数据库系统原理与应用教程(066)—— MySQL 练习题:操作题 71-81(十):连接查询
71、连接查询(1)
数据表:Customers,包含顾客名称:cust_name、顾客 id:cust_id,表中数据如下:
cust_id | cust_name |
cust10 | andy |
cust1 | ben |
cust2 | tony |
cust22 | tom |
cust221 | an |
cust2217 | hex |
数据表:Orders(订单信息表),包含订单号:order_num,顾客 id:cust_id,表中数据如下:
order_num | cust_id |
a1 | cust10 |
a2 | cust1 |
a3 | cust2 |
a4 | cust22 |
a5 | cust221 |
a7 | cust2217 |
【问题】编写 SQL 语句,查询 Customers 表中的顾客名称(cust_name)和 Orders 表中的相关订单号(order_num),并按顾客名称再按订单号对结果进行升序排序。查询结果如下:
cust_name | order_num |
an | a5 |
andy | a1 |
ben | a2 |
hex | a7 |
tom | a4 |
tony | a3 |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `Orders`;
CREATE TABLE IF NOT EXISTS `Orders`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
cust_id VARCHAR(255) NOT NULL COMMENT '顾客id'
);
INSERT `Orders` VALUES ('a1','cust10'),('a2','cust1'),('a3','cust2'),('a4','cust22'),('a5','cust221'),('a7','cust2217');
DROP TABLE IF EXISTS `Customers`;
CREATE TABLE IF NOT EXISTS `Customers`(
cust_id VARCHAR(255) NOT NULL COMMENT '客户id',
cust_name VARCHAR(255) NOT NULL COMMENT '客户姓名'
);
INSERT `Customers` VALUES ('cust10','andy'),('cust1','ben'),('cust2','tony'),('cust22','tom'),('cust221','an'),('cust2217','hex');
*/
解答:
/*
select c.cust_name, o.order_num
from Customers c join Orders o on c.cust_id = o.cust_id
order by c.cust_name, o.order_num;
*/
mysql> select c.cust_name, o.order_num
-> from Customers c join Orders o on c.cust_id = o.cust_id
-> order by c.cust_name, o.order_num;
+-----------+-----------+
| cust_name | order_num |
+-----------+-----------+
| an | a5 |
| andy | a1 |
| ben | a2 |
| hex | a7 |
| tom | a4 |
| tony | a3 |
+-----------+-----------+
6 rows in set (0.00 sec)
72、连接查询(2)
数据表:Customers,包含顾客名称:cust_name,顾客id:cust_id,表中数据如下:
cust_id | cust_name |
cust10 | andy |
cust1 | ben |
cust2 | tony |
cust22 | tom |
cust221 | an |
cust2217 | hex |
数据表:Orders(订单信息表),包含订单号:order_num,顾客id:cust_id,表中数据如下:
order_num | cust_id |
a1 | cust10 |
a2 | cust1 |
a3 | cust2 |
a4 | cust22 |
a5 | cust221 |
a7 | cust2217 |
数据表:OrderItems,包含商品订单号:order_num,订购数量:quantity,单价:item_price,表中数据如下:
order_num | quantity | item_price |
a1 | 1000 | 10 |
a2 | 200 | 10 |
a3 | 10 | 15 |
a4 | 25 | 50 |
a5 | 15 | 25 |
a7 | 7 | 7 |
【问题】查询顾客名称、订单号以及每个订单的总价(OrderTotal),并按顾客名称再按订单号对结果进行升序排序。查询结果如下:
cust_name | order_num | OrderTotal |
an | a5 | 375 |
andy | a1 | 10000 |
ben | a2 | 2000 |
hex | a7 | 49 |
tom | a4 | 1250 |
tony | a3 | 150 |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `Orders`;
CREATE TABLE IF NOT EXISTS `Orders`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
cust_id VARCHAR(255) NOT NULL COMMENT '顾客id'
);
INSERT `Orders` VALUES ('a1','cust10'),('a2','cust1'),('a3','cust2'),('a4','cust22'),('a5','cust221'),('a7','cust2217');
DROP TABLE IF EXISTS `Customers`;
CREATE TABLE IF NOT EXISTS `Customers`(
