CREATE TABLE student(
sno VARCHAR(10) PRIMARY KEY,
sname VARCHAR(20),
sage NUMERIC(2),
ssex VARCHAR(5)
);
CREATE TABLE teacher(
tno VARCHAR(10) PRIMARY KEY,
tname VARCHAR(20)
);
CREATE TABLE course(
cno VARCHAR(10),
cname VARCHAR(20),
tno VARCHAR(20),
CONSTRAINT pk_course PRIMARY KEY (cno,tno)
);
CREATE TABLE sc(
sno VARCHAR(10),
cno VARCHAR(10),
score NUMERIC(4,2),
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',59.9);
commit;
练习：
注意：以下练习中的数据是根据初始化到数据库中的数据来写的SQL 语句，请大家务必注意。
1、查询“c001”课程比“c002”课程成绩高的所有学生的学号；
2、查询平均成绩大于60 分的同学的学号和平均成绩；
3、查询所有同学的学号、姓名、选课数、总成绩；
4、查询姓“刘”的老师的个数；
5、查询没学过“谌燕”老师课的同学的学号、姓名；
6、查询学过“c001”并且也学过编号“c002”课程的同学的学号、姓名
7、查询学过“谌燕”老师所教的课的同学的学号、姓名；
我的思路：1、找到该老师的工号
2、找到该老师所教的课程的号码
3、在分数表中找到有这些课程号码的学生的号码
4、在学生表中找到相应的学生
具体sql如下：
select *
from student
where sno in
(select distinct (s.sno)
from sc s
where s.cno in
(select b.cno
from course b
where b.tno in
(select a.tno from teacher a where a.tname = '谌燕')))
答案思路：使用连接一环套一环往后走，直至最后的条件是老师名字
8、查询课程编号“c002”的成绩比课程编号“c001”课程低的所有同学的学号、姓名；
我的思路：1、两个数据c002的成绩和c001的成绩同出一表，所以使用内连接
2、给定两个表不同的条件然后查询即可
具体sql如下：select *
from sc a
join sc b
on a.sno = b.sno
join student s
on a.sno = s.sno
where a.cno = 'c001'
and b.cno = 'c002'
and a.score > b.score
答案思路：即我的思路
9、查询所有课程成绩小于60 分的同学的学号、姓名；
我的思路：1、查询出分数小小于60的学生号码并且使用内连接查询出学生姓名
具体sql如下：
select distinct (a.sno), s.sname
from sc a
join student s
on a.sno = s.sno
where a.score < 60
答案思路：即我的思路
10、查询没有学全所有课的同学的学号、姓名；
我的思路：1、在分数表中使用groupby查询每个学生所学习的科目的数量
2、找到学习数量等于总数量的学生学号
3、查询student表，排除第二步得到的学号即可
具体sql如下：select *
from student d
where d.sno not in
(select sno
from (select a.sno, count(a.cno) cnum from sc a group by a.sno) b
where b.cnum = (select count(*) from course))
答案思路：第一种思路：使用左连接把学生表和成绩表连接起来，然后使用分组统计出每个学生所学的科目数量，
分组统计完成之后，再选出所学科目数小于总数量的学生
第二种思路：使用笛卡尔成绩和minus取差集，这个集合里面的数据就是某位学生和他没有学过的课程，妙哉
回过头看，我写的sql是什么狗屁东西
11、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；
我的思路：使用内连接拼接sc表，然后选择a表中是s001而b表中不是s001的即可得到选修了与s001同样课程之人
具体sql如下：select s.sno, s.sname
from student s
where s.sno in (select distinct (b.sno)
from sc a
join sc b
on a.cno = b.cno
where a.sno = 's001'
and b.sno != 's001')
and s.sno != 's001'
答案思路： 整体思路与我一致，只是采用的不同的连接方式
12、查询至少学过学号为“s001”同学所有一门课的其他同学学号和姓名；
我的思路：没读懂题目，这题跟上一个有什么区别吗？
答案思路：将sc表与student表进行左连接，其中一条筛选条件为连接表的课程编号要在s001同学学过的课程编号中
13、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；
我的思路：首先找到该老师所教的课程的编号，然后进行更改
具体sql如下：update sc c
set c.score =
(select avg(e.score) from sc e where c.cno = e.cno group by e.cno)
where c.cno in (select a.cno
from course a
join teacher b
on a.tno = b.tno
where b.tname = '谌燕')
答案思路：
14、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；
我的思路：不会做
答案思路：答案是错的
可能的方向:
select t.sno, WMSYS.WM_CONCAT(t.cno) TIME
From (select * from sc order by cno desc) t
GROUP BY t.sNO
15、删除学习“谌燕”老师课的SC 表记录；
我的思路：找到该老师所教授课程的编号，然后去sc表中删除即可
具体sql如下：delete from sc c
where c.cno in (select a.cno
from course a
join teacher b
on a.tno = b.tno
where b.tname = '谌燕')
答案思路：跟我一样
16、向SC 表中插入一些记录，这些记录要求符合以下条件：没有上过编号“c002”课程的同学学号、“c002”号课的平均成绩；
我的思路：不会写
答案思路：需要知道向sc表中插入的数据都是从哪里来的，然后再考虑怎么去组合这些数据
学号是从学生表中来（排除那些有成绩的），cno是常量，平均分是求出来的，也是常量
