数据库编程基本练习题

1、用一条SQL语句查询出每门课都大于80分的学生姓名

准备数据的sql代码：

create table score(
id int primary key auto_increment,
name varchar(20),
subject varchar(20),
score int);

insert into score values
(null,'张三','语文',81),
(null,'张三','数学',75),
(null,'李四','语文',76),
(null,'李四','数学',90),
(null,'王五','语文',81),
(null,'王五','数学',100),
(null,'王五 ','英语',90);

答案：

select distinct name from score where name not in (select distinct name from score where score<=80)

2、每个月份的发生额都比101科目多的科目

请用SQL语句实现：从TestDB数据表中查询出所有月份的发生额都比101科目相应月份的发生额高的科目。请注意：TestDB中有很多科目，都有1－12月份的发生额。 AccID：科目代码，Occmonth：发生额月份，DebitOccur：发生额。数据库名：JcyAudit，数据集：Select * from TestDB

准备数据的sql代码：

drop table if exists TestDB;
create table TestDB(
id int primary key auto_increment,
AccID varchar(20), 
Occmonth date, 
DebitOccur bigint);

insert into TestDB values
(null,'101','1988-1-1',100),
(null,'101','1988-2-1',110),
(null,'101','1988-3-1',120),
(null,'101','1988-4-1',100),
(null,'101','1988-5-1',100),
(null,'101','1988-6-1',100),
(null,'101','1988-7-1',100),
(null,'101','1988-8-1',100);

--复制上面的数据，故意把第一个月份的发生额数字改小一点
insert into TestDB values
(null,'102','1988-1-1',90),
(null,'102','1988-2-1',110),
(null,'102','1988-3-1',120),
(null,'102','1988-4-1',100),
(null,'102','1988-5-1',100),
(null,'102','1988-6-1',100),
(null,'102','1988-7-1',100),
(null,'102','1988-8-1',100);

--复制最上面的数据，故意把所有发生额数字改大一点
insert into TestDB values
(null,'103','1988-1-1',150),
(null,'103','1988-2-1',160),
(null,'103','1988-3-1',180),
(null,'103','1988-4-1',120),
(null,'103','1988-5-1',120),
(null,'103','1988-6-1',120),
(null,'103','1988-7-1',120),
(null,'103','1988-8-1',120);

--复制最上面的数据，故意把所有发生额数字改大一点
insert into TestDB values
(null,'104','1988-1-1',130),
(null,'104','1988-2-1',130),
(null,'104','1988-3-1',140),
(null,'104','1988-4-1',150),
(null,'104','1988-5-1',160),
(null,'104','1988-6-1',170),
(null,'104','1988-7-1',180),
(null,'104','1988-8-1',140);

--复制最上面的数据，故意把第二个月份的发生额数字改小一点
insert into TestDB values
(null,'105','1988-1-1',100),
(null,'105','1988-2-1',80),
(null,'105','1988-3-1',120),
(null,'105','1988-4-1',100),
(null,'105','1988-5-1',100),
(null,'105','1988-6-1',100),
(null,'105','1988-7-1',100),
(null,'105','1988-8-1',100);

答案：

select distinct AccID from TestDB
where AccID not in
(select TestDB.AccIDfrom TestDB,
(select * from TestDB where AccID='101') as db101
where TestDB.Occmonth=db101.Occmonth 
and TestDB.DebitOccur<=db101.DebitOccur
);

3、统计每年每月的信息

year      month      amount
1991       1          1.1  
1991       2          1.2 
1991       3          1.3 
1991       4          1.4 
1992       1          2.1 
1992       2          2.2 
1992       3          2.3 
1992       4          2.4 
查成这样一个结果： 
year   m1      m2      m3      m4 
1991   1.1     1.2     1.3     1.4 
1992   2.1     2.2     2.3     2.4

准备sql语句：

drop table if exists sales;
create table sales(
id int auto_increment primary key,
year varchar(10), 
month varchar(10),
amount float(2,1));

insert into sales values
(null,'1991','1',1.1),
(null,'1991','2',1.2),
(null,'1991','3',1.3),
(null,'1991','4',1.4),
(null,'1992','1',2.1),
(null,'1992','2',2.2),
(null,'1992','3',2.3),
(null,'1992','4',2.4);

答案：

select sales.year,
(select t.amount from sales as t where t.month='1' and t.year = sales.year) as 'm1',
(select t.amount from sales as t where t.month='2' and t.year = sales.year) as 'm2',
(select t.amount from sales as t where t.month='3' and t.year = sales.year) as 'm3',
(select t.amount from sales as t where t.month='4' and t.year = sales.year) as 'm4'
from sales group by year

