题目
给定 3 个表： salesperson， company， orders。
输出所有表 salesperson 中，没有向公司 'RED' 销售任何东西的销售员。
表： salesperson
+----------+------+--------+-----------------+-----------+
| sales_id | name | salary | commission_rate | hire_date |
+----------+------+--------+-----------------+-----------+
| 1 | John | 100000 | 6 | 4/1/2006 |
| 2 | Amy | 120000 | 5 | 5/1/2010 |
| 3 | Mark | 65000 | 12 | 12/25/2008|
| 4 | Pam | 25000 | 25 | 1/1/2005 |
| 5 | Alex | 50000 | 10 | 2/3/2007 |
+----------+------+--------+-----------------+-----------+
表 salesperson 存储了所有销售员的信息。每个销售员都有一个销售员编号 sales_id 和他的名字 name 。
表： company
+---------+--------+------------+
| com_id | name | city |
+---------+--------+------------+
| 1 | RED | Boston |
| 2 | ORANGE | New York |
| 3 | YELLOW | Boston |
| 4 | GREEN | Austin |
+---------+--------+------------+
表 company 存储了所有公司的信息。每个公司都有一个公司编号 com_id 和它的名字 name 。
表： orders
+----------+------------+---------+----------+--------+
| order_id | order_date | com_id | sales_id | amount |
+----------+------------+---------+----------+--------+
| 1 | 1/1/2014 | 3 | 4 | 100000 |
| 2 | 2/1/2014 | 4 | 5 | 5000 |
| 3 | 3/1/2014 | 1 | 1 | 50000 |
| 4 | 4/1/2014 | 1 | 4 | 25000 |
+----------+----------+---------+----------+--------+
表 orders 存储了所有的销售数据，包括销售员编号 sales_id 和公司编号 com_id 。
输出
+------+
| name |
+------+
| Amy |
| Mark |
| Alex |
+------+
根据表 orders 中的订单 '3' 和 '4' ，容易看出只有 'John' 和 'Pam' 两个销售员曾经向公司 'RED' 销售过。
生成数据
CREATE TABLE salesperson (sales_id INT,NAME VARCHAR(25),salary INT,commission_rate INT,hire_date DATE);
INSERT INTO salesperson VALUES(1,'John',100000,6,'2006-04-01');
INSERT INTO salesperson VALUES (2,'Amy',120000,5,'2010-05-01');
INSERT INTO salesperson VALUES (3,'Mark',65000,6,'2008-12-25');
INSERT INTO salesperson VALUES (4,'Pam',25000,25,'2005-01-01');
INSERT INTO salesperson VALUES (5,'Alex',50000,10,'2007-02-03');
CREATE TABLE company(com_id INT, NAME VARCHAR(20),city VARCHAR(20));
INSERT INTO company VALUES (1,'Red','Boston');
INSERT INTO company VALUES (2,'ORANGE','New York');
INSERT INTO company VALUES (3,'YELLO','Boston');
INSERT INTO company VALUES (4,'GREEN','Austin');
CREATE TABLE orders1 (order_id INT,DATE DATE,com_id INT,sales_id INT,amount INT);
INSERT INTO orders1 VALUES (1,'2014-01-01',3,4,100000);
INSERT INTO orders1 VALUES (2,'2014-02-01',4,5,5000);
INSERT INTO orders1 VALUES (3,'2014-03-01',1,1,50000);
INSERT INTO orders1 VALUES (4,'2014-04-01',1,4,25000);
解答
先选出公司 'RED' 的 com_id
select com_id
from company
where name = 'RED'
再查找给 'RED' 的 com_id销售过的 sales_id
select sales_id
from orders
where com_id = (select com_id
from company
where name = 'RED')
再选出不在上边sales_id 的姓名即可
select name
from salesperson
where sales_id not in (select sales_id
from orders
where com_id = (select com_id
from company
where name = 'RED'));
别的方法
方法一、集合差法
用全部的销售员的集合减去向RED销售过的销售员。
SELECT sp.`name`
FROM salesperson AS sp
LEFT JOIN (SELECT sales_id
FROM orders1
WHERE com_id = (SELECT com_id
FROM company
WHERE NAME = 'RED')) tmp
ON sp.`sales_id` = tmp.`sales_id`
WHERE tmp.`sales_id` IS NULL;
方法二、
每个人销售产品给公司。找出每个人业务往来的公司。
select *
FROM
salesperson AS SP LEFT JOIN Orders AS O
ON(SP.sales_id = O.sales_id)
left join Company AS C
ON(C.com_id = O.com_id)
注意：存在某些人没有与任何公司有业务往来。也存在某些公司没有订单。所以用left join保留没有业务的人。
对人分组，统计每个人与red公司相关的订单数。并找出订单数为0的人。
SELECT max(SP.NAME) AS `name`
FROM
salesperson AS SP LEFT JOIN Orders AS O
ON(SP.sales_id = O.sales_id)
left join Company AS C
ON(C.com_id = O.com_id)
GROUP BY SP.sales_id
HAVING SUM(If(C.NAME='RED',1,0)) = 0