MySQL练习题和代码附录
2024-04-15 22:21:11
文章目录
- 建立表格和插入数据
- 练习题和答案
- 我的练习记录
- 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
建立表格和插入数据
--建表
--学生表
CREATE TABLE Student(s_id VARCHAR(20), s_name VARCHAR(20) NOT NULL DEFAULT '', /*--名字*/s_birth VARCHAR(20) NOT NULL DEFAULT '', /*--出生日期*/s_sex VARCHAR(20) NOT NULL DEFAULT '', /*--性别*/PRIMARY KEY(s_id)
);--课程表
CREATE TABLE Course(c_id VARCHAR(20), /*--课程代号*/c_name VARCHAR(20) NOT NULL DEFAULT '', /*--课程名字*/t_id VARCHAR(20) NOT NULL, /*--老师学号*/PRIMARY KEY(c_id)
);--教师表
CREATE TABLE Teacher(t_id VARCHAR(20), /*--老师学号*/t_name VARCHAR(20) NOT NULL DEFAULT '', /*--老师姓名*/PRIMARY KEY(t_id)
);--成绩表
CREATE TABLE Score(s_id VARCHAR(20), /*--学生学号*/c_id VARCHAR(20), /*--课程代号*/s_score INT(3), /*--学生成绩*/PRIMARY KEY(s_id,c_id)
);--插入学生表的测试数据
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'); /*--数学老师*/
insert into Teacher values('02' , '李四'); /*--语文老师*/
insert into Teacher values('03' , '王五'); /*--英语老师*/--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);练习题和答案
-- 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 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'left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.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 a.* ,b.s_score as 01_score,c.s_score as 02_score from student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from student b join score a on b.s_id = a.s_idGROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)>=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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from student b left join score a on b.s_id = a.s_idGROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)<60union
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);-- 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 a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from student a left join score b on a.s_id=b.s_idGROUP BY a.s_id,a.s_name;-- 6、查询"李"姓老师的数量
select count(t_id) from teacher where t_name like '李%';-- 7、查询学过"张三"老师授课的同学的信息
select a.* from student a join score b on a.s_id=b.s_id where b.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 * from student c where c.s_id not in(select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in(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 a.* from student a,score b,score c where a.s_id = b.s_id and a.s_id = c.s_id and b.c_id='01' and c.c_id='02';-- 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 a.* from student a where a.s_id in (select s_id from score where c_id='01' ) and a.s_id not in(select s_id from score where c_id='02')
-- 11、查询没有学全所有课程的同学的信息
select s.* from student s where s.s_id in(select s_id from score where s_id not in(select a.s_id from score a join score b on a.s_id = b.s_id and b.c_id='02'join score c on a.s_id = c.s_id and c.c_id='03'where a.c_id='01'))
-- 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 * from student where s_id in(select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.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 select a.* from student a where a.s_id in(select distinct s_id from score where s_id!='01' and c_id in(select c_id from score where s_id='01')group by s_id having count(1)=(select count(1) from score where s_id='01'));
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select a.s_name from student a where a.s_id not in (select s_id from score where c_id = (select c_id from course where t_id =(select t_id from teacher where t_name = '张三')) group by s_id);-- 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 a.s_id,a.s_name,ROUND(AVG(b.s_score)) from student a left join score b on a.s_id = b.s_idwhere a.s_id in(select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2)GROUP BY a.s_id,a.s_name-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息
select a.*,b.c_id,b.s_score from student a,score b where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.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
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;-- 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
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,ROUND(100*(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 else 0 end)),2) as 中等率,ROUND(100*(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 else 0 end)),2) as 优良率,ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name
-- 19、按各科成绩进行排序，并显示排名(实现不完全)
-- mysql没有rank函数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 scorefrom (select s_id,c_id,s_score from score WHERE c_id='01' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)sunionselect a.s_id,a.c_id,@i:=@i +1 as i,@k:=(case when @score=a.s_score then @k else @i end) as rank,@score:=a.s_score as scorefrom (select s_id,c_id,s_score from score WHERE c_id='02' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)sunionselect a.s_id,a.c_id,@i:=@i +1 as i,@k:=(case when @score=a.s_score then @k else @i end) as rank,@score:=a.s_score as scorefrom (select s_id,c_id,s_score from score WHERE c_id='03' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