4、显示文章标题，发帖人、最后回复时间

表：id,title,postuser,postdate,parentid

准备sql语句：

drop table if exists articles;

create table articles(
id int auto_increment primary key,
title varchar(50), 
postuser varchar(10), 
postdate datetime,
parentid int references articles(id)
);

insert into articles values
(null,'第一条','张三','1998-10-10 12:32:32',null),
(null,'第二条','张三','1998-10-10 12:34:32',null),
(null,'第一条回复1','李四','1998-10-10 12:35:32',1),
(null,'第二条回复1','李四','1998-10-10 12:36:32',2),
(null,'第一条回复2','王五','1998-10-10 12:37:32',1),
(null,'第一条回复3','李四','1998-10-10 12:38:32',1),
(null,'第二条回复2','李四','1998-10-10 12:39:32',2),
(null,'第一条回复4','王五','1998-10-10 12:39:40',1);

答案：

select zhu.title,zhu.postuser,
(select max(fu.postdate) from articles as fu where zhu.id = fu.parentid) as postdata
from articles as zhu 
where zhu.parentid is null

5、删除除了id号不同,其他都相同的学生冗余信息

学生表 如下: 
id号         学号          姓名       课程编号       课程名称       分数  
 1         2005001        张三        0001          数学         69  
 2         2005002        李四        0001          数学         89 
 3         2005001        张三        0001          数学         69

准备数据：
create table student2(
id int auto_increment primary key,
code varchar(20),
name varchar(20)
);

insert into student2 values
(null,'2005001','张三'),
(null,'2005002','李四'),
(null,'2005001','张三');

答案：

delete from student2 
where id
not in(select min(id) from (select * from student2) as t group by t.name);

6、航空网的几个航班查询题：

实验环境：
create table city(
cityID int auto_increment primary key,
cityName varchar(20));

create table flight (
flightID int auto_increment primary key,
StartCityID int references city(cityID),
endCityID  int references city(cityID),
StartTime timestamp);

//航班本来应该没有日期部分才好，但是下面的题目当中涉及到了日期
insert into city values
(null,'北京'),
(null,'上海'),
(null,'广州');

insert into flight values
(null,1,2,'9:37:23'),
(null,1,3,'9:37:23'),
(null,1,2,'10:37:23'),
(null,2,3,'10:37:23');

(1) 查询起飞城市是北京的所有航班，按到达城市的名字排序

答案：

select * from flight as f,city as c
where f.endCityID = c.cityID
and StartCityID = (select cityID from city where cityName = "北京")
order by endCityID desc

(2)查询北京到上海的所有航班纪录（起飞城市，到达城市，起飞时间，航班号）

答案：

select c1.cityName as '起飞城市',c2.cityName as '到大城市',f.StartTime as '到大城市',f.flightID as '航班号'
from city as c1,city as c2,flight f
where c1.cityID = f.StartCityID
and c2.cityID = f.endCityID
and c1.cityName = "北京"
and c2.cityName = "上海"

(3)查询具体某一天（2005-5-8）的北京到上海的的航班次数

答案：

select count(*) 
from city as c1,city as c2,flight as f
where c1.cityID =  f.StartCityID
and c2.cityID = f.endCityID
and c1.cityName = "北京"
and c2.cityName = "上海" 
and 查帮助获得的某个日期处理函数(startTime) like '2005-5-8%'

7、查出比经理薪水还高的员工信息：

准备数据：
Drop table if not exists employees;
create table employees(
id int primary key auto_increment,
name varchar(50),
salary int,
managerid int references employees(id));

insert into employees values 
(null,' lhm',10000,null), 
(null,' zxx',15000,1),
(null,'flx',9000,1),
(null,'tg',10000,2),
(null,'wzg',10000,3);

Wzg大于flx,lhm大于zxx

答案：

select e.* from employees as m,employees as e 
where m.id = e.managerid
and m.salary < e.salary

8、求出小于45岁的各个老师所带的大于12岁的学生人数

实验数据：
drop table if exists tea_stu;
drop table if exists teacher;
drop table if exists student;

create table teacher(
teaID int primary key,
name varchar(50),
age int);

create table student(
stuID int primary key,
name varchar(50),
age int);

create table tea_stu(
teaID int references teacher(teaID),
stuID int references student(stuID));

insert into teacher values
(1,'zxx',45), 
(2,'lhm',25) , 
(3,'wzg',26) , 
(4,'tg',27);

insert into student values
(1,'wy',11), 
(2,'dh',25) , 
(3,'ysq',26) , 
(4,'mxc',27);

insert into tea_stu values
(1,1), 
(1,2), 
(1,3);

insert into tea_stu values
(2,2), 
(2,3), 
(2,4);

 insert into tea_stu values
(3,3), 
(3,4), 
(3,1);

insert into tea_stu values
(4,4), 
(4,1), 
(4,2), 
(4,3);

