sql数据库查询

-- 2

select A.student_id,A.num as 生物,B.num as 物理 from
(select student_id,num,cname from score left join course on course_id = cid where cname = ‘生物‘) as A
left join
(select student_id,num,cname from score left join course on course_id = cid where cname = ‘物理‘) as B
on A.student_id = B.student_id  where A.num > if(isnull(B.num),0,B.num )

-- 3
-- select student_id as 学号,avg(num) as 平均成绩 from score group by  student_id having avg(num) >60;

-- 4

有问题的代码,不知道什么问题
select student_id,count(course_id),group_concat(sname) from
(select * from score left join student on score.student_id = student.sid) as A
group by A.student_id

 正确代码
select student_id,sname,课程数,总成绩 from
(select student_id,count(course_id) as 课程数,sum(num) as 总成绩 from score group by student_id) as A
left join
student on A.student_id = student.sid

-- 5
-- select count(tid) from teacher where tname like ‘李%‘

-- 6

select student_id,GROUP_CONCAT(DISTINCT sname),GROUP_CONCAT(DISTINCT tname)  from
(select * from
	(select student_id,course_id,num,sname from score left join student on score.student_id = student.sid) as A
	left join
	course on A.course_id = course.cid)
as B
left join
teacher on B.teacher_id = teacher.tid  group by student_id  HAVING  not FIND_IN_SET(‘李平老师‘,GROUP_CONCAT(DISTINCT tname));

-- 7

select student_id,sname from
(select student_id,GROUP_CONCAT(course_id) as group_course from score  group by student_id HAVING FIND_IN_SET(1,group_course) and  FIND_IN_SET(2,group_course)) as A
left JOIN
student on A.student_id = student.sid

-- 8
--
方法1,有问题,不能自己写李平的课程是2 4
select  student_id,GROUP_CONCAT(DISTINCT sname),GROUP_CONCAT( tname),GROUP_CONCAT( DISTINCT course_id) from
(select * from
	(select student_id,course_id,num,sname from score left join student on score.student_id = student.sid) as A
	left join
	course on A.course_id = course.cid)
as B
left join
teacher on B.teacher_id = teacher.tid  group by student_id  HAVING  FIND_IN_SET(‘2‘,GROUP_CONCAT(DISTINCT course_id)) and FIND_IN_SET(‘4‘,GROUP_CONCAT(DISTINCT course_id))

 方法2
SELECT * from
(select student_id,GROUP_CONCAT(course_id ORDER BY course_id ASC) as course_student,GROUP_CONCAT(DISTINCT sname),GROUP_CONCAT(DISTINCT tname) as tname   from
(select * from
	(select student_id,course_id,num,sname from score left join student on score.student_id = student.sid) as A
	left join
	course on A.course_id = course.cid)
as B
left join
teacher
on B.teacher_id = teacher.tid  WHERE tname = ‘李平老师‘  GROUP BY student_id) as C
left JOIN
(select tname,GROUP_CONCAT(cid ORDER BY cid ASC) as course_teacher from course INNER JOIN teacher on course.teacher_id = teacher.tid where tname = ‘李平老师‘ GROUP BY tname) as D
on C.tname = D.tname  WHERE course_student = course_teacher

 方法3
select score.student_id,student.sname,count(score.course_id) as 选课总数,GROUP_CONCAT(DISTINCT tc2_tw) as 讲课总数 from
score inner join student
		inner join
	(select teacher.tname as tc2_tn,course.cid as tc2_tcid,tc.td as tc2_tw from course inner join teacher inner join
		(select teacher.tname as tn,count(course.cid) as td from course inner join teacher on course.teacher_id = teacher.tid group by tn having tn=‘李平老师‘) as tc
	 on course.teacher_id = teacher.tid and tc.tn = teacher.tname)
	as tc2
	on score.course_id = tc2.tc2_tcid and score.student_id = student.sid
group by score.student_id  HAVING  选课总数 = 讲课总数

 方法三拆分1
李平老师教的课程数
select teacher.tname as tn,count(course.cid) as td from
course
inner join
teacher on course.teacher_id = teacher.tid group by tn having tn=‘李平老师‘ and tc.tn = teacher.tname