-- 20、查询学生的总成绩并进行排名
select a.s_id,@i:=@i+1 as i,@k:=(case when @score=a.sum_score then @k else @i end) as rank,@score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,(select @k:=0,@i:=0,@score:=0)s
-- 21、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,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course aleft join score b on a.c_id=b.c_id left join teacher c on a.t_id=c.t_idGROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;
-- 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select d.*,c.排名,c.s_score,c.c_id from (select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01' )cleft join student d on c.s_id=d.s_idwhere 排名 BETWEEN 2 AND 3UNIONselect d.*,c.排名,c.s_score,c.c_id from (select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02' )cleft join student d on c.s_id=d.s_idwhere 排名 BETWEEN 2 AND 3UNIONselect d.*,c.排名,c.s_score,c.c_id from (select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03' )cleft join student d on c.s_id=d.s_idwhere 排名 BETWEEN 2 AND 3;-- 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 distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score aleft join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比from score GROUP BY c_id)b on a.c_id=b.c_idleft join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比from score GROUP BY c_id)c on a.c_id=c.c_idleft join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比from score GROUP BY c_id)d on a.c_id=d.c_idleft join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比from score GROUP BY c_id)e on a.c_id=e.c_idleft join course f on a.c_id = f.c_id
-- 24、查询学生平均成绩及其名次 select a.s_id,@i:=@i+1 as '不保留空缺排名',@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',@avg_score:=avg_s as '平均分'from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;
-- 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_scoregroup by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3ORDER BY a.c_id,a.s_score DESC-- 26、查询每门课程被选修的学生数 select c_id,count(s_id) from score a GROUP BY c_id-- 27、查询出只有两门课程的全部学生的学号和姓名 select s_id,s_name from student where s_id in(select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);-- 28、查询男生、女生人数 select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex-- 29、查询名字中含有"风"字的学生信息select * from student where s_name like '%风%';-- 30、查询同名同性学生名单，并统计同名人数 select a.s_name,a.s_sex,count(*) from student a JOIN student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sexGROUP BY a.s_name,a.s_sex-- 31、查询1990年出生的学生名单select s_name from student where s_birth like '1990%'-- 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 c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC-- 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 a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score aleft join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85-- 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 a.s_name,b.s_score from score b LEFT JOIN student a on a.s_id=b.s_id where b.c_id=(select c_id from course where c_name ='数学') and b.s_score<60-- 35、查询所有学生的课程及分数情况； select a.s_id,a.s_name,SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',SUM(b.s_score) as '总分'from student a left join score b on a.s_id = b.s_id left join course c on b.c_id = c.c_id GROUP BY a.s_id,a.s_name-- 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 a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_idleft join student a on a.s_id=c.s_id where c.s_score>=70-- 37、查询不及格的课程select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_idwhere a.s_score<60 --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 a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_idwhere a.c_id = '01' and a.s_score>80-- 39、求每门课程的学生人数 select count(*) from score GROUP BY c_id;-- 40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩-- 查询老师id select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'-- 查询最高分（可能有相同分数）select MAX(s_score) from score where c_id='02'-- 查询信息select a.*,b.s_score,b.c_id,c.c_name from student aLEFT JOIN score b on a.s_id = b.s_idLEFT JOIN course c on b.c_id=c.c_idwhere b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')and b.s_score in (select MAX(s_score) from score where c_id='02')
-- 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 select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score
