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sql复杂查询例题

题源
题源答案看的有点费劲，自己做了一遍，但我自己做的也不能保证正确或者最简单，欢迎指正！
复杂查询的方法就是把题干不断分解，理出1，2，3步骤，从小到大逐步查出需要的内容，嵌套子查询一定要从儿子往爸爸写呀！

建表和插数据

/***学生表***/create table student(
sno varchar(10) primary key,  -- 学号
sname varchar(20),            -- 姓名
sage int2,                -- 年龄
ssex varchar(5)               -- 性别
);
/***教师表***/
create table teacher(
tno varchar(10) primary key,  -- 教师编号
tname varchar(20)             -- 教师姓名
);
/***课程表***/
create table course(
cno varchar(10),              -- 课程编号
cname varchar(20),            -- 课程名
tno varchar(20),              -- 教师编号
constraint pk_course primary key (cno,tno)
);
/***成绩表***/
create table sc(
sno varchar(10),              -- 学号
cno varchar(10),              -- 课程编号
score int4,             -- 成绩
constraint pk_sc primary key (sno,cno)
);
#/*******初始化学生表的数据******/
insert into student values ('s001','张三',23,'男');
insert into student values ('s002','李四',23,'男');
insert into student values ('s003','吴鹏',25,'男');
insert into student values ('s004','琴沁',20,'女');
insert into student values ('s005','王丽',20,'女');
insert into student values ('s006','李波',21,'男');
insert into student values ('s007','刘玉',21,'男');
insert into student values ('s008','萧蓉',21,'女');
insert into student values ('s009','陈萧晓',23,'女');
insert into student values ('s010','陈美',22,'女');
commit;
#/******************初始化教师表***********************/
insert into teacher values ('t001', '刘阳');
insert into teacher values ('t002', '谌燕');
insert into teacher values ('t003', '胡明星');
commit;
#/***************初始化课程表****************************/
insert into course values ('c001','J2SE','t002');
insert into course values ('c002','Java Web','t002');
insert into course values ('c003','SSH','t001');
insert into course values ('c004','Oracle','t001');
insert into course values ('c005','SQL SERVER 2005','t003');
insert into course values ('c006','C#','t003');
insert into course values ('c007','JavaScript','t002');
insert into course values ('c008','DIV+CSS','t001');
insert into course values ('c009','PHP','t003');
insert into course values ('c010','EJB3.0','t002');
commit;
#/***************初始化成绩表***********************/
insert into sc values ('s001','c001',78.9);
insert into sc values ('s002','c001',80.9);
insert into sc values ('s003','c001',81.9);
insert into sc values ('s004','c001',60.9);
insert into sc values ('s001','c002',82.9);
insert into sc values ('s002','c002',72.9);
insert into sc values ('s003','c002',81.9);
insert into sc values ('s001','c003',59);