方法三拆分2
李平老师教的课程号和课程数
select teacher.tname as tc2_tn, course.cid as tc2_tcid, tc.td as tc2_tw from
		course inner join teacher inner join
		(select teacher.tname as tn,count(course.cid) as td from course inner join teacher on course.teacher_id = teacher.tid group by tn having tn=‘李平老师‘) as tc
	 on course.teacher_id = teacher.tid and tc.tn = teacher.tname

-- 9

select student_id,sname from
(select A.student_id  from
(SELECT * from score where course_id = 1) as A
left JOIN
(SELECT * from score where course_id = 2) as B
on A.student_id = B.student_id where A.num > B.num)
as C
left JOIN
student on C.student_id = student.sid

-- 10
SELECT student_id,sname from
(select student_id from score where num <60 GROUP BY student_id) as A
left JOIN
student
on A.student_id = student.sid

-- 11

select student_id,sname from
(select * from
(SELECT student_id,GROUP_CONCAT(course_id ORDER BY course_id) as 选的课程 from score GROUP BY student_id) as A
inner JOIN
(select GROUP_CONCAT(cid ORDER BY cid ASC) as 全部课程 from course) as B
where 选的课程 != 全部课程) as C
left JOIN
student  on C.student_id = student.sid

-- 12
 有问题的代码
SELECT score.student_id from
(select student_id,GROUP_CONCAT(course_id)as a  from score GROUP BY student_id HAVING student_id = 1) as A
inner JOIN
score where course_id in (1,2,4)  GROUP BY score.student_id
-- 这里的a 是字符串 ,而in方法只能用列表,mysql 又不能将字符串转化为列表,不得不说这个mysql是真难用
-- where a = ‘1,2,4‘
-- where course_id in (1,2,4)

SELECT student_id,sname from
(SELECT student_id  from
(select course_id as A_course from score where student_id = 1) as A
inner JOIN
score where course_id = A_course GROUP BY student_id) as B
LEFT JOIN
student
on B.student_id = student.sid 

-- 13
 不知道是不是审题错了,和12题基本一样
SELECT student_id,sname from
(SELECT student_id  from
(select course_id as A_course from score where student_id = 1) as A
inner JOIN
score where course_id = A_course GROUP BY student_id) as B
LEFT JOIN
student
on B.student_id = student.sid where student_id !=1

-- 14

SELECT student_id,sname from
(SELECT student_id from
(select student_id as a_id, GROUP_CONCAT(course_id ORDER BY course_id ASC)as a  from score GROUP BY student_id HAVING student_id = 2) as A
inner JOIN
(select student_id,GROUP_CONCAT(course_id ORDER BY course_id ASC) as b from score GROUP BY student_id) as B
where a = b and student_id != 2)
as C
LEFT JOIN
student
on C.student_id = student.sid

-- 15
不知道这样写为什么不行,难道是主键不能删除
delete  from score where sid in
(select sid from score
left join
(select * from course left join teacher on course.teacher_id = teacher.tid) as A
on score.course_id = A.cid
where tname = ‘李平老师‘)

delete from score where course_id in
 ( select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.tname = ‘李平老师‘ )

-- 16

INSERT into score(student_id,course_id,num)
SELECT student_id,2,AVG(num) from score
where student_id not in
(select student_id from score where course_id = 2)
GROUP BY student_id

-- 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

select A.student_id as 学号,生物,物理,体育,美术,平均分  from
(select student_id,num as 生物 from score
inner JOIN
course on score.course_id = course.cid  where cname = ‘生物‘) as A
inner JOIN
(select student_id,num as 物理 from score
inner JOIN
course on score.course_id = course.cid  where cname = ‘物理‘) as B
inner join
(select student_id,num as 体育 from score
inner JOIN
course on score.course_id = course.cid  where cname = ‘体育‘) as C
inner join
(select student_id,num as 美术 from score
inner JOIN
course on score.course_id = course.cid  where cname = ‘美术‘) as D
inner join
(select student_id,AVG(num) as 平均分 from score GROUP BY student_id) as E

on A.student_id = B.student_id and A.student_id = C.student_id and A.student_id = D.student_id and A.student_id = E.student_id ORDER BY 平均分 ASC