-- 42、查询每门功成绩最好的前两名 -- 牛逼的写法select a.s_id,a.c_id,a.s_score from score awhere (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.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 select c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC
-- 44、检索至少选修两门课程的学生学号 select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2
-- 45、查询选修了全部课程的学生信息 select * from student where s_id in( select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))
--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select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') - (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as agefrom student;-- 47、查询本周过生日的学生select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))-- 48、查询下周过生日的学生select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)-- 49、查询本月过生日的学生select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)-- 50、查询下月过生日的学生select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)我的练习记录
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 select a.*, b.*, c.*from student as ajoin score as bon a.s_id = b.s_id and b.c_id = '01' join score as c on a.s_id = c.s_idand c.c_id = '02'or c.c_id = NULLwhere b.s_score > c.s_score-- 备注：
-- 1 首先应该把所有的东西都连接起来，并选取出来
-- 2 然后再去增加判断条件
2、查询"01"课程比"02"课程成绩低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
select a.* , b.* , c.* from student as a join score as b on a.s_id = b.s_idand b.c_id = '01'left join score as con a.s_id = c.s_idand c.c_id = '02'or c.c_id = nullwhere b.s_score < c.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
-- 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 a.s_id , a.s_name , round(avg(b.s_score),2) as avg_scorefrom student as a join score as b on a.s_id = b.s_idgroup by a.s_id , a.s_namehaving avg(b.s_score)>=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)
-- 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 a.* , round(avg(b.s_score),2) as avg_scorefrom student as a join score as bon a.s_id = b.s_idgroup by a.s_id,a.s_namehaving avg(b.s_score)<60
-- or b.s_score is null
-- 注意，在group by中是不能直接判断是否为非空的
union
select a.*, 0 as avg_scorefrom student as a where a.s_id not in (select s_id from score)-- 最后的几个句子堪称精髓，用一个子查询来判断是否为NULL
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
-- 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(b.c_id), sum(b.s_score)from student as a join score as b on a.s_id = b.s_idgroup by a.s_id , a.s_name
6、查询"李"姓老师的数量
-- 6、查询"李"姓老师的数量
select count(t_name)from teacher
-- 姓李的老师，就是开头是李，使用字符串通配符来表示where t_name like '李%'
7、查询学过"张三"老师授课的同学的信息
-- 7、查询学过"张三"老师授课的同学的信息
select a.* , b.s_score , c.c_name , d.t_namefrom student as ajoin score as b on a.s_id = b.s_idjoin course as c on b.c_id = c.c_idjoin teacher as d on c.t_id = d.t_idwhere d.t_name = '张三'
-- 下面也是可以的
-- where d.t_name like '张三'-- 本题目的关键是用join连接和on判断连接条件
或者使用子查询
select a.*from student as a where a.s_id in (select s_id from score where c_id = (select c_id from course where t_id = (select t_id from teacher where t_name = '张三' )))
-- 注意写法，先写子查询
-- 然后一个个拷贝到外层循环
-- 如果子查询返回多个值，使用in，如果返回一个值，直接使用=进行判断8、查询没学过"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
select a.*from student as a where a.s_id not in (select s_id from student where s_id in(select s_id from score where c_id = (select c_id from course where t_id = (select t_id from teacher where t_name = '张三') )))-- 首先应该查到所有选了张三老师的同学
-- 然后使用not in 去做排除
当然，还是使用连接更加简单易读
select * from student where s_id not in (select a.s_idfrom student as ajoin score as b on a.s_id = b.s_idjoin course as c on b.c_id = c.c_idjoin teacher as d on c.t_id = d.t_idwhere d.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
mysql不支持intersect
MySQL交集INTERSECT运算符 - MySQL教程™
https://www.yiibai.com/mysql/sql-union-mysql.html
-- 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 a.* from student as ajoin score as b on a.s_id = b.s_idjoin score as c on a.s_id = c.s_idwhere b.c_id = '01' and c.c_id = '02'
-- 注意，where中有两个条件，一定要写在一起！
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
-- 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 distinct a.*, b.c_id, c.c_idfrom student as a join score as b on a.s_id = b.s_idjoin score as c on a.s_id = c.s_idwhere b.c_id = '01'and c.c_id <> '02'
11、查询没有学全所有课程的同学的信息
-- 11、查询没有学全所有课程的同学的信息
select s.*from student as swhere s.s_id not in (select a.s_idfrom student as a join score as b on a.s_id = b.s_idand b.c_id = '01'join score as c on a.s_id = c.s_idand c.c_id = '02'join score as d on a.s_id = d.s_idand d.c_id = '03')
-- 思路：
-- 首先建立一个子查询，检索出学全了的学生
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
-- 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
-- 首先查询出学号为01学生学的课程作为一个子查询
select distinct b.*from student as b join score as c on b.s_id = c.s_idand c.c_id in (select a.c_id as course_setfrom score as a where a.s_id = '07')
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