUP BY s_id HAVING COUNT(*)=(SELECT COUNT(*) FROM score WHERE s_id='01');13.查询没学过’张三’老师讲授的任何一门课程的学生姓名（重点，同47题）
#1.连接表score和表course和表teacher查询出学过'张三'老师课程的学生编号 SELECT sc.s_id FROM score sc INNER JOIN course co ON sc.c_id=co.`c_id` INNER JOIN teacher te ON co.`t_id`=te.t_id WHERE te.t_name='张三'; #2.在表student查询出不包含步骤1的学号 SELECT st.s_name FROM student st WHERE st.s_id NOT IN ( SELECT sc.s_id FROM score sc INNER JOIN course co ON sc.c_id=co.`c_id` INNER JOIN teacher te ON co.`t_id`=te.t_id WHERE te.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（重点）
#1.在score表中查询两门及其以上不及格课程的同学的学号 SELECT *,AVG(sc.s_score) ag FROM score sc WHERE sc.s_score<60 GROUP BY sc.s_id HAVING COUNT(sc.c_id)>=2; #2.将表student与步骤1连接起来 SELECT st.s_id,st.s_name,a.ag FROM student st INNER JOIN ( SELECT *,AVG(sc.s_score) ag FROM score sc WHERE sc.s_score<60 GROUP BY sc.s_id HAVING COUNT(sc.c_id)>=2 )a ON st.s_id=a.s_id; #方法二： SELECT st.s_id,st.s_name,AVG(sc.s_score) FROM score sc INNER JOIN student st ON st.s_id=sc.s_id WHERE sc.s_score<60 GROUP BY sc.s_id HAVING COUNT(sc.s_score)>=2;16.检索’01’课程分数小于60,按分数降序排列的学生信息（和34题重复，不重点）
SELECT * FROM student st INNER JOIN score sc ON st.s_id=sc.s_id WHERE c_id='01' AND s_score<60 ORDER BY 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（重点，与35一样）（挺难的）
SELECT sc.s_id,st.s_name, MIN(CASE WHEN c_id='02' THEN s_score ELSE NULL END) '数学', #此处的min和max没有别的意思，只是为了将筛选的值取出 MAX(CASE WHEN c_id='01' THEN s_score ELSE NULL END) '语文', MAX(CASE WHEN c_id='03' THEN s_score ELSE NULL END) '英语', AVG(sc.s_score) '平均成绩' FROM score sc INNER JOIN student st ON st.s_id=sc.s_id GROUP BY sc.s_id ORDER BY AVG(sc.s_score) 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（超级重点，超级好的题目）
SELECT sc.c_id,co.c_name,MAX(sc.s_score),MIN(sc.s_score),AVG(sc.s_score), AVG(CASE WHEN sc.s_score>=60 THEN 1 ELSE 0 END) '及格率', AVG(CASE WHEN sc.s_score>=70 AND sc.s_score<80 THEN 1 ELSE 0 END)'中等率', AVG(CASE WHEN sc.s_score>=80 AND sc.s_score<90 THEN 1 ELSE 0 END)'优良率', AVG(CASE WHEN sc.s_score>=90 THEN 1 ELSE 0 END)'优秀率' FROM score sc INNER JOIN course co ON sc.c_id=co.c_id GROUP BY sc.c_id;19.按各科成绩进行排序，并显示排名,score重复时保留名次空缺（重点）（不会做，难度爆炸）
#orcale SELECT s_id,c_id,s_score,row_number() over (ORDER BY s_score DESC) FROM score; #mysql SELECT a.c_id, a.s_id, a.s_score, COUNT(b.s_score)+1 AS rank FROM score AS a LEFT JOIN score AS b ON a.s_score<b.s_score AND a.c_id = b.c_id GROUP BY a.c_id, a.s_id,a.s_score ORDER BY a.c_id, rank ASC;20.查询学生的总成绩并进行排名（不重点）
SELECT sc.s_id,SUM(sc.s_score) FROM score sc GROUP BY sc.s_id ORDER BY SUM(sc.s_score)DESC;21.查询不同老师所教不同课程平均分从高到低显示（不重点）
SELECT te.t_id,te.t_name,co.c_name,AVG(sc.s_score) FROM score sc INNER JOIN course co ON sc.c_id=co.`c_id` INNER JOIN teacher te ON co.`t_id`=te.t_id GROUP BY sc.c_id ORDER BY AVG(sc.s_score)DESC;22.查询所有课程的成绩第二名到第三名的学生信息及该课程成绩（重要，与25类似）
SELECT * FROM (SELECT st.*,c_id,s_score,row_number() over(PARTITION BY c_id ORDER BY s_score DESC)m FROM score sc INNER JOIN student st ON sc.s_id=st.s_id) a WHERE m IN (2,3);23.使用分段[100-85],[85-70],[70-60],[<60]来统计各科成绩，分别统计各分数段人数、课程ID和课程名称（重点，类似18题）