-- 18
-- select course_id,max(num),min(num) from score group by course_id

-- 19

SELECT C.course_id,平均分,及格率 from
(select course_id,avg(num) as 平均分,GROUP_CONCAT(num) from score group by course_id) as C
LEFT JOIN
(select A.course_id,及格, 总数,及格/总数 as 及格率 from
(SELECT course_id,count(num) as 及格 from score where num >=60 GROUP BY course_id) as A
left JOIN
(SELECT course_id,count(num) as 总数 from score GROUP BY course_id) as B
on A.course_id = B.course_id) as D
on C.course_id = D.course_id  ORDER BY 平均分 ASC, 及格率 DESC

-- 20

select 平均分,tname from
(select course_id,avg(num) as 平均分 from score GROUP BY course_id ORDER BY 平均分 DESC  ) as A
LEFT JOIN
(select * from course LEFT JOIN teacher on course.teacher_id = teacher.tid ) as B
on A.course_id = B.cid ORDER BY 平均分 DESC

-- 21
-- 还没做出来,有问题
--  select course_id,GROUP_CONCAT(num),GROUP_CONCAT(student_id ORDER BY num DESC ) from score GROUP BY course_id

-- 22
-- select course_id,COUNT(student_id) from score GROUP BY course_id

-- 23

select A.student_id,sname from
(select student_id from score GROUP BY student_id  HAVING count(course_id) = 1) as A
LEFT JOIN
student
on A.student_id = student.sid

-- 24
 方法一
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

-- 方法二
-- select gender,count(1) from student group by gender

-- 25
-- select * from student where sname like ‘张%‘

-- 26
-- select sname,count(1) from student GROUP BY sname 

-- 27
-- 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

SELECT student_id,sname,avg from
(select student_id,avg(if(isnull(num),0,num)) as avg from score GROUP BY student_id HAVING avg > 85) as A
LEFT JOIN
student on A.student_id = student.sid

-- 29

select student_id,sname,num from
(select * from
score left JOIN
course on score.course_id = course.cid where cname = ‘生物‘ and num < 60) as A
left join student on A.student_id = student.sid

-- 30

SELECT A.student_id,sname from
(select student_id from score where course_id = 3 and num >80) as A
LEFT JOIN
student on A.student_id = student.sid

-- 31

select count(1) from
(select student_id,count(1) from score  GROUP BY student_id) as A

-- 32

select student_id,sname from
(select student_id,num from
(select * from score
LEFT JOIN
course on score.course_id = course.cid) as A
LEFT JOIN
teacher on A.teacher_id = teacher.tid where tname = ‘刘海燕老师‘ ORDER BY num desc LIMIT 1) as B
LEFT JOIN
student on B.student_id = student.sid

-- 33

select cname,count  from
(select course_id,count(1) as count from score GROUP BY course_id) as A
left join
course on A.course_id = course.cid

-- 34
-- 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
--  有问题
-- select course_id,GROUP_CONCAT(sid ORDER BY num desc) from score GROUP BY course_id 

-- 36
-- select student_id from score GROUP BY student_id HAVING count(course_id) > 1

-- 37

select course_id,GROUP_CONCAT(student_id ORDER BY student_id) as st from score GROUP BY course_id
 HAVING st=
(select GROUP_CONCAT( DISTINCT student_id ORDER BY student_id) from score )

-- 38

SELECT A.student_id,sname from
(select * from score where student_id not in
(select DISTINCT student_id from score
INNER JOIN
course
INNER JOIN
teacher on score.course_id = course.cid AND course.teacher_id = teacher.tid  WHERE tname = ‘李平老师‘)
) as A
LEFT JOIN
student on A.student_id = student.sid

-- 39

select A.student_id,avg from
(select student_id ,count(1) as count from score  where num<60  GROUP BY student_id HAVING count >1) as A
LEFT JOIN
(select student_id,avg(num) as avg from score GROUP BY student_id) as B
on A.student_id = B.student_id

-- 40
-- SELECT student_id from score where course_id= 4 and num < 60 ORDER BY num desc 

-- 41
-- delete from score where course_id = 1 and student_id = 2

21 35题 一个问题,没有解决,没做出来

　　

原文地址：https://www.cnblogs.com/perfey/p/9902215.html

时间： 2024-08-05 07:03:35

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