答案：

select teacher.name,count(student.name) as count from teacher,student,tea_stu
where teacher.teaID = tea_stu.teaID
and student.stuID = tea_stu.stuID
and teacher.age < 45
and student.age > 12
group by teacher.name

9、一个用户表中有一个积分字段，假如数据库中有100多万个用户，若要在每年第一天凌晨将积分清零，你将考虑什么，你将想什么办法解决?

alter table drop column score;

alter table add colunm score int;

可能会很快，但是需要试验，试验不能拿真实的环境来操刀，并且要注意，

这样的操作时无法回滚的，在我的印象中，只有inert update delete等DML语句才能回滚，

对于create table,drop table ,alter table等DDL语句是不能回滚。

解决方案一，
update user set score=0;

解决方案二，假设上面的代码要执行好长时间，超出我们的容忍范围，那我就
alter table user drop column score;alter table user add column score int。

下面代码实现每年的那个凌晨时刻进行清零

Runnable runnable =
      new Runnable(){
            public void run(){
                  clearDb();
                  schedule(this,new Date(new Date().getYear()+1,0,0));
            }          
       };
schedule(runnable,new Date(new Date().getYear()+1,0,1));

10、用一条SQL 语句查询出每门课都大于80 分的学生姓名

name   course   grade 
张三       语文       81 
张三       数学       75 
李四       语文       76 
李四       数学       90 
王五       语文       81 
王五       数学      100 
王五       英语       90

答案：

select name from table group by name having min(grade) > 80

11. 现有学生表如下:

自动编号         学号     姓名      课程编号       课程名称       分数 
1            2005001    张三        0001         数学          69 
2            2005002    李四        0001         数学          89 
3            2005001    张三        0001         数学          69

删除除了自动编号不同, 其他都相同的学生冗余信息

答案：

select 自动编号 from 表 where 自动编号 not in 
(select min(自动编号) from 表 group by 学号,姓名,课程编号,课程名称,分数)

12、一个叫 team 的表，里面只有一个字段name, 一共有4 条纪录，分别是a,b,c,d, 对应四个球对，现在四个球对进行比赛，用一条sql 语句显示所有可能的比赛组合

select a.name as '主队',b.name as '客队' from team as a,team as b
where a.name < b.name

13、查询” 01 “课程比” 02 “课程成绩高的学生的信息及课程分数

实验数据：

学生表：
create table Student(
Sid varchar(6), 
Sname varchar(10), 
Sage datetime, 
Ssex varchar(10));
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女')

成绩表
create table SC(
Sid varchar(10),
Cid varchar(10), 
score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98)

课程表
create table Course(
Cid varchar(10),
Cname varchar(10),
Tid varchar(10));
insert into Course values
('01' , '语文' , '02');
insert into Course values
('02' , '数学' , '01');
insert into Course values
('03' , '英语' , '03')

教师表
create table Teacher(
Tid varchar(10),
Tname varchar(10));
insert into Teacher values
('01' , '张三');
insert into Teacher values
('02' , '李四');
insert into Teacher values
('03' , '王五')

14、根据（13题）表查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩

答案：

select s.Sid,s.Sname,avg(score) as a from Student as s,SC
where s.Sid = SC.Sid 
group by s.Sid having a>60

15、根据（13题）表查询在 SC 表存在成绩的学生信息

答案：

select s.* from Student as s,SC where s.Sid = SC.Sid group by Sid

16、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )

答案：

select s.Sid,s.Sname,count(SC.Cid),sum(score) from Student as s left join SC on s.Sid = SC.Sid group by s.Sid

17、查有成绩的学生信息

答案：

select s.Sid,S.Sname,count(score),sum(score),
sum(case when SC.Cid=01 then score else null end) as score_1,
sum(case when SC.Cid=02 then score else null end) as score_2,
sum(case when SC.Cid=03 then score else null end) as score_3
from Student as s,SC,Course where s.Sid = SC.Sid
and SC.Cid = Course.Cid group by s.Sid

18、查询「李」姓老师的数量

答案：

select count(*) from Teacher where Tname like '李%'

19、查询学过「张三」老师授课的同学的信息

select s.* from Student as s,SC,course,Teacher
where s.Sid=SC.Sid 
and Course.Cid=SC.Cid
and Teacher.Tid=Course.Tid
and Teacher.Tname = '张三'
group by s.Sid

20、查询没有学全所有课程的同学的信息

select * from Student where Sid  in 
(select Sid from SC group by Sid having count(cid) < 3)