SELECT co.c_id,co.c_name, COUNT(CASE WHEN sc.s_score<60 THEN 1 ELSE NULL END)'低于60的人数', COUNT(CASE WHEN sc.s_score>=60 AND sc.s_score<70 THEN 1 ELSE NULL END)'60-70的人数', COUNT(CASE WHEN sc.s_score>=70 AND sc.s_score<85 THEN 1 ELSE NULL END)'70-85的人数', COUNT(CASE WHEN sc.s_score>=85 AND sc.s_score<=100 THEN 1 ELSE NULL END)'85-100的人数' FROM score sc INNER JOIN course co ON sc.c_id=co.c_id GROUP BY sc.c_id;24.查询学生平均成绩及其名次（同19题，重点）
SELECT s_id,AVG(s_score),row_number()over(ORDER BY AVG(s_score)DESC) FROM score GROUP BY s_id;25.查询各科成绩前三名的记录（不考虑成绩并列情况）（重点，类似22题）
SELECT c_id, MAX(CASE WHEN m=1 THEN s_score ELSE NULL END)'第一', MAX(CASE WHEN m=2 THEN s_score ELSE NULL END)'第二', MAX(CASE WHEN m=3 THEN s_score ELSE NULL END)'第三' FROM (SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,c_id,s_score,row_number() over(PARTITION BY c_id ORDER BY s_score DESC)m FROM score sc INNER JOIN student st ON sc.s_id=st.s_id) s WHERE m IN (1,2,3) GROUP BY c_id;26.查询每门课程被选修的学生数（不重点）
SELECT sc.c_id,COUNT(sc.s_id) FROM score sc GROUP BY sc.c_id;27.查询出只有两门课程的全部学生的学号和姓名（不重点）
SELECT st.s_id,st.s_name FROM score sc INNER JOIN student st ON sc.s_id=st.s_id GROUP BY sc.s_id HAVING COUNT(sc.c_id)=2;28.查询男生、女生人数（不重点）
SELECT s_sex,COUNT(s_id) FROM student GROUP BY s_sex;29.查询名字中含有’风’字的学生信息（不重点）
SELECT * FROM student st INNER JOIN score sc ON sc.s_id=st.s_id WHERE st.s_name LIKE '%风%';31.查询1990年出生的学生名单（重点）
SELECT * FROM student st WHERE st.s_birth LIKE '%1990%'; #或者 SELECT * FROM student st WHERE YEAR(st.s_birth)=1990;32.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（不重要）
SELECT st.s_id,st.s_name,AVG(sc.s_score) FROM student st INNER JOIN score sc ON st.s_id=sc.s_id GROUP BY sc.s_id HAVING AVG(sc.s_score)>=85;33.查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排序（不重要）
SELECT c_id,AVG(sc.s_score) FROM score sc GROUP BY sc.c_id ORDER BY AVG(sc.s_score),sc.c_id DESC;34.查询课程名称为’数学’，且分数低于60的学生姓名和分数（不重点）
SELECT st.s_name,sc.s_score FROM student st INNER JOIN score sc ON st.s_id=sc.s_id INNER JOIN course co ON sc.c_id=co.c_id WHERE co.c_name='数学' AND sc.s_score<60;35.查询所有学生的课程及分数情况（重点）**
#自己写的 SELECT st.s_id,st.s_name, MIN(CASE WHEN c_id='01' THEN sc.s_score ELSE NULL END)'语文', MIN(CASE WHEN c_id='02' THEN sc.s_score ELSE NULL END)'数学', MIN(CASE WHEN c_id='03' THEN sc.s_score ELSE NULL END)'英语' FROM student st INNER JOIN score sc ON st.s_id=sc.s_id GROUP BY st.s_id; #方法二 SELECT st.s_id,st.s_name, (SELECT s_score FROM score sc WHERE sc.c_id='01'AND sc.s_id=st.s_id)'语文', (SELECT s_score FROM score sc WHERE sc.c_id='02'AND sc.s_id=st.s_id)'数学', (SELECT s_score FROM score sc WHERE sc.c_id='03'AND sc.s_id=st.s_id)'英语' FROM student st INNER JOIN score sc ON st.s_id=sc.s_id GROUP BY sc.s_id;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（重点）
#注：不用group by SELECT st.s_id,st.s_name,co.c_name,sc.s_score FROM student st INNER JOIN score sc ON st.s_id=sc.s_id INNER JOIN course co ON sc.c_id=co.c_id WHERE sc.s_score>70;37.查询不及格的课程并按课程号从大到小排列（不重点）