17、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
我的思路：使用聚合函数直接查询
具体sql如下：select a.cno, max(score), min(score) from sc a group by a.cno
答案思路：跟我一样
18、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
我的思路：找到这三个数据然后排序
具体sql如下：select a.cno,
round(avg(a.score), 2) avgsc,
(select count(b.score) from sc b where b.score>60 and b.cno = a.cno)/ count(a.score) bili
from sc a
group by a.cno
order by avgsc,bili desc
答案思路：基本跟我相同，但是其找及格率的时候使用casewhen
19、查询不同老师所教不同课程平均分从高到低显示
我的思路：最开始题意理解错误，因为我想到了一种情况，就是一门课可以有不同的老师去教，
但是在本练习中没有存在这样的情况，所以本题就相当于是简单地按科目分类，按平均分排序而已
具体sql如下：select a.cno, avg(a.score), max(b.tno), max(c.tname)
from sc a
join course b
on a.cno = b.cno
join teacher c
on b.tno = c.tno
group by a.cno
order by avg(a.score) desc
答案思路：只是采用的连接方式不同而已
20、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
我的思路：本来以为这是一个很简单的casewhen应用，但是我之前没有使用过casewhen计数，所以出错了，
百度了一下，有两种写法，可以体会一下这两种写法的不同。
具体sql如下：select a.cno,
max(a.cname),
count(case when b.score>85 then 1 else null end),
count(case when b.score <= 85 and b.score>70 then 1 else null end) ,
count(case when b.score <= 70 and b.score>60 then 1 else null end) ,
count(case when b.score <60 then 1 else null end)
from course a
join sc b
on a.cno = b.cno
group by a.cno;
select a.cno,
max(a.cname),
sum(case when b.score>85 then 1 else 0 end),
sum(case when b.score <= 85 and b.score>70 then 1 else 0 end) ,
sum(case when b.score <= 70 and b.score>60 then 1 else 0 end) ,
sum(case when b.score <60 then 1 else 0 end)
from course a
join sc b
on a.cno = b.cno
group by a.cno;
答案思路：使用的是sum，也是这么个思路
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
我的思路：不会做，按班级和成绩排序之后不知道该怎么办了。
答案思路：使用分区函数partition by和row_number函数。其中row_number函数和rank()函数的区别在于
是否考虑了相同分数学生的排名。
22、查询每门课程被选修的学生数
我的思路：很简单的一个sql，关联聚合就行了
具体sql如下：select a.cno, count(b.score)
from course a
left join sc b
on a.cno = b.cno
group by a.cno
order by a.cno
答案思路：答案比我还简单，直接从sc里面去查
23、查询出只选修了一门课程的全部学生的学号和姓名
我的思路：简单的sql，关联聚合即可
具体sql如下：select a.sno, a.sname ,count(b.score)
from student a
left join sc b
on a.sno = b.sno
group by a.sno ,a.sname
having count(b.score) = 1
答案思路：差不多跟我一样，但是写的比我好
24、查询男生、女生人数
我的思路：分两种写法，横着写和竖着写。
具体sql如下：select count(case a.ssex
when '男' then
1
else
null
end) as 男,
count(case a.ssex
when '女' then
1
else
null
end) as 女
from student a;
select a.ssex,count(*) from student a group by a.ssex;
答案思路：是我的第二种写法
25、查询姓“张”的学生名单
select * from student a where a.sname like '张%'
26、查询同名同性学生名单，并统计同名人数
我的思路：也是两种写法，一种直接按名字分组，找大于1的；另外一种是使用连接找名字相同但是学号不同的
具体sql如下：
select a.sname, count(a.sno)/2
from student a
join student b
on a.sno != b.sno
and a.sname = b.sname
group by a.sname;
select a.sname,count(a.sno) from student a group by a.sname having count(a.sno) >1
答案思路：是我下面这个sql的思路
27、1981 年出生的学生名单(注：Student 表中Sage 列的类型是number)
我的思路：这个题主要考察对日期的获取
具体sql如下：
select * from student a where a.sage=(to_number(to_char(sysdate,'yyyy'))-1981)
答案思路：跟我差不多（oracle中应该有自动类型转换机制，因为就算我上面的sql中没有to_number那一步，也能执行且正确）
28、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
我的思路：使用课程表左连接成绩表，然后按课程号分组，按成绩和课程号排序即可。
具体sql：select a.cno, avg(b.score)
from course a
