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此50题参考出处，题目一样，代码有出入,因为题目比较有意思，都自己做了一次。
https://zhuanlan.zhihu.com/p/50662216
— 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。 — 方法1 select * from score a inner join score b on (a.s_id = b.s_id and a.c_id=’01’ and b.c_id=’02’ and a.s_score>b.s_score); — 方法2 select * from score a inner join score b where (a.s_id = b.s_id and a.c_id=’01’ and b.c_id=’02’ and a.s_score>b.s_score); — 方法3 select * from (select * from score where c_id=’01’) as a inner join (select * from score where c_id=’02’) as b on a.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 — 方法1 select s_id,avg(s_score) as avgs from score group by s_id HAVING avg(s_score)>60; — 方法2 select a.s_id,a.avgs from (select s_id,avg(s_score) as avgs from score group by s_id) as a where a.avgs >60;
— 3、查询所有学生的学号、姓名、选课数、总成绩 select stu.s_id,stu.s_name,count(stu.s_id),sum(sc.s_score) from student as stu inner join score as sc on sc.s_id =stu.s_id group by (s_id);
— 4、查询姓“张”的老师的个数 select count(t_id) from teacher where t_name like “张%”;
— 5.查询没学过“张三”老师课的学生的学号、姓名(重点)sc.s_id,stu.s_name
— 错误： select * from score as sc inner join student stu on stu.s_id=sc.s_id where sc.c_id not in (select c_id from course where t_id= (select t_id from teacher where t_name=’张三’)) group by sc.s_id;
— 正确1： select s_id, s_name from student as c where c.s_id not in (select sc.s_id from score as sc inner join student stu on stu.s_id=sc.s_id where sc.c_id in (select c_id from course where t_id= (select t_id from teacher where t_name=’张三’)));
— 正确2 推荐： select s_id, s_name from Student where s_id not in (select s_id from Score join Course on Score.c_id = Course.c_id join Teacher on Course.t_id = Teacher.t_id where t_name = ‘张三’);
— 理解过程及注意1： — 在一对多的情况下，一定不能够直接not in某两个表里笛卡尔集的数据，要先找出唯一对应的，即学过张三的课的学生，再找没有学过的 — 1、select t_id from teacher where t_name=’张三’ 查张三的id — 2、select c_id from course where t_id=(查张三的id) 查课程的id — 错误：3、select * from score as sc inner join student stu on stu.s_id=sc.s_id where sc.c_id not in (查课程的id) group by sc.s_id — 正确3：select sc.s_id from score as sc inner join student stu on stu.s_id=sc.s_id where sc.c_id in(查课程的id) 找出学过张三的课的学生 — 正确4：select * from student as c where c.s_id not in (找出学过张三的课的学生)
— 理解过程及注意2：推荐 — 使找出学过的id，再找没学过的id — 1、Score 有学生id和课程id,并作为主键，可以找出唯一学过张三id的所有学生 — select s_id from Score join Course on Score.c_id = Course.c_id — join Teacher on Course.t_id = Teacher.t_id — where t_name = ‘张三’ — 2、使用not in
— 6、查询学过“张三”老师所教的所有课的同学的学号、姓名 — 方法1 select * from score join course on score.c_id = course.c_id join teacher on teacher.t_id =course.t_id join student on student.s_id=score.s_id where teacher.t_name=’张三’;
— 方法2 select s_id, s_name from Student where s_id in (select s_id from Score join Course on Score.c_id = Course.c_id join Teacher on Course.t_id = Teacher.t_id where t_name = ‘张三’);
— 7、查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名 select student.s_id,student.s_name from (select s_id from score where score.c_id=’01’) as a join (select s_id from score where score.c_id=’02’) as b on a.s_id=b.s_id join student where student.s_id=a.s_id;
select student.s_id,student.s_name from student where s_id in (select s_id from score where score.c_id=’01’) and s_id in (select s_id from score where score.c_id=’02’);
— 8、查询课程编号为“02”的总成绩 select sum(s_score) from Score where c_id = ’02’;
— 9、查询所有课程成绩小于60分的学生的学号、姓名 — select student.s_id,student.s_name from score join student on student.s_id=score.s_id group by (score.s_id) having max(score.s_score)<60 –这个会缺少没有成绩的学生,当然，题目也没有明确 — 方式1 select student.s_id,student.s_name from student where student.s_id not in (select s_id from score group by (score.s_id) having max(score.s_score)>=60);
— 方式2 select s_id, s_name from Student where s_id NOT IN (SELECT s_id FROM Score where s_score >=60);
— 10、查询没有学全所有课的学生的学号、姓名（题目没有明确要不要完全没学的学生）
select s_id, s_name from Student where s_id NOT IN (SELECT s_id FROM Score group by s_id having count(*)=3); — 含那个没成绩的学生
— 方式1 select s_id, s_name from Student where s_id IN (SELECT s_id FROM Score group by s_id having count(*)<3); — 不含那个没成绩的学生
— 方式2 SELECT student.s_id,student.s_name FROM Score join student on student.s_id=score.s_id group by score.s_id having count(*)<3; — 不含那个没成绩的学生