21、查询至少有一门课与学号为” 01 “的同学所学相同的同学的信息

select s.* from SC,Student as s where Cid in 
(select Cid from SC where Sid = '01')
and s.Sid = SC.Sid
group by Sid

22、查询没学过”张三”老师讲授的任一门课程的学生姓名

select Sname from Student where Sname not in
(select s.Sname from Teacher as t,SC,Course,Student as s
where t.Tid = Course.Tid
and s.Sid=SC.Sid
and SC.Cid=Course.Cid
and t.Tname = '张三')

23、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

select s.Sid,s.Sname,avg(score) from Student as s,SC
where s.Sid = SC.Sid
and SC.score < 60
group by s.Sid 
having count(score > 2)

24、检索” 01 “课程分数小于 60，按分数降序排列的学生信息

select s.*,SC.score from Student as s,SC 
where Cid = '01'
and s.Sid=SC.Sid 
and score < 60
group by score desc

25、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select Sid,
sum(case when Cid=01 then score else null end) as score_01,
sum(case when Cid=02 then score else null end) as score_02,
sum(case when Cid=03 then score else null end) as score_03,
avg(score)
from SC group by Sid
order by avg(score) desc

26、查询各科成绩最高分、最低分和平均分，以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率(及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90）。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 c.cid as 课程号, c.cname as 课程名称, count(*) as 选修人数,
max(score) as 最高分, min(score) as 最低分, avg(score) as 平均分,
sum(case when score >= 60 then 1 else 0 end)/count(*) as 及格率,
sum(case when score >= 70 and score < 80 then 1 else 0 end)/count(*) as 中等率,
sum(case when score >= 80 and score < 90 then 1 else 0 end)/count(*) as 优良率,
sum(case when score >= 90 then 1 else 0 end)/count(*) as 优秀率
from sc, course c
where c.cid = sc.cid
group by c.cid
order by count(*) desc, c.cid asc

27、统计各科成绩各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0] 及所占百分比

select c.Cid as 课程编号, c.Cname as 课程名称, A.*
from course as c,
(select Cid,
sum(case when score >= 85 then 1 else 0 end)/count(*) as 100_85,
sum(case when score >= 70 and score < 85 then 1 else 0 end)/count(*) as 85_70,
sum(case when score >= 60 and score < 70 then 1 else 0 end)/count(*) as 70_60,
sum(case when score < 60 then 1 else 0 end)/count(*) as 60_0
from SC group by Cid) as A
where c.Cid = A.Cid

28、查询出只选修两门课程的学生学号和姓名

select s.Sid,s.Sname,count(sc.Cid) 
from Student as s,SC as sc
where s.Sid = sc.Sid 
group by s.Sid having count(sc.Cid) = 2

29、查询名字中含有「风」字的学生信息

select Sname from Student where Sname like '%风%'

30、查询 1990 年出生的学生名单

select * from Student where Sage like('1990%')

31、成绩不重复，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

select s.* from Student as s,Teacher as t,SC as sc,Course as c
where s.Sid=sc.Sid
and c.Tid=t.Tid
and sc.Cid = c.Cid
and t.Tname='张三'
order by sc.score desc limit 0,1

32、成绩有重复的情况下，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

select s.* from Student as s,SC
where SC.Sid=s.Sid 
and SC.Cid = (select c.Cid from Teacher as t,Course as c
where t.Tid = c.Tid
and t.Tname = '张三')
and SC.score =
(
select sc.score from Student as s,Teacher as t,SC as sc,Course as c
where s.Sid=sc.Sid
and c.Tid=t.Tid
and sc.Cid = c.Cid
and t.Tname='张三'
order by sc.score desc limit 0,1
)

33、查询各学生的年龄，只按年份来算

select Sname,year(now())-year(Sage) as age from Student

34、按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

select Sname,year(now())-year(Sage)-1 as age from Student

35、查询本周过生日的学生

select Sname from Student where week(now()) = week(Sage)

36、查询下周过生日的学生

select Sname from Student where (week(now())+1) = week(Sage)

37、查询本月过生日的学生

select Sname from Student where month(now()) = month(Sage)

38、查询下月过生日的学生

select Sname from Student where (month(now())+1) = month(Sage)

39、查询和” 01 “号的同学学习的课程完全相同的其他同学的信息

select * from Student where sid in
(select sid from SC where cid in
(select sc.Cid from SC where sc.Sid='01') 
and sid <>'01'
group by sid 
having count(cid) >= 3
)

posted @ 2019-06-27 19:57 追风的小蚂蚁阅读(3077) 评论(0) 编辑收藏举报

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