cust_id VARCHAR(255) NOT NULL COMMENT '客户id',
cust_name VARCHAR(255) NOT NULL COMMENT '客户姓名'
);
INSERT `Customers` VALUES ('cust10','andy'),('cust1','ben'),('cust2','tony'),('cust22','tom'),('cust221','an'),('cust2217','hex');
DROP TABLE IF EXISTS `OrderItems`;
CREATE TABLE IF NOT EXISTS `OrderItems`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
quantity INT(16) NOT NULL COMMENT '商品数量',
item_price INT(16) NOT NULL COMMENT '商品价格'
);
INSERT `OrderItems` VALUES ('a1',1000,10),('a2',200,10),('a3',10,15),('a4',25,50),('a5',15,25),('a7',7,7);
*/
解答:
/*
select cust_name, o.order_num,
sum(oi.quantity * oi.item_price) OrderTotal
from Customers c join Orders o on c.cust_id = o.cust_id
join OrderItems oi on o.order_num = oi.order_num
group by cust_name, o.order_num
order by cust_name, o.order_num;
*/
mysql> select cust_name, o.order_num,
-> sum(oi.quantity * oi.item_price) OrderTotal
-> from Customers c join Orders o on c.cust_id = o.cust_id
-> join OrderItems oi on o.order_num = oi.order_num
-> group by cust_name, o.order_num
-> order by cust_name, o.order_num;
+-----------+-----------+------------+
| cust_name | order_num | OrderTotal |
+-----------+-----------+------------+
| an | a5 | 375 |
| andy | a1 | 10000 |
| ben | a2 | 2000 |
| hex | a7 | 49 |
| tom | a4 | 1250 |
| tony | a3 | 150 |
+-----------+-----------+------------+
6 rows in set (0.00 sec)
73、连接查询(3)
数据表:OrderItems,表中数据如下:
prod_id | order_num |
BR01 | a0001 |
BR01 | a0002 |
BR02 | a0003 |
BR02 | a0013 |
数据表:Orders,表中数据如下:
order_num | cust_id | order_date |
a0001 | cust10 | 2022-01-01 00:00:00 |
a0002 | cust1 | 2022-01-01 00:01:00 |
a0003 | cust1 | 2022-01-02 00:00:00 |
a0013 | cust2 | 2022-01-01 00:20:00 |
【问题】编写 SQL 语句,查询哪些订单购买了 prod_id 为 “BR01” 的产品,显示每个产品对应的顾客 ID(cust_id)和订单日期(order_date),按订购日期对结果进行升序排序。查询结果如下:
cust_id | order_date |
cust10 | 2022-01-01 00:00:00 |
cust1 | 2022-01-01 00:01:00 |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `OrderItems`;
CREATE TABLE IF NOT EXISTS `OrderItems`(
prod_id VARCHAR(255) NOT NULL COMMENT '产品id',
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号'
);
INSERT `OrderItems` VALUES ('BR01','a0001'),('BR01','a0002'),('BR02','a0003'),('BR02','a0013');
DROP TABLE IF EXISTS `Orders`;
CREATE TABLE IF NOT EXISTS `Orders`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
cust_id VARCHAR(255) NOT NULL COMMENT '顾客id',
order_date TIMESTAMP NOT NULL COMMENT '下单时间'
);
INSERT `Orders` VALUES ('a0001','cust10','2022-01-01 00:00:00'),('a0002','cust1','2022-01-01 00:01:00'),('a0003','cust1','2022-01-02 00:00:00'),('a0013','cust2','2022-01-01 00:20:00');
*/
解答:
/*
-- 使用子查询
select cust_id, order_date
from Orders where order_num in
(select order_num from OrderItems where prod_id = 'BR01')
order by order_date;
-- 使用连接查询
select o.cust_id, o.order_date
from Orders o join OrderItems oi on o.order_num = oi.order_num