— 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 — 方式一 select student.s_id,student.s_name from score join student on student.s_id=score.s_id where score.c_id in (select c_id from score where s_id=’01’) and score.s_id !=’01’ group by score.s_id; — 方式2 select distinct a.s_id,a.s_name from Student a join Score b on a.s_id= b.s_id where c_id in (select c_id from Score where s_id = ’01’) and a.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
select sc.s_id,count(sc.s_id) from score as sc where sc.s_id <>’01’ and sc.c_id in (select c_id from score where s_id=’01’) group by sc.s_id having count(sc.s_id)=(select COUNT(*) from score where s_id=’01’) and (select COUNT(*) from score where s_id=’01’)=(select COUNT(*) from score where s_id=sc.s_id)
— 以下是作者提供的答案，个人认为不正确。例如，把01换成07,没有考虑多出来的情况 select s_id from Score where c_id in (select c_id from Score where s_id=’01’) and s_id <> ’01’ group by s_id having count(c_id)=(select count(c_id) from Score where s_id=’01’);
— 13.把“Score”表中“张三”老师教的课的成绩都更改为此课程的平均成绩（难点）
— 方式1 update score as a join (select Score.c_id,avg(s_score) as t from score where score.c_id=(select c_id from course where t_id=(select t_id from teacher where t_name=’张三’)) group by c_id) as b on a.c_id=b.c_id set a.s_score=t;
— 方式2 UPDATE Score AS a JOIN ( SELECT AVG( s_score ) AS t, Score.c_id FROM Score JOIN Course ON Score.c_id = Course.c_id JOIN Teacher ON Teacher.t_id = Course.t_id WHERE t_name = ‘张三’ GROUP BY c_id ) AS b ON a.c_id = b.c_id SET a.s_score = b.t;
— 14、查询和“02”号的同学学习的课程完全相同的其他同学学号和姓名(同12题）
— 15、删除学习“张三”老师课的SC表记录
delete from Score where c_id =(select c_id from Course join Teacher on Course.t_id=Teacher.t_id where t_name =’张三’)
— 16.检索”01″课程分数小于60，按分数降序排列的学生信息 select * from score join student on student.s_id=score.s_id where score.s_score<60 and score.c_id=’01’ order by score.s_score desc
— 17、按平均成绩从高到低显示所有学生的“数据库”（c_id=’04’）、“企业管理”（c_id=’01’）、“英语”（c_id=’06’）三门的课程成绩，按如下形式显示：学生ID，数据库，企业管理，英语，有效课程数，有效平均分
select s_id as ‘学生ID’, (case when c_id=’04’ then s_score else NULL end) as ‘数据库’, (case when c_id=’01’ then s_score else NULL end) as ‘企业管理’, (case when c_id=’06’ then s_score else NULL end) as ‘英语’, count(c_id) as ‘有效课程数’, avg(s_score) as ‘有效平均分’ from Score group by s_id order by avg(s_score) DESC;
— 18、查询各科成绩最高和最低的分： 以如下的形式显示：课程ID，最高分，最低分
select c_id,max(s_score),min(s_score) from score group by c_id
— 19、按各科平均成绩从低到高和及格率的百分数从高到低排列，以如下形式显示：课程号,课程名,平均成绩,及格百分数 select course.c_id,course.c_name,avg(sc.s_score),concat((select count(*) from score as sc1 where sc1.s_score>=60 and sc1.c_id=sc.c_id)/count(*)*100,”%”) from score as sc join course on course.c_id=sc.c_id group by sc.c_id order by avg(sc.s_score)
— 21、查询不同老师所教不同课程平均分从高到低显示
select course.c_id,teacher.t_name,course.c_name,avg(sc.s_score),concat((select count(*) from score as sc1 where sc1.s_score>=60 and sc1.c_id=sc.c_id)/count(*)*100,”%”) from score as sc join course on course.c_id=sc.c_id join teacher on teacher.t_id=course.t_id group by sc.c_id order by avg(sc.s_score);
— 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
select sc.c_id,cu.c_name, sum(case when sc.s_score>=85 and sc.s_score <=100 then 1 else 0 end) ‘[100-85]’, sum(case when sc.s_score>=70 and sc.s_score <=85 then 1 else 0 end) ‘[85-70]’, sum(case when sc.s_score>=60 and sc.s_score <=70 then 1 else 0 end) ‘[70-60]’, sum(case when sc.s_score>=0 and sc.s_score <60 then 1 else 0 end) ‘[<60]’ from score as sc join course as cu on sc.c_id=cu.c_id group by c_id
— 24、查询学生平均成绩及其名次 — 作者提示要点：算出在所有同学中有几个同学的平均分高于某个ID,然后 1，就是名次 — 方式1：推荐 select table1.s_id as ‘学号’,avg as ‘平均成绩’ ,@a:=@a 1 as ‘名次 ‘ from (select *,avg(sc.s_score) as avg from score as sc group by sc.s_id order by avg(sc.s_score) desc) as table1,(select @a:=0) as b — 方式2: SELECT s_id as ‘学号’,平均成绩, (SELECT COUNT(*) FROM(SELECT s_id,AVG(s_score)AS ‘平均成绩’ FROM Score GROUP BY s_id)AS b WHERE b.平均成绩>a.平均成绩) 1 as RANK FROM (select s_id,avg(S_score) as 平均成绩 from Score group by s_id)AS a order by 平均成绩 desc;
— 25、查询各科成绩前三名的记录（不考虑成绩并列情况）
select sc1.c_id,sc1.s_id,sc1.s_score from score as sc1 where (select count(*) from score as sc2 where sc1.s_score < sc2.s_score and sc1.c_id=sc2.c_id)<=2 ORDER BY sc1.c_id ASC,sc1.s_score DESC;