left join sc b
on a.cno = b.cno
group by a.cno
order by avg(b.score), a.cno desc
答案思路：只使用了成绩表
29、查询平均成绩大于85 的所有学生的学号、姓名和平均成绩
我的思路：连接聚合即可
具体sql如下：select a.sno, max(b.sname), avg(a.score)
from sc a
join student b
on a.sno = b.sno
group by a.sno
having avg(a.score) > 85
答案思路：跟我差不多
30、查询课程名称为“数据库”，且分数低于60 的学生姓名和分数
我的思路：简单sql
具体sql如下：select b.cname, a.score
from sc a
join course b
on a.cno = b.cno
where b.cname = 'SSH'
and a.score < 60
答案思路：差不多一样
31、查询所有学生的选课情况；
我的思路：以分数表为中介，连接学生表和课程表进行查询
具体sql如下；select a.sno, a.sname, b.cno, c.cname
from student a
left join sc b
on a.sno = b.sno
left join course c
on b.cno = c.cno
order by a.sno
答案思路：采用的连接方式跟我不一样，使用sc表作为中介的方式跟我一样。（现在我怎么老是想用左连接啊）
32、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；
我的思路：简单的三表联查
具体sql如下：select c.sname, b.cname, a.score
from sc a
join course b
on a.cno = b.cno
join student c
on a.sno = c.sno
where a.score > 70
答案思路：跟我的连接方式不同而已
33、查询不及格的课程，并按课程号从大到小排列
我的思路：这个题目没说清楚
具体sql如下：select * from sc a where a.score<60 order by a.cno desc
34、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；
我的思路：简单查询
具体sql如下：select a.sno, b.sname, a.score
from sc a
join student b
on a.sno = b.sno
where a.score > 80
and a.cno = 'c001'
答案思路： 跟我一样
35、求选了课程的学生人数
sql：select count(distinct sno) from sc
36、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
我的思路：这个题目需要到四个表中去取数据，要找最高的成绩，就需要用到排列，就是之前所学的rank或者row_number
按学科分集合，按分数排列，取第一个
具体sql如下：select *
from (select e.sname,
a.score,
a.cno,
row_number() over(partition by a.cno order by a.score desc) seq
from sc a
join course b
on a.cno = b.cno
join teacher c
on b.tno = c.tno
join student e
on a.sno = e.sno
where c.tname = '谌燕')
where seq = 1
答案思路：找到最高分数，以最高分数作为条件去找学生
37、查询各个课程及相应的选修人数
我的思路：简单的左连接查询
具体sql如下：select a.cno, count(b.sno)
from course a
left join sc b
on a.cno = b.cno
group by a.cno
order by a.cno
答案思路：简单查询
38、查询不同课程成绩相同的学生的学号、课程号、学生成绩
我的思路：自连接加条件即可
具体sql如下：select a.*
from sc a, sc b
where a.sno = b.sno
and a.cno != b.cno
and a.score = b.score
答案思路：跟我一样
39、查询每门功课成绩最好的前两名
我的思路：使用partitionby分组
具体sql：select *
from (select a.*,
row_number() over(partition by a.cno order by a.score desc) seq
from sc a) b
where b.seq <= 2
答案思路：跟我一样
40、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
我的思路：连接分组排序
具体sql：select a.cno, count(b.sno)
from course a
left join sc b
on a.cno = b.cno
group by a.cno
having count(b.sno) > 0
order by count(b.sno) desc, a.cno
答案思路：给我一样
41、检索至少选修两门课程的学生学号
我的思路：简单分组
具体sql如下：select a.sno from sc a group by a.sno having count(a.cno) >=2
答案思路：跟我一样
42、查询全部学生都选修的课程的课程号和课程名
我的思路：简单连接查询
具体sql如下：select a.cno, max(b.cname)
from sc a
join course b
on a.cno = b.cno
group by a.cno
having count(a.sno) = (select count(*) from student)
答案思路：答案错了
43、查询没学过“谌燕”老师讲授的任一门课程的学生姓名
我的思路：采用反证，将所有学过该老师课的学生都排除
具体sql如下：select *
from student a
where a.sno not in (select distinct b.sno
from sc b
where b.cno in (select c.cno
from course c
join teacher e
on e.tno = c.tno
where e.tname = '谌燕'))
答案思路：思路一样，细节略有出入
44、查询两门以上不及格课程的同学的学号及其平均成绩
我的思路：简单查询
具体sql如下：select a.sno, avg(a.score)
from sc a
group by a.sno
having count(case when a.score < 60 then 1 else null end) > 2