where oi.prod_id = 'BR01'
order by o.order_date;
*/
-- 使用子查询
mysql> select cust_id, order_date
-> from Orders where order_num in
-> (select order_num from OrderItems where prod_id = 'BR01')
-> order by order_date;
+---------+---------------------+
| cust_id | order_date |
+---------+---------------------+
| cust10 | 2022-01-01 00:00:00 |
| cust1 | 2022-01-01 00:01:00 |
+---------+---------------------+
2 rows in set (0.00 sec)
-- 使用连接查询
mysql> select o.cust_id, o.order_date
-> from Orders o join OrderItems oi on o.order_num = oi.order_num
-> where oi.prod_id = 'BR01'
-> order by o.order_date;
+---------+---------------------+
| cust_id | order_date |
+---------+---------------------+
| cust10 | 2022-01-01 00:00:00 |
| cust1 | 2022-01-01 00:01:00 |
+---------+---------------------+
2 rows in set (0.00 sec)
74、连接查询(4)
数据表:OrderItems,表中数据如下:
prod_id | order_num |
BR01 | a0001 |
BR01 | a0002 |
BR02 | a0003 |
BR02 | a0013 |
数据表:Orders,表中数据如下:
order_num | cust_id | order_date |
a0001 | cust10 | 2022-01-01 00:00:00 |
a0002 | cust1 | 2022-01-01 00:01:00 |
a0003 | cust1 | 2022-01-02 00:00:00 |
a0013 | cust2 | 2022-01-01 00:20:00 |
数据表:Customers,表中数据如下:
cust_id | cust_email |
cust10 | cust10@cust.com |
cust1 | cust1@cust.com |
cust2 | cust2@cust.com |
【问题】查询购买 prod_id 为 “BR01” 的产品的所有顾客的电子邮件。查询结果如下:
cust_email |
cust10@cust.com |
cust1@cust.com |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `OrderItems`;
CREATE TABLE IF NOT EXISTS `OrderItems`(
prod_id VARCHAR(255) NOT NULL COMMENT '产品id',
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号'
);
INSERT `OrderItems` VALUES ('BR01','a0001'),('BR01','a0002'),('BR02','a0003'),('BR02','a0013');
DROP TABLE IF EXISTS `Orders`;
CREATE TABLE IF NOT EXISTS `Orders`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
cust_id VARCHAR(255) NOT NULL COMMENT '顾客id',
order_date TIMESTAMP NOT NULL COMMENT '下单时间'
);
INSERT `Orders` VALUES ('a0001','cust10','2022-01-01 00:00:00'),('a0002','cust1','2022-01-01 00:01:00'),('a0003','cust1','2022-01-02 00:00:00'),('a0013','cust2','2022-01-01 00:20:00');
DROP TABLE IF EXISTS `Customers`;
CREATE TABLE IF NOT EXISTS `Customers`(
cust_id VARCHAR(255) NOT NULL COMMENT '顾客id',
cust_email VARCHAR(255) NOT NULL COMMENT '顾客email'
);
INSERT `Customers` VALUES ('cust10','cust10@cust.com'),('cust1','cust1@cust.com'),('cust2','cust2@cust.com');
*/
解答:
/*
-- 使用子查询
select cust_email from Customers where cust_id in
(select cust_id from Orders where order_num in
(select order_num from OrderItems where prod_id = 'BR01')
);
-- 使用连接查询
select c.cust_email
from Customers c join Orders o on c.cust_id = o.cust_id
join OrderItems oi on o.order_num = oi.order_num
where prod_id = 'BR01';
*/
-- 使用子查询
mysql> select cust_email from Customers where cust_id in
-> (select cust_id from Orders where order_num in