— 26、查询每门课程被选修的学生数
select c_id,count(*) from score group by c_id
— 27、查询出只选修了两门课程的全部学生的学号和姓名
select score.s_id,student.s_name from score join student on student.s_id=score.s_id group by s_id HAVING count(*)=2
— 28、查询男生、女生人数 select s_sex,count(*) 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、查询同名同姓学生名单并统计同名人数 — 方式1: select * from student as s1 where (select count(*) from student s2 where s2.s_name=s1.s_name and s1.s_id<>s2.s_id) — 方式2: select * from student as s1 join student as s2 on s1.s_id<>s2.s_id and s1.s_name=s2.s_name — 方式3:推荐 select s_name,count(*) from student GROUP BY s_name HAVING count(*)>1
— 31、1990年出生的学生名单（注：Student表中s_birth列的类型是datetime） select * from student where s_birth like “1990%”
select * from student where year(s_birth)=1990
— 32、查询平均成绩大于85的所有学生的学号、姓名和平均成绩 select student.s_id,student.s_name,avg(s_score) from student join score on student.s_id=score.s_id group by (score.s_id) HAVING avg(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，按课程号降序排列
select c_id,avg(s_score) as avg from score group by c_id order by avg asc,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
select student.s_name,score.s_score from score join course on score.c_id=course.c_id join student on student.s_id=score.s_id where course.c_name=’数学’ and score.s_score<60;
— 35、查询所有学生的选课情况
select student.s_name,course.c_name from student join score on score.s_id=student.s_id join course on course.c_id=score.c_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 select s_name as 姓名,c_name as 课程名称,s_score as 分数 from student join score on student.s_id=score.s_id join course on course.c_id=score.c_id where s_score>70
— 37、查询不及格的课程并按课程号从大到小排列 select * from score where s_score<60 order by c_id
— 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 — 方式一: select * from student join score on student.s_id=score.s_id where s_score>80 and c_id=’03’ — 方式二: select * from student where s_id in (select s_id from score where s_score>80 and c_id=’03’)
— 39、查询选了课程的学生人数 — 方式一:推荐 select count(distinct s_id) from score — 方式二: select count(*) from (select count(*) from score group by s_id) as b
— 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 — 方式1 select s_name,max(s_score) from teacher join course on teacher.t_id = course.t_id join score on score.c_id=course.c_id join student on student.s_id=score.s_id where t_name=’张三’ group by (teacher.t_id)
— 方式2 select s_name,s_score from teacher join course on teacher.t_id = course.t_id join score on score.c_id=course.c_id join student on student.s_id=score.s_id where t_name=’张三’ order by s_score desc limit 1
— 41、查询各个课程及相应的选修人数 select score.c_id,course.c_name,count(*) from score join course on score.c_id=course.c_id group by c_id
— 42、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 sc1.s_id,sc1.c_id,sc1.s_score from score as sc1 join score as sc2 on sc1.s_id=sc2.s_id and sc1.c_id<>sc2.c_id and sc1.s_score=sc2.s_score;
— 43、查询每门课程成绩最好的前两名 同25题 select * from score as sc1 where (select count(*) from score as sc2 where sc2.s_score>sc1.s_score and sc1.c_id=sc2.c_id)<2 order BY sc1.c_id
— 44、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序，若人数相同，按课程号升序排序
select c_id,count(*) from score group by c_id having count(*)>5 order by count(*) desc,c_id asc
— 查询至少选修两门课程的学生学号 select s_id,count(*) from score group by s_id having count(*)>=2
— 46、查询选修了全部课程的学生信息
select s_id,count(*) from score group by s_id having count(*)=(select count(*) from course)
— 47、查询没学过“张三”老师讲授的任一门课程的学生姓名
select * from student where s_id not in ( select score.s_id from teacher join course on course.t_id=teacher.t_id join score on score.c_id=course.c_id join student on student.s_id=score.s_id where teacher.t_name=’张三’ group by score.s_id)
— 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 s_id,avg(s_score) from score where s_score<60 group by s_id
— 49、检索课程编号为“01”且分数小于60的学生学号，结果按按分数降序排列 select s_id from Score where c_id=’01’ and s_score <60 order by s_score DESC;
— 50、删除学生编号为“02”的课程编号为“01”的成绩 delete from score where s_id=’02’ and c_id=’01’
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