2、查詢“生物”課程比“物理”課程成績高的所有學生的學號；
思路：
獲取所有有生物課程的人（學號，成績） - 臨時表
獲取所有有物理課程的人（學號，成績） - 臨時表
根據【學號】連接兩個臨時表：
學號 物理成績 生物成績

然後再進行篩選

select A.student_id,sw,ty from

(select student_id,num as sw from score left join course on score.course_id = course.cid where course.cname = ‘生物‘) as A

left join

(select student_id,num as ty from score left join course on score.course_id = course.cid where course.cname = ‘體育‘) as B

on A.student_id = B.student_id where sw > if(isnull(ty),0,ty);

3、查詢平均成績大於60分的同學的學號和平均成績；
思路：
根據學生分組，使用avg獲取平均值，通過having對avg進行篩選

select student_id,avg(num) from score group by student_id having avg(num) > 60

4、查詢所有同學的學號、姓名、選課數、總成績；

select score.student_id,sum(score.num),count(score.student_id),student.sname
from
score left join student on score.student_id = student.sid
group by score.student_id

5、查詢姓“李”的老師的個數；
select count(tid) from teacher where tname like ‘李%‘

select count(1) from (select tid from teacher where tname like ‘李%‘) as B

6、查詢沒學過“葉平”老師課的同學的學號、姓名；
思路：
先查到“李平老師”老師教的所有課ID
獲取選過課的所有學生ID
學生表中篩選
select * from student where sid not in (
select DISTINCT student_id from score where score.course_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname = ‘李平老師‘
)
)

7、查詢學過“001”並且也學過編號“002”課程的同學的學號、姓名；
思路：
先查到既選擇001又選擇002課程的所有同學
根據學生進行分組，如果學生數量等於2表示，兩門均已選擇

select student_id,sname from

(select student_id,course_id from score where course_id = 1 or course_id = 2) as B

left join student on B.student_id = student.sid group by student_id HAVING count(student_id) > 1


8、查詢學過“葉平”老師所教的所有課的同學的學號、姓名；

同上，只不過將001和002變成 in (葉平老師的所有課)

9、查詢課程編號“002”的成績比課程編號“001”課程低的所有同學的學號、姓名；
同第1題


10、查詢有課程成績小於60分的同學的學號、姓名；

select sid,sname from student where sid in (
select distinct student_id from score where num < 60
)

11、查詢沒有學全所有課的同學的學號、姓名；
思路：
在分數表中根據學生進行分組，獲取每一個學生選課數量
如果數量 == 總課程數量，表示已經選擇了所有課程

select student_id,sname
from score left join student on score.student_id = student.sid
group by student_id HAVING count(course_id) = (select count(1) from course)


12、查詢至少有一門課與學號為“001”的同學所學相同的同學的學號和姓名；
思路：
獲取 001 同學選擇的所有課程
獲取課程在其中的所有人以及所有課程
根據學生篩選，獲取所有學生信息
再與學生表連接，獲取姓名

select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id

13、查詢至少學過學號為“001”同學所有課的其他同學學號和姓名；
先找到和001的學過的所有人
然後個數 ＝ 001所有學科 ＝＝》 其他人可能選擇的更多

select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) ＝ (select count(course_id) from score where student_id = 1)

14、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；

個數相同
002學過的也學過

select student_id,sname from score left join student on score.student_id = student.sid where student_id in (
select student_id from score where student_id != 1 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
) and course_id in (select course_id from score where student_id = 1) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)


15、刪除學習“葉平”老師課的score表記錄；

delete from score where course_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = ‘葉平‘
)

16、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；
思路：
由於insert 支持
inset into tb1(xx,xx) select x1,x2 from tb2;
所有，獲取所有沒上過002課的所有人，獲取002的平均成績

insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2)
from student where sid not in (
select student_id from score where course_id = 2
)

17、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；
select sc.student_id,
(select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,
(select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,
(select num from score left join course on score.course_id = course.cid where course.cname = "體育" and score.student_id=sc.student_id) as ty,
count(sc.course_id),
avg(sc.num)
from score as sc
group by student_id desc

18、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, max(num) as max_num, min(num) as min_num from score group by course_id;

19、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；
思路：case when .. then
select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;

20、課程平均分從高到低顯示（現實任課老師）；

select avg(if(isnull(score.num),0,score.num)),teacher.tname from course
left join score on course.cid = score.course_id
left join teacher on course.teacher_id = teacher.tid

group by score.course_id


21、查詢各科成績前三名的記錄:(不考慮成績並列情況)
select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
(
select
sid,
(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 3,1) as second_num
from
score as s1
) as T
on score.sid =T.sid
where score.num <= T.first_num and score.num >= T.second_num

22、查詢每門課程被選修的學生數；

select course_id, count(1) from score group by course_id;

23、查詢出只選修了一門課程的全部學生的學號和姓名；
select student.sid, student.sname, count(1) from score

left join student on score.student_id = student.sid

group by course_id having count(1) = 1


24、查詢男生、女生的人數；
select * from
(select count(1) as man from student where gender=‘男‘) as A ,
(select count(1) as feman from student where gender=‘女‘) as B

25、查詢姓“張”的學生名單；
select sname from student where sname like ‘張%‘;

26、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;

27、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;

28、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;

29、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 = ‘生物‘

30、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

31、求選了課程的學生人數

select count(distinct student_id) from score

select count(c) from (
select count(student_id) as c from score group by student_id) as A

32、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；

select sname,num from score
left join student on score.student_id = student.sid
where score.course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname=‘張磊老師‘) order by num desc limit 1;

33、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;


34、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;

35、查詢每門課程成績最好的前兩名；

select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
(
select
sid,
(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
) as T
on score.sid =T.sid
where score.num <= T.first_num and score.num >= T.second_num

36、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

37、查詢全部學生都選修的課程的課程號和課程名；
select course_id,count(1) from score group by course_id having count(1) = (select count(1) from student);

38、查詢沒學過“葉平”老師講授的任一門課程的學生姓名；
select student_id,student.sname from score
left join student on score.student_id = student.sid
where score.course_id not in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname = ‘張磊老師‘
)
group by student_id

39、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,count(1) from score where num < 60 group by student_id having count(1) > 2

40、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;

41、刪除“002”同學的“001”課程的成績；
delete from score where course_id = 1 and student_id = 2

MySQL--題庫解析