SELECT st.s_id,st.s_name,sc.c_id,co.c_name,sc.s_score FROM score sc INNER JOIN course co ON sc.c_id=co.c_id INNER JOIN student st ON st.s_id=sc.s_id WHERE sc.s_score<60 ORDER BY sc.c_id DESC;38.查询课程编号为03且课程成绩在80分以上的学生的学号和姓名（不重要）
SELECT st.s_id,st.s_name,sc.s_score FROM student st INNER JOIN score sc ON st.s_id=sc.s_id WHERE sc.c_id='03' AND sc.s_score>80;39.求每门课程的学生人数（不重要）
SELECT sc.c_id,COUNT(sc.s_id) FROM score sc GROUP BY sc.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姓名及其成绩（重要top）
SELECT st.s_id,st.s_name,sc.s_score FROM student st INNER JOIN score sc ON st.s_id=sc.s_id INNER JOIN course co ON sc.c_id=co.`c_id` INNER JOIN teacher te ON co.`t_id`=te.t_id WHERE te.t_name='张三' ORDER BY sc.s_score DESC LIMIT 1;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（重点）
SELECT a.s_id,a.c_id,b.c_id,c.c_id,a.s_score,b.s_score,c.s_score FROM (SELECT s_id,c_id,s_score FROM score WHERE c_id='01') a INNER JOIN (SELECT s_id,c_id,s_score FROM score WHERE c_id='02') b ON a.s_id=b.s_id INNER JOIN (SELECT s_id,c_id,s_score FROM score WHERE c_id='03') c ON b.s_id=c.s_id WHERE a.s_score=b.s_score AND b.s_score=c.s_score;42.查询每门课程成绩最好的前两名（同22和25题）
SELECT st.s_id,c_id, MAX(CASE WHEN m=1 THEN s_score ELSE NULL END)'第一', MAX(CASE WHEN m=2 THEN s_score ELSE NULL END)'第二' FROM (SELECT st.s_id,st.s_name,st.s_birth,st.s_sex,c_id,s_score,row_number() over(PARTITION BY c_id ORDER BY s_score DESC)m FROM score sc INNER JOIN student st ON sc.s_id=st.s_id) s WHERE m IN (1,2) GROUP BY 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（不重要）
SELECT sc.c_id,COUNT(sc.s_id) FROM score sc GROUP BY sc.c_id HAVING COUNT(sc.s_id)>5 ORDER BY COUNT(sc.s_id) DESC,sc.c_id ASC;44.检索至少选修两门课程的学生学号（不重要）
SELECT sc.s_id,COUNT(sc.c_id) FROM score sc GROUP BY sc.s_id HAVING COUNT(sc.c_id)>=2;45.查询选修了全部课程的学生信息（重点）
#step1.查询一共有多少门课程 SELECT COUNT(co.c_id) FROM course co; #step2.查询选修课程数=step1的学生信息 SELECT st.s_id,st.s_name,COUNT(sc.c_id) FROM student st INNER JOIN score sc ON st.s_id=sc.s_id GROUP BY st.s_id HAVING COUNT(sc.c_id)=( SELECT COUNT(co.c_id) FROM course co );46.查询各学生的年龄（精确到月份）（重点记，注意区分datediff和timestampdiff的区别）
SELECT s_id,s_birth, TIMESTAMPDIFF(MONTH,s_birth,'2020-8-10') 年龄（月） FROM student; SELECT DATEDIFF('2020-4-12','2020-3-12'); #返回的是天数47.查询没学过’张三’老师讲授的任何一门课程的学生姓名（重点、再做一次）
#step1.查询学过'张三'老师课程的学生编号 SELECT sc.s_id FROM course co INNER JOIN teacher te ON co.`t_id`=te.t_id INNER JOIN score sc ON sc.c_id=co.c_id WHERE te.t_name='张三'; #step2.查询出没学过'张三'老师讲授的任何一门课程的学生姓名 SELECT DISTINCT st.s_id FROM student st WHERE st.s_id NOT IN( SELECT sc.s_id FROM course co INNER JOIN teacher te ON co.`t_id`=te.t_id INNER JOIN score sc ON sc.c_id=co.c_id WHERE te.t_name='张三' );48.查询两门以上不及格课程的同学的学号及其平均成绩
SELECT sc.s_id,AVG(sc.s_score) FROM score sc WHERE sc.s_score<60 GROUP BY sc.s_id HAVING COUNT(sc.c_id)>2;49.查询本周过生日的同学
SELECT * FROM student st WHERE WEEK(st.s_birth,1)=WEEK(DATE(NOW()),1); #1表示指定星期一为一周的第一天50.查询下周过生日的学生
SELECT * FROM student WHERE WEEK(s_birth,1)=WEEK(DATE(NOW()),1)+1;51.查询本月过生日的学生
SELECT * FROM student WHERE MONTH(s_birth)=MONTH(DATE(NOW()));52.查询下月过生日的学生
SELECT * FROM student WHERE MONTH(s_birth)=MONTH(DATE(NOW()))+1;