具体题目及个人答案

#查询“c001”课程比“c002”课程成绩高的所有学生的学号
select *
from sc a,sc b 
where a.cno='c001' and b.cno='c002' and a.score>b.score#查询平均成绩大于60 分的同学的学号和平均成绩【每位同学的平均成绩指某个同学所有课程的平均成绩
select avg(score),sno
from sc
group by sno
having avg(score)>=60#查询所有同学的学号、姓名、选课数、总成绩；
select s.sno,s.sname,count(sc.cno),sum(sc.score)
from student s inner join sc on s.sno=sc.sno
group by s.sno#查询姓“刘”的老师的个数
select count(tno) from teacher where tname like '刘%'#####查询没学过“谌燕”老师课的同学的学号、姓名【先找出谌燕老师的工号，再找出该工号没教的课程，再找出学这些课程的学生学号
select s.sname,s.sno 
from student s inner join sc on s.sno=sc.sno 
where sc.cno in(select cno from course where tno not in (select tno from teacher where tname='谌燕'))#####查询学过“c001”并且也学过编号“c002”课程的同学的学号、姓名【先查出学过c001课程的学生，再从中查出学过c002的
select distinct s001.sname,s001.sno 
from sc,(select s.sname,s.sno from student s join sc on s.sno=sc.sno where sc.cno='c001') s001
where sc.cno='c002'#查询学过“谌燕”老师所教的所有课的同学的学号、姓名
select distinct s.sname,s.sno 
from student s inner join sc on s.sno=sc.sno 
where sc.cno in              #sc表中的cno逐个对比，所以会产生重复结果，记得去重(select cno from course where tno in (select tno from teacher where tname='谌燕'))#查询所有课程成绩小于60 分的同学的学号、姓名【先查出大于60的成绩涉及哪些学生，再去掉这些学生
select s.sno,s.sname
from student s 
where s.sno not in(
select s.sno
from student s join sc on s.sno=sc.sno
where sc.score>=60
)#查询没有学全所有课的同学的学号、姓名；【先查出总课程数，再根据学生分组，找出每组课程数小于总课程数的学生
select s.sno,s.sname,count(sc.cno) 
from student s left join sc on s.sno=sc.sno   #sc表中可能有学生没出现，所以用左连接
group by s.sno 
having count(sc.cno)<(select count(c.cno) from course c )#查询至少有一门课与学号为“s001”的同学所学相同的同学的学号和姓名；【先查s001学生学的所有课程号，再找出选课在这个范围的学生，排除s001!!!
select distinct s.sname,s.sno 
from student s join sc on s.sno=sc.sno
where sc.cno in(select sc.cno from sc where sc.sno='s001') and s.sno !='s001'  #sc表中的cno逐个对比，所以会产生重复结果，记得去重#把“SC”表中“谌燕”老师教的课的成绩都更改为此课程的平均成绩；【先查谌燕”老师的工号，再根据工号去查对应课程的平均成绩，把这个查询结果当作一个表赋值给原表
update sc inner join   #内连接可以不加on条件
(select sc.cno,avg(score) as newScore from sc join course c on sc.cno=c.cno where c.tno=(select tno from teacher where tname='谌燕') group by sc.cno
) newSc 
set sc.score=newSc.newScore 
where sc.cno=newSc.cno#查询和“s001”号的同学学习的课程完全相同的其他同学学号和姓名；
#【先查s001学的全部课程，再查出没学过这些课程的学生，再查出只选了s001学的课程的学生（课程数不确定），再看这些人学习课程数是否等于s001学的课程数.排除s001
select sno 
from sc 
where sno !='s001' and sno not in(select sno from sc where cno not in(select cno from sc where sno='s001'))
group by sno
having count(*)=(select count(*) from sc where sno='s001')#删除学习“谌燕”老师课的SC 表记录；【1.谌燕对应课程号 2.删除sc表中对应课程号记录
delete from sc where cno in(select cno from course c join teacher t on c.tno=t.tno where t.tname='谌燕')#查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分
select cno,max(score),min(score)
from sc
group by cno
#按各科 平均成绩 从 低到高 和 及格率的百分数 从高到低顺序【使用case when求每科及格人数
select cno,avg(score) 平均成绩,sum(case when score>=60 then 1 else 0 end)/count(1) 及格率
from sc
group by cno
order by 平均成绩,及格率 desc####查询各科成绩前三名的记录:(不考虑成绩并列情况)【mysql版本问题，没运行成功，但我觉得sql没毛病。不考虑并列用row_number，考虑并列用rank/dense_rank
select * from(select sno,cno,score,row_number() over(partition by cno order by score desc) numfrom sc) sc1
where sc1.num<=3#查询每门课程被选修的学生数
select count(sno),cno
from sc 
group by cno#查询出只选修了一门课程的全部学生的学号和姓名