-> (select order_num from OrderItems where prod_id = 'BR01')
-> );
+-----------------+
| cust_email |
+-----------------+
| cust10@cust.com |
| cust1@cust.com |
+-----------------+
2 rows in set (0.00 sec)
-- 使用连接查询
mysql> select c.cust_email
-> from Customers c join Orders o on c.cust_id = o.cust_id
-> join OrderItems oi on o.order_num = oi.order_num
-> where prod_id = 'BR01';
+-----------------+
| cust_email |
+-----------------+
| cust10@cust.com |
| cust1@cust.com |
+-----------------+
2 rows in set (0.00 sec)
75、连接查询(5)
数据表:OrderItems,表中数据如下:
order_num | item_price | quantity |
a1 | 10 | 105 |
a2 | 1 | 1100 |
a2 | 1 | 200 |
a4 | 2 | 1121 |
a5 | 5 | 10 |
a2 | 1 | 19 |
a7 | 7 | 5 |
数据表:Orders,表中数据如下:
order_num | cust_id |
a1 | cust10 |
a2 | cust1 |
a3 | cust2 |
a4 | cust22 |
a5 | cust221 |
a7 | cust2217 |
数据表:Customers,表中数据如下:
cust_id | cust_name |
cust10 | andy |
cust1 | ben |
cust2 | tony |
cust22 | tom |
cust221 | an |
cust2217 | hex |
【问题】编写 SQL 语句,查询订单总价不小于 1000 的客户名称和总额(item_price * quantity),按总额对结果进行排序。查询结果如下:
cust_name | total_price |
andy | 1050 |
ben | 1319 |
tom | 2242 |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `OrderItems`;
CREATE TABLE IF NOT EXISTS `OrderItems`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
item_price INT(16) NOT NULL COMMENT '售出价格',
quantity INT(16) NOT NULL COMMENT '商品数量'
);
INSERT `OrderItems` VALUES ('a1',10,105),('a2',1,1100),('a2',1,200),('a4',2,1121),('a5',5,10),('a2',1,19),('a7',7,5);
DROP TABLE IF EXISTS `Customers`;
CREATE TABLE IF NOT EXISTS `Customers`(
cust_id VARCHAR(255) NOT NULL COMMENT '客户id',
cust_name VARCHAR(255) NOT NULL COMMENT '客户姓名'
);
INSERT `Customers` VALUES ('cust10','andy'),('cust1','ben'),('cust2','tony'),('cust22','tom'),('cust221','an'),('cust2217','hex');
DROP TABLE IF EXISTS `Orders`;
CREATE TABLE IF NOT EXISTS `Orders`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
cust_id VARCHAR(255) NOT NULL COMMENT '顾客id'
);
INSERT `Orders` VALUES ('a1','cust10'),('a2','cust1'),('a3','cust2'),('a4','cust22'),('a5','cust221'),('a7','cust2217');
*/
解答:
/*
select c.cust_name,
sum(oi.item_price * oi.quantity) total_price
from OrderItems oi join Orders o on oi.order_num = o.order_num
join Customers c on o.cust_id = c.cust_id
group by c.cust_name
having total_price >= 1000
order by total_price;
*/
mysql> select c.cust_name,
-> sum(oi.item_price * oi.quantity) total_price
-> from OrderItems oi join Orders o on oi.order_num = o.order_num
-> join Customers c on o.cust_id = c.cust_id
-> group by c.cust_name
-> having total_price >= 1000
-> order by total_price;
+-----------+-------------+
| cust_name | total_price |
+-----------+-------------+
| andy | 1050 |
| ben | 1319 |
| tom | 2242 |
+-----------+-------------+
3 rows in set (0.00 sec)
76、连接查询(6)
数据表:Customers,表中数据如下:
cust_id | cust_name |