答案思路：答案没我写的好，使用了两次分组
45、检索“c004”课程分数小于60，按分数降序排列的同学学号
我的思路；简单查询
具体sql如下：select *
from sc a
where a.score < 90
and a.cno = 'c004'
order by a.sno desc
46、删除“s002”同学的“c001”课程的成绩
我的思路：直接删除，简单sql
具体sql如下；delete from sc a where a.sno='s002' and a.cno='c001'
答案:
select a.* from
(select * from sc a where a.cno='c001') a,
(select * from sc b where b.cno='c002') b
where a.sno=b.sno and a.score > b.score;
select * from sc a
where a.cno='c001'
and exists(select * from sc b where b.cno='c002' and a.score>b.score
and a.sno = b.sno)
select sno,avg(score) from sc group by sno having avg(score)>60;
select a.*,s.sname from (select sno,sum(score),count(cno) from sc group by sno) a ,student s where a.sno=s.sno
select count(*) from teacher where tname like '刘%';
select a.sno,a.sname from student a
where a.sno
not in
(select distinct s.sno
from sc s,
(select c.*
from course c ,
(select tno
from teacher t
where tname='谌燕')t
where c.tno=t.tno) b
where s.cno = b.cno )
select * from student st where st.sno not in
(select distinct sno from sc s join course c on s.cno=c.cno
join teacher t on c.tno=t.tno where tname='谌燕')
select st.* from sc a
join sc b on a.sno=b.sno
join student st
on st.sno=a.sno
where a.cno='c001' and b.cno='c002' and st.sno=a.sno;
select st.* from student st join sc s on st.sno=s.sno
join course c on s.cno=c.cno
join teacher t on c.tno=t.tno
where t.tname='谌燕'
select * from student st
join sc a on st.sno=a.sno
join sc b on st.sno=b.sno
where a.cno='c002' and b.cno='c001' and a.score < b.score
select st.*,s.score from student st
join sc s on st.sno=s.sno
join course c on s.cno=c.cno
where s.score <60
select stu.sno,stu.sname,count(sc.cno) from student stu
left join sc on stu.sno=sc.sno
group by stu.sno,stu.sname
having count(sc.cno)<(select count(distinct cno)from course)
===================================
select * from student where sno in
(select sno from
(select stu.sno,c.cno from student stu
cross join course c
minus
select sno,cno from sc)
)
===================================
select st.* from student st,
(select distinct a.sno from
(select * from sc) a,
(select * from sc where sc.sno='s001') b
where a.cno=b.cno) h
where st.sno=h.sno and st.sno<>'s001'
select * from sc
left join student st
on st.sno=sc.sno
where sc.sno<>'s001'
and sc.cno in
(select cno from sc
where sno='s001')
update sc c set score=(select avg(c.score) from course a,teacher b
where a.tno=b.tno
and b.tname='谌燕'
and a.cno=c.cno
group by c.cno)
where cno in(
select cno from course a,teacher b
where a.tno=b.tno
and b.tname='谌燕')
select* from sc where sno<>'s001'
minus
(
select* from sc
minus
select * from sc where sno='s001'
)
delete from sc
where sc.cno in
(
select cno from course c
left join teacher t on c.tno=t.tno
where t.tname='谌燕'
)
insert into sc (sno,cno,score)
select distinct st.sno,sc.cno,(select avg(score)from sc where cno='c002')
from student st,sc
where not exists
(select * from sc where cno='c002' and sc.sno=st.sno) and sc.cno='c002';
select cno ,max(score),min(score) from sc group by cno;
select cno,avg(score),sum(case when score>=60 then 1 else 0 end)/count(*)
