Oracle数据库实验四查询实验三（Oracle 11g）

实验要求：

1、在fruits表和suppliers表之间使用等值连接查询。（ s_id ，s_name，f_name, f_price ），

2、在fruits表和suppliers表之间，使用INNER JOIN语法进行内连接查询（ s_id ，s_name，f_name, f_price ），

SELECT suppliers.s_id, s_name,f_name, f_price FROM fruits INNER JOIN suppliers ON fruits.s_id = suppliers.s_id;

3、．LEFT JOIN左连接

首先创建表orders，SQL语句如下：

CREATE TABLE orders

(

o_num number(9) NOT NULL,

o_date date NOT NULL,

c_id number(9) NOT NULL,

PRIMARY KEY (o_num)

) ;

插入需要演示的数据，SQL语句如下：

INSERT INTO orders(o_num, o_date, c_id)VALUES(30001, '01-9月-2008', 10001);

INSERT INTO orders(o_num, o_date, c_id)VALUES (30002, '12-9月-2008', 10003) ;

INSERT INTO orders(o_num, o_date, c_id)VALUES (30003, '30-9月-2008', 10004) ;

INSERT INTO orders(o_num, o_date, c_id)VALUES (30004, '03-10月-2008', 10005) ;

INSERT INTO orders(o_num, o_date, c_id)VALUES (30005, '08-10月-2008', 10001) ;

在customers表和orders表中，查询所有客户，包括没有订单的客户

4、RIGHT JOIN右连接

在customers表和orders表中，查询所有订单，包括没有客户的订单

5、复合条件连接查询

在customers表和orders表中，使用INNER JOIN语法查询customers表中ID为10001的客户的订单信息

SELECT customers.c_id, orders.o_num FROM customers INNER JOIN orders ON customers.c_id = orders.c_id AND customers.c_id = 10001;

6、在fruits表和suppliers表之间，使用INNER JOIN语法进行内连接查询，并对查询结果排序

> SELECT suppliers.s_id, s_name,f_name, f_price FROM fruits INNER JOIN suppliers ON fruits.s_id = suppliers.s_id

7、下面定义两个表tb1和tb2：

CREATE table tbl1 ( num1 INT NOT NULL);

CREATE table tbl2 ( num2 INT NOT NULL);

分别向两个表中插入数据：

I INSERT INTO tbl1 values(1);

INSERT INTO tbl1 values(5);

INSERT INTO tbl1 values(13);

INSERT INTO tbl1 values(27);

INSERT INTO tbl2 values(6);

INSERT INTO tbl2 values(14);

INSERT INTO tbl2 values(11);

INSERT INTO tbl2 values(20);

8、返回tbl2表的所有num2列，然后将tbl1中的num1的值与之进行比较，只要大于num2的任何1个值，即为符合查询条件的结果。

9、返回tbl1表中比tbl2表num2 列所有值都大的值

10、在orderitems表中查询f_id为c0的订单号，并根据订单号查询具有订单号的客户c_id

11、在suppliers表中查询s_city等于“Tianjin”的供应商s_id，然后在fruits表中查询所有该供应商提供的水果的种类

12、查询所有价格小于9的水果的信息，查询s_id等于101和103所有的水果的信息，使用UNION连接查询结果

13、查询所有价格小于9的水果的信息，查询s_id等于101和103的所有水果的信息，使用UNION ALL连接查询结果

create table student(

sno varchar2(10) primary key,

sname varchar2(20),

sage number(2),

ssex varchar2(5)

);

create table teacher(

tno varchar2(10) primary key,

tname varchar2(20)

);

create table course(

cno varchar2(10),

cname varchar2(20),

tno varchar2(20),

constraint pk_course primary key (cno,tno)

);

create table sc(

sno varchar2(10),

cno varchar2(10),

score number(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');

commit;

14、查询“c001”课程比“c002”课程成绩高的所有学生的学号；

15、查询平均成绩大于60 分的同学的学号和平均成绩；

16、查询所有同学的学号、姓名、选课数、总成绩；

17、查询姓“刘”的老师的个数

18、查询没学过“谌燕”老师课的同学的学号、姓名；

代码如下：

select fruits.s_id,suppliers.s_name,fruits.f_name,fruits.f_price

from fruits,suppliers

where fruits.s_id=suppliers.s_id;

SELECT suppliers.s_id, s_name,f_name, f_price

FROM fruits INNER JOIN suppliers ON fruits.s_id=suppliers.s_id;

select customers.c_id

from customers left outer join orders on (customers.c_id=orders.c_id);

select orders.o_num

from customers right outer join orders on (customers.c_id=orders.c_id);

SELECT customers.c_id, orders.o_num

FROM customers INNER JOIN orders ON customers.c_id = orders.c_id AND customers.c_id = 10001;

SELECT suppliers.s_id, s_name,f_name, f_price

FROM fruits INNER JOIN suppliers ON fruits.s_id = suppliers.s_id;

CREATE table tbl1 ( num1 INT NOT NULL);

CREATE table tbl2 ( num2 INT NOT NULL);

INSERT INTO tbl1 values(1);

INSERT INTO tbl1 values(5);

INSERT INTO tbl1 values(13);

INSERT INTO tbl1 values(27);

INSERT INTO tbl2 values(6);

INSERT INTO tbl2 values(14);

INSERT INTO tbl2 values(11);

INSERT INTO tbl2 values(20);

select num1

from tbl1

where num1>any(

select num2

from tbl2

);

select num1

from tbl1

where num1>all(

select num2

from tbl2

);

10.

select c_id

from orders

where o_num in

(

select o_num

from orderitems

where f_id='c0'

);

11.

select f_id

from fruits

where s_id in

(

select s_id

from suppliers

where s_city='Tianjin' or s_city='tianjin'

);

12.

select *

from fruits

where f_price<9

union

select *

from fruits

where s_id=101 and s_id=103;

13.

select *

from fruits

where f_price<9

union all

select *

from fruits

where s_id=101 and s_id=103;

14.

select a.sno

from sc a,sc b

where a.sno=b.sno and a.cno='c001' and b.cno='c002' and a.score>b.score;

15.

select sno,avg(score)

from sc

group by sno

having avg(score)>60;

16.

select a.sno,b.sname,count(a.cno),sum(a.score)

from student b left outer join sc a on a.sno=b.sno

group by a.sno,b.sname;

17.

select count(tname)

from teacher

where tname like '刘%';

18.

select sno,sname

from student

where sno in

(

select sno

from sc

where cno in

(

select cno

from course

where tno in

(

select tno

from teacher

where tname!='谌燕'

)

);