cust10 | andy |
cust1 | ben |
cust2 | tony |
cust22 | tom |
cust221 | an |
cust2217 | hex |
数据表:Orders,表中数据如下:
order_num | cust_id |
a1 | cust10 |
a2 | cust1 |
a3 | cust2 |
a4 | cust22 |
a5 | cust221 |
a7 | cust2217 |
【问题】编写 SQL语句,查询每个顾客的名称和所有的订单号,根据顾客姓名 cust_name 升序排序。查询结果如下:
cust_name | order_num |
an | a5 |
andy | a1 |
ben | a2 |
hex | a7 |
tom | a4 |
tony | a3 |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `Customers`;
CREATE TABLE IF NOT EXISTS `Customers`(
cust_id VARCHAR(255) NOT NULL COMMENT '客户id',
cust_name VARCHAR(255) NOT NULL COMMENT '客户姓名'
);
INSERT `Customers` VALUES ('cust10','andy'),('cust1','ben'),('cust2','tony'),('cust22','tom'),('cust221','an'),('cust2217','hex');
DROP TABLE IF EXISTS `Orders`;
CREATE TABLE IF NOT EXISTS `Orders`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
cust_id VARCHAR(255) NOT NULL COMMENT '顾客id'
);
INSERT `Orders` VALUES ('a1','cust10'),('a2','cust1'),('a3','cust2'),('a4','cust22'),('a5','cust221'),('a7','cust2217');
*/
解答:
/*
select c.cust_name, o.order_num
from Customers c join Orders o on c.cust_id = o.cust_id
order by c.cust_name;
*/
mysql> select c.cust_name, o.order_num
-> from Customers c join Orders o on c.cust_id = o.cust_id
-> order by c.cust_name;
+-----------+-----------+
| cust_name | order_num |
+-----------+-----------+
| an | a5 |
| andy | a1 |
| ben | a2 |
| hex | a7 |
| tom | a4 |
| tony | a3 |
+-----------+-----------+
6 rows in set (0.00 sec)
77、连接查询(7)
数据表:Orders,表中数据如下:
order_num | cust_id |
a1 | cust10 |
a2 | cust1 |
a3 | cust2 |
a4 | cust22 |
a5 | cust221 |
a7 | cust2217 |
数据表:Customers,表中数据如下:
cust_id | cust_name |
cust10 | andy |
cust1 | ben |
cust2 | tony |
cust22 | tom |
cust221 | an |
cust2217 | hex |
cust40 | ace |
【问题】检索每个顾客的名称和所有的订单号,列出所有的顾客,即使他们没有下过订单。查询结果如下:
cust_name | order_num |
ace | NULL |
an | a5 |
andy | a1 |
ben | a2 |
hex | a7 |
tom | a4 |
tony | a3 |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `Customers`;
CREATE TABLE IF NOT EXISTS `Customers`(
cust_id VARCHAR(255) NOT NULL COMMENT '客户id',
cust_name VARCHAR(255) NOT NULL COMMENT '客户姓名'
);
INSERT `Customers` VALUES ('cust10','andy'),('cust1','ben'),('cust2','tony'),('cust22','tom'),('cust221','an'),('cust2217','hex'),('cust40','ace');
DROP TABLE IF EXISTS `Orders`;
CREATE TABLE IF NOT EXISTS `Orders`(
order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',
cust_id VARCHAR(255) NOT NULL COMMENT '顾客id'
);
INSERT `Orders` VALUES ('a1','cust10'),('a2','cust1'),('a3','cust2'),('a4','cust22'),('a5','cust221'),('a7','cust2217');
*/
解答:
/*
select c.cust_name, o.order_num
from Customers c left join Orders o on c.cust_id = o.cust_id
order by c.cust_name;
*/
mysql> select c.cust_name, o.order_num
-> from Customers c left join Orders o on c.cust_id = o.cust_id
-> order by c.cust_name;
+-----------+-----------+
