1、查詢沒有學全所有課的同學的學號、姓名；
# 先統計一共有多少門課程
select count(cid) from course;
# 查看每個學生選擇的課程書
select count(course_id) from score group by student_id;
# 查詢所學課程數小於總課程數的學生學號
select student_id
from (select count(course_id) c_course_id,student_id from score group by student_id) t1
where t1.c_course_id < (select count(cid) from course) ;
# 查詢沒有學全所有課的同學的學號、姓名；
select sid,sname from student where sid in (
select student_id from (select count(course_id) c_course_id,student_id from score group by student_id
) t1 where t1.c_course_id < (select count(cid) from course)
) ;
2、查詢和“002”號的同學學習的課程完全相同的其他同學學號和姓名；
# 先查詢2號同學學了哪些課程
select * from score where student_id =2;
# 找到學習了2號同學沒學習課程的所有同學（找到所有和2號同學學習的課程不一樣的同學）
select student_id from score where course_id not in (select course_id from score where student_id=2)
# 找到score表中所有的學生並且把 2號同學 以及（和2號同學學習的課程不一樣的同學）排除出去
select student_id from score where student_id not in (select student_id from score where course_id not in (select course_id from score where student_id=2)) and student_id !=2
# 對剩余的和2號同學所選課程沒有不同的同學所選課程數進行統計，如果和2號同學的課程數相同，就是選擇了相同的課程
select student_id from score where student_id not in (
select student_id from score where course_id not in (select course_id from score where student_id=2)
) and student_id !=2
group by student_id
having count(course_id)= (select count(course_id) from score where student_id=2);
3、刪除學習“葉平”老師課的SC(score)表記錄；
# 先查出李平老師的id
select tid from teacher where tname = '李平老師';
# 查看李平老師所教授的課程
select cid from course where teacher_id = (select tid from teacher where tname = '李平老師’);
# 查看李平老師所教課程的成績數據
select * from score where course_id in (select cid from course where teacher_id = (select tid from teacher where tname = '李平老師'));
# 執行刪除命令
delete from score where course_id in (select cid from course where teacher_id = (select tid from teacher where tname = '李平老師'));
4、向SC表中插入一些記錄，這些記錄要求符合以下條件：①沒有上過編號“002”課程的同學學號；②插入“002”號課程的平均成績；
# 先找尋上過2號課程的同學
select student_id from score where course_id = 2;
# 再找到沒上過2號課程的所有同學
select * from student where sid not in (select student_id from score where course_id = 2);
# 計算出學習2號課程的同學的平均成績
select avg(num) from score where course_id = 2 group by course_id;
# 用笛卡爾積將上述兩個表拼起來
select * from (select sid from student where sid not in (select student_id from score where course_id = 2)) t1,(select avg(num) from score where course_id = 2 group by course_id) t2;
# 向SC表中插入記錄
insert into score (course_id,student_id,num) select 2,t1.sid,t2.avg_num from (select sid from student where sid not in (select student_id from score where course_id = 2)) t1,(select avg(num) avg_num from score where course_id = 2 group by course_id) t2;
5、按平均成績從低到高顯示所有學生的“語文”、“數學”、“英語”三門的課程成績，按如下形式顯示： 學生ID,語文,數學,英語,有效課程數,有效平均分；
# 查看每個學生的數學成績
select student_id,num from score where course_id = (select cid from course where cname = '數學');
# 查看每個學生的語文成績
select student_id,num from score where course_id = (select cid from course where cname = '語文');
# 查看每個學生的英語成績
select student_id,num from score where course_id = (select cid from course where cname = '英語');
# 查看每個學生的平均成績
select student_id,avg(num),count(num) from score group by student_id;
# 將上面的幾張表拼接起來,為了生成所有學生的信息，用student表作為左連接的第一張表
select sid 學生ID,t2.num 語文,t1.num 數學, t3.num 英語,t4.count_course 有效課程數,t4.avg_num 有效平均分 from student
left join (select student_id,num from score where course_id = (select cid from course where cname = '數學')) t1
on student.sid = t1.student_id
left join (select student_id,num from score where course_id = (select cid from course where cname = '語文')) t2
on student.sid = t2.student_id
left join (select student_id,num from score where course_id = (select cid from course where cname = '英語')) t3
on student.sid = t3.student_id
left join (select student_id,avg(num) avg_num,count(num) count_course from score group by student_id) t4
on student.sid = t4.student_id
6、查詢各科成績最高和最低的分：以如下形式顯示：課程ID，最高分，最低分；
select course_id 課程ID,max(num) 最高分,min(num) 最低分 from score group by course_id;
7、按各科平均成績從低到高和及格率的百分數從高到低順序；
# 方法1:
# 先求平均成績
select course_id,avg(num) from score group by course_id;
# 解決計算各科及格率的問題
所有及格的人/所有人數
select t1.course_id,t1.count1/t2.count2 from
(select course_id,count(course_id) count1 from score where num>60 group by course_id) t1
left join
(select course_id,count(course_id) count2 from score group by course_id) t2
on t1.course_id = t2.course_id;
# 根據上述內容進行表的拼接
select t_out1.course_id,t_out1.avgnum, t_out2.pass_per from
(select course_id,avg(num) avgnum from score group by course_id ) t_out1
left join
(select t1.course_id,t1.count1/t2.count2 pass_per from
(select course_id,count(course_id) count1 from score where num>60 group by course_id) t1
left join
(select course_id,count(course_id) count2 from score group by course_id) t2
on t1.course_id = t2.course_id) t_out2
on t_out1.course_id = t_out2.course_id
# 加上排序
select t_out1.course_id,t_out1.avgnum, t_out2.pass_per from (select course_id,avg(num) avgnum from score group by course_id ) t_out1 left join (select t1.course_id,t1.count1/t2.count2 pass_per from (select course_id,count(course_id) count1 from score where num>60 group by course_id) t1 left join (select course_id,count(course_id) count2 from score group by course_id) t2 on t1.course_id = t2.course_id) t_out2 on t_out1.course_id = t_out2.course_id order by avgnum ,pass_per desc;