as 及格率
from sc group by cno
order by avg(score) , 及格率desc
select max(t.tno),max(t.tname),max(c.cno),max(c.cname),c.cno,avg(score) from sc , course c,teacher t
where sc.cno=c.cno and c.tno=t.tno
group by c.cno
order by avg(score) desc
select sc.cno,c.cname,
sum(case when score between 85 and 100 then 1 else 0 end) AS "[100-85]",
sum(case when score between 70 and 85 then 1 else 0 end) AS "[85-70]",
sum(case when score between 60 and 70 then 1 else 0 end) AS "[70-60]",
sum(case when score <60 then 1 else 0 end) AS "[<60]"
from sc, course c
where sc.cno=c.cno
group by sc.cno ,c.cname;
select * from
(select sno,cno,score,row_number()over(partition by cno order by score desc) rn from sc)
where rn<4
select cno,count(sno)from sc group by cno;
select sc.sno,st.sname,count(cno) from student st
left join sc
on sc.sno=st.sno
group by st.sname,sc.sno having count(cno)=1;
select ssex,count(*)from student group by ssex;
select * from student where sname like '张%';
select sname,count()from student group by sname having count()>1;
select sno,sname,sage,ssex from student t where to_char(sysdate,'yyyy')-sage =1988
select cno,avg(score) from sc group by cno order by avg(score)asc,cno desc;
select st.sno,st.sname,avg(score) from student st
left join sc
on sc.sno=st.sno
group by st.sno,st.sname having avg(score)>85;
select sname,score from student st,sc,course c
where st.sno=sc.sno and sc.cno=c.cno and c.cname='Oracle' and sc.score<60
select st.sno,st.sname,c.cname from student st,sc,course c
where sc.sno=st.sno and sc.cno=c.cno;
select st.sname,c.cname,sc.score from student st,sc,course c
where sc.sno=st.sno and sc.cno=c.cno and sc.score>70
select sc.sno,c.cname,sc.score from sc,course c
where sc.cno=c.cno and sc.score<60 order by sc.cno desc;
select st.sno,st.sname,sc.score from sc,student st
where sc.sno=st.sno and cno='c001' and score>80;
select count(distinct sno) from sc;
select st.sname,score from student st,sc ,course c,teacher t
where
st.sno=sc.sno and sc.cno=c.cno and c.tno=t.tno
and t.tname='谌燕' and sc.score=
(select max(score)from sc where sc.cno=c.cno)
select cno,count(sno) from sc group by cno;
select a.* from sc a ,sc b where a.score=b.score and a.cno<>b.cno
select * from (
select sno,cno,score,row_number()over(partition by cno order by score desc) my_rn from sc t
)
where my_rn<=2
select cno,count(sno) from sc group by cno
having count(sno)>10
order by count(sno) desc,cno asc;
select sno from sc group by sno having count(cno)>1;
||
select sno from sc group by sno having count(sno)>1;
select distinct(c.cno),c.cname from course c ,sc
where sc.cno=c.cno
||
select cno,cname from course c
where c.cno in
(select cno from sc group by cno)
select st.sname from student st
where st.sno not in
(select distinct sc.sno from sc,course c,teacher t
where sc.cno=c.cno and c.tno=t.tno and t.tname='谌燕')
select sno,avg(score)from sc
where sno in
(select sno from sc where sc.score<60
group by sno having count(sno)>1
) group by sno
select sno from sc where cno='c004' and score<90 order by score desc;
delete from sc where sno='s002' and cno='c001';