| cust_name | order_num |
+-----------+-----------+
| ace | NULL |
| an | a5 |
| andy | a1 |
| ben | a2 |
| hex | a7 |
| tom | a4 |
| tony | a3 |
+-----------+-----------+
7 rows in set (0.00 sec)
78、连接查询(8)
数据表:Products,表中数据为:
prod_id | prod_name |
a0001 | egg |
a0002 | sockets |
a0013 | coffee |
a0003 | cola |
a0023 | soda |
数据表:OrderItems,表中数据为:
prod_id | order_num |
a0001 | a105 |
a0002 | a1100 |
a0002 | a200 |
a0013 | a1121 |
a0003 | a10 |
a0003 | a19 |
a0003 | a5 |
【问题】编写 SQL 语句,查询产品名称(prod_name)和与之相关的订单号(order_num)的列表,并按照产品名称升序排序。查询结果如下:
prod_name | order_num |
coffee | a1121 |
cola | a5 |
cola | a19 |
cola | a10 |
egg | a105 |
sockets | a200 |
sockets | a1100 |
soda | NULL |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `Products`;
CREATE TABLE IF NOT EXISTS `Products` (
`prod_id` VARCHAR(255) NOT NULL COMMENT '产品 ID',
`prod_name` VARCHAR(255) NOT NULL COMMENT '产品名称'
);
INSERT INTO `Products` VALUES ('a0001','egg'),
('a0002','sockets'),
('a0013','coffee'),
('a0003','cola'),
('a0023','soda');
DROP TABLE IF EXISTS `OrderItems`;
CREATE TABLE IF NOT EXISTS `OrderItems`(
prod_id VARCHAR(255) NOT NULL COMMENT '产品id',
order_num VARCHAR(255) NOT NULL COMMENT '商品数量'
);
INSERT `OrderItems` VALUES ('a0001','a105'),('a0002','a1100'),('a0002','a200'),('a0013','a1121'),('a0003','a10'),('a0003','a19'),('a0003','a5');
*/
解答:
/*
select p.prod_name, oi.order_num
from Products p left join OrderItems oi on p.prod_id = oi.prod_id
order by p.prod_name;
*/
mysql> select p.prod_name, oi.order_num
-> from Products p left join OrderItems oi on p.prod_id = oi.prod_id
-> order by p.prod_name;
+-----------+-----------+
| prod_name | order_num |
+-----------+-----------+
| coffee | a1121 |
| cola | a10 |
| cola | a19 |
| cola | a5 |
| egg | a105 |
| sockets | a1100 |
| sockets | a200 |
| soda | NULL |
+-----------+-----------+
8 rows in set (0.00 sec)
79、连接查询(9)
数据表:Products,表中数据如下:
prod_id | prod_name |
a0001 | egg |
a0002 | sockets |
a0013 | coffee |
a0003 | cola |
a0023 | soda |
数据表:OrderItems,表中数据如下:
prod_id | order_num |
a0001 | a105 |
a0002 | a1100 |
a0002 | a200 |
a0013 | a1121 |
a0003 | a10 |
a0003 | a19 |
a0003 | a5 |
【问题】查询产品名称(prod_name)和每一项产品的总订单数,并按产品名称升序排序。查询结果如下:
prod_name | orders |
coffee | 1 |
cola | 3 |
egg | 1 |
sockets | 2 |
soda | 0 |
解答:
/*
select p.prod_name, count(order_num) orders
from Products p left join OrderItems oi on p.prod_id = oi.prod_id
group by p.prod_name
order by p.prod_name;
*/
mysql> select p.prod_name, count(order_num) orders
-> from Products p left join OrderItems oi on p.prod_id = oi.prod_id
-> group by p.prod_name
-> order by p.prod_name;
+-----------+--------+
| prod_name | orders |
+-----------+--------+
| coffee | 1 |
| cola | 3 |
| egg | 1 |
| sockets | 2 |
| soda | 0 |
+-----------+--------+
5 rows in set (0.00 sec)
80、连接查询(10)
数据表:Vendors,表中数据如下:
vend_id |
a0002 |
a0013 |
a0003 |
a0010 |
数据表:Products,表中数据如下:
vend_id | prod_id |
a0001 | egg |
a0002 | prod_id_iphone |
a00113 | prod_id_tea |
a0003 | prod_id_vivo phone |
a0010 | prod_id_huawei phone |
【问题】列出供应商(Vendors 表中的 vend_id)及可供产品的数量(包括没有产品的供应商),根据 vend_id 升序排序。查询结果如下:
vend_id | prod_id |
a0002 | 1 |
a0003 | 1 |
a0010 | 1 |
a0013 | 0 |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `Vendors`;
CREATE TABLE IF NOT EXISTS `Vendors` (
`vend_id` VARCHAR(255) NOT NULL COMMENT 'vend名称'
);
INSERT INTO `Vendors` VALUES ('a0002'),
('a0013'),
('a0003'),
('a0010');
DROP TABLE IF EXISTS `Products`;
CREATE TABLE IF NOT EXISTS `Products` (
`vend_id` VARCHAR(255) NOT NULL COMMENT '产品 ID',
`prod_id` VARCHAR(255) NOT NULL COMMENT '产品名称'
);
INSERT INTO `Products` VALUES ('a0001','egg'),
('a0002','prod_id_iphone'),
('a00113','prod_id_tea'),
('a0003','prod_id_vivo phone'),
('a0010','prod_id_huawei phone');
*/
解答:
/*
select v.vend_id, count(prod_id) prod_id
from Vendors v left join Products p on v.vend_id = p.vend_id
group by v.vend_id
order by v.vend_id;
*/
mysql> select v.vend_id, count(prod_id) prod_id
-> from Vendors v left join Products p on v.vend_id = p.vend_id
-> group by v.vend_id
-> order by v.vend_id;
+---------+---------+
| vend_id | prod_id |
+---------+---------+
| a0002 | 1 |
| a0003 | 1 |
| a0010 | 1 |
| a0013 | 0 |
+---------+---------+
4 rows in set (0.00 sec)
81、union 联合查询
数据表:OrderItems,表中数据如下:
prod_id | quantity |
a0001 | 105 |
a0002 | 100 |
a0002 | 200 |
a0013 | 1121 |
a0003 | 10 |
a0003 | 19 |
a0003 | 5 |
BNBG | 10002 |
【问题】查询数量(quantity)为 100 的产品信息和产品 id(prod_id)以【BNBG】开头的产品信息,按产品 id 对结果进行升序排序。查询结果如下:
prod_id | quantity |
a0002 | 100 |
BNBG | 10002 |
表结构及数据如下:
/*
DROP TABLE IF EXISTS `OrderItems`;
CREATE TABLE IF NOT EXISTS `OrderItems`(
prod_id VARCHAR(255) NOT NULL COMMENT '产品id',
quantity VARCHAR(255) NOT NULL COMMENT '商品数量'
);
INSERT `OrderItems` VALUES ('a0001',105),('a0002',100),('a0002',200),('a0013',1121),('a0003',10),('a0003',19),('a0003',5),('BNBG',10002);
*/
解答:
/*
-- 使用 union
select * from OrderItems where quantity = 100
union
select * from OrderItems where prod_id like 'BNBG%'
order by prod_id;
-- 使用 or
select * from OrderItems
where quantity = 100 or prod_id like 'BNBG%'
order by prod_id;
*/
-- 使用 union
mysql> select * from OrderItems where quantity = 100
-> union
-> select * from OrderItems where prod_id like 'BNBG%'
-> order by prod_id;
+---------+----------+
| prod_id | quantity |
+---------+----------+
| a0002 | 100 |
| BNBG | 10002 |
+---------+----------+
2 rows in set (0.03 sec)
-- 使用 or 构造查询条件
mysql> select * from OrderItems
-> where quantity = 100 or prod_id like 'BNBG%'
-> order by prod_id;
+---------+----------+
| prod_id | quantity |
+---------+----------+
| a0002 | 100 |
| BNBG | 10002 |
+---------+----------+
2 rows in set (0.00 sec)