# 方法2
# 使用case when直接計算合格率
select
sum(case when num>60 then 1 else 0 end)/count(course_id)
from score group by course_id
# 加上課程id和平均值
select course_id,avg(num),
sum(case when num>60 then 1 else 0 end)/count(course_id)
from score group by course_id
# 加上排序
select course_id,avg(num) avgnum,
sum(case when num>60 then 1 else 0 end)/count(course_id) pass_per
from score group by course_id
order by avgnum ,pass_per desc;
8、查詢各科成績前三名的記錄:(不考慮成績並列情況)
select
t1.sid,t1.student_id,t1.course_id,t1.num from score t1
left join
(
select sid,course_id,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0, 1) as first_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1, 1) as second_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 2, 1) as third_num
from score as s1
) t2
on t1.sid = t2.sid
where t1.num = t2.first_num or t1.num = t2.second_num or t1.num = t2.third_num;
9、查詢每門課程被選修的學生數；
select course_id,count(course_id) from score group by course_id;
10、查詢同名同姓學生名單，並統計同名人數；
select sname,count(1) as count from student group by sname;
11、查詢每門課程的平均成績，結果按平均成績升序排列，平均成績相同時，按課程號降序排列；
select course_id,avg(if(isnull(num), 0 ,num)) as avg from score group by course_id order by avg asc,course_id desc;
12、查詢平均成績大於85的所有學生的學號、姓名和平均成績；
select student_id,sname, avg(if(isnull(num), 0 ,num)) from score left join student on score.student_id = student.sid group by student_id;
13、查詢課程名稱為“數學”，且分數低於60的學生姓名和分數；
select student.sname,score.num from score
left join course on score.course_id = course.cid
left join student on score.student_id = student.sid
where score.num < 60 and course.cname = '數學'
14、查詢課程編號為003且課程成績在80分以上的學生的學號和姓名；
select * from score where score.student_id = 3 and score.num > 80
15、求選了課程的學生人數
select sid,sname from student where sid not in (select student_id from score group by student_id);
16、查詢選修“楊艷”老師所授課程的學生中，成績最高的學生姓名及其成績；
# 先找到“楊艷”老師的教師id
select tid from teacher where tname = '楊艷';
# 再找到楊艷老師教的所有課程
select cid from course where teacher_id in (select tid from teacher where tname = '楊艷');
# 再找到楊艷老師教的所有課程的最高分
select max(num) from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老師'));
# 再找到楊艷老師教的所有課程的最高分對應的學生
select distinct student_id,num from score
where num = (select max(num) from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老師')))
and course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老師'));
# 找到學生的姓名
select student.sname,t1.num from(
select distinct student_id,num from score
where num = (select max(num) from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老師')))
and course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老師'))
) t1
left join
student
on
t1.student_id = student.sid;
17、查詢各個課程及相應的選修人數；
select course.cname,count(1) from score
left join course on score.course_id = course.cid
group by course_id;
18、查詢不同課程但成績相同的學生的學號、課程號、學生成績；
select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;
19、查詢每門課程成績最好的前兩名；
先查詢每條數據對應學科成績的第一名和第二名，這里必須要保留所有的s1，以便后續進行連表查詢
select sid,course_id,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0, 1) as first_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1, 1) as second_num
from score as s1
按照sid連表，把學生的成績和對應的第一名、第二名成績連起來
select
* from score t1
left join
(
select sid,course_id,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0, 1) as first_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1, 1) as second_num
from score as s1
) t2
on t1.sid = t2.sid
判斷如果學生的成績是第一名、第二名的成績，那么就符合條件，顯示學生的id、學科和成績
select
t1.sid,t1.student_id,t1.course_id,t1.num from score t1
left join
(
select sid,course_id,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0, 1) as first_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1, 1) as second_num
from score as s1
) t2
on t1.sid = t2.sid
where t1.num = t2.first_num or t1.num = t2.second_num;
20、檢索至少選修兩門課程的學生學號；
select student_id from score group by student_id having count(student_id) > 1;
21、查詢全部學生都選修的課程的課程號和課程名；
# 先查看一共有多少學生
select count(sid) from student;
# 查看哪一門課選秀的學生個數和學生的總個數相等
select course_id from score group by course_id having count(student_id) = (select count(sid) from student);
22、查詢沒學過“葉平”老師講授的任一門課程的學生姓名；
# 先查看要查找老師的id
select tid from teacher where tname = '李平老師';
# 查看該老師交了哪些課程
select cid from course where teacher_id in (select tid from teacher where tname = '李平老師')
# 看看有多少學生學習了該老師的課程
select distinct student_id from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老師'));
# 把不在上表中的學生姓名查出來
select sname from student where sid not in (select distinct student_id from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老師')));
23、查詢兩門以上不及格課程的同學的學號及其平均成績；
select student_id,avg(num) from score where num<60 group by student_id having count(num)>=2;
24、檢索“004”課程分數小於60，按分數降序排列的同學學號；
select student_id from score where num< 60 and course_id = 4 order by num desc;
25、刪除“002”同學的“001”課程的成績；
delete from score where course_id = 1 and student_id = 2