select sc.sno,s.sname,count(sc.cno)
from sc join student s on s.sno=sc.sno
group by sc.sno
having count(sc.cno)=1#查询男生、女生人数
select sum(case when ssex='男' then 1 else 0 end) 男生数,sum(case when ssex='女' then 1 else 0 end)
from student#查询姓“张”的学生名单
select *
from student
where sname like '张%'#查询同名同性学生名单，并统计同名人数【统计每个名字出现次数,根据名字和性别分组
select sname,ssex,count(sno)
from student
group by sname,ssex
having count(sno)>1#1997 年出生的学生名单(注：Student 表中Sage 列的类型是number)
select *
from student 
where year(now())-sage=1997#查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列
select cno,avg(score)
from sc
group by cno
order by avg(score),cno desc#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 sc.sno,s.sname,avg(sc.score)
from sc join student s on s.sno=sc.sno
group by sc.sno
having avg(sc.score)>85#查询课程名称为“SSH”，且分数低于60 的学生姓名和分数【三表连接
select s.sname,sc.score
from student s join sc on s.sno=sc.sno join course c on c.cno=sc.cno
where c.cname='SSH' and sc.score<60#查询所有学生的选课情况【所有学生，用左连接
select s.sno,s.sname,c.cno,c.cname
from student s left join sc on s.sno=sc.sno left join course c on c.cno=sc.cno#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【任何一门？至少一门的意思？
select s.sname,c.cname,sc.score
from student s join sc on s.sno=sc.sno join course c on c.cno=sc.cno
where sc.score>70#查询不及格的课程，并按课程号从大到小排列【课程咋不及格？这啥题目啊摔！
select cno
from sc
where score<60
order by cno desc#查询课程编号为c001 且课程成绩在80 分以上的学生的学号和姓名
select sc.sno,s.sname
from sc join student s on s.sno=sc.sno
where cno='c001' and score>80#求选了课程的学生人数【统计sc表不重复的sno个数
select count(sno)
from (select distinct sno from sc) sc1#查询选修“谌燕”老师所授课程的学生中，成绩最高的学生姓名及其成绩【1.该老师教课程号 2.上述课程号对应学生和成绩 3.对成绩降序排序，最后只查出第一条记录
select s.sname,sc.score
from sc join student s on s.sno=sc.sno
where sc.cno in(select cno from course c join teacher t on t.tno=c.tno where t.tname='谌燕')
order by sc.score desc
limit 1#查询各个课程及相应的选修人数
select cno,count(sno)
from sc
group by cno#查询每门功课成绩最好的前两名【一用rownum就报错，但我真没看出来错哪了......
select * from (select  row_number()over(partition by cno order by score desc) rn,cno,sno,scorefrom sc)t
where t.rn < 3#统计每门课程的学生选修人数（超过10 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 cno,count(sno)
from sc
group by cno
having count(sno)>10
order by count(sno) desc,cno#检索至少选修两门课程的学生学号
select sno,count(cno)
from sc
group by sno
having count(cno)>=2#查询全部学生都选修的课程的课程号和课程名
select cno,count(sno)
from sc
group by cno
having count(sno)=(select count(1)from student)#查询没学过“谌燕”老师讲授的任一门课程的学生姓名
##错误写法，只要学生选的课存在不是该老师教的，就会查出来
select s.sname
from sc join student s on s.sno=sc.sno
where sc.cno not in(select cno from course c join teacher t on t.tno=c.tno where t.tname='谌燕')  ##正确写法:查出该老师教的课，再查上这些课的学生，再取补集
select s.sname
from sc join student s on s.sno=sc.sno
where sc.sno not in(select sno from sc where cno in(select cno from course c join teacher t on t.tno=c.tno where t.tname='谌燕') 
)###查询两门以上不及格课程的同学的学号及其平均成绩【单表别名自连接（我自己瞎起的名字，懒得查专业术语）
select sno,avg(score)
from sc sc1
where (select count(*) from sc sc2 where sc1.sno=sc2.sno and score<60)>=2
group by sno#检索“c004”课程分数小于60，按分数降序排列的同学学号
select sno
from sc
where cno='c004' and score<60
order by sno desc#删除“s002”同学的“c001”课程的成绩
delete from sc
where sno='s002' and cno='c001'