如何根据第三张表过滤左连接的结果答案

【问题标题】：How to filter results of left join based on third table如何根据第三张表过滤左连接的结果
【发布时间】：2016-01-13 23:09:46
【问题描述】：

在这里扩展这个问题：Efficient way to select records missing in another table，假设我有以下 3 个表（原始示例，请参见下面更好的示例）：

request: (id, name)
response: (id, request_id, provider_id, resource)
provider: (id, name)

首先，我想检索所有不存在结果的请求。这很容易，使用LEFT JOIN。我也想要任何尚未针对给定provider 执行的requests。 简单的英语：我想要所有没有结果的请求或只有其他提供商的结果。

这是一个更好的例子：

customer: (id, email)
vehicle: (id, modelname)
testdrive: (id, vehicle_id REFERENCES vehicle(id), customer_id REFERENCES customer(id)

假设 Bob、Alice、Mary、Fred 和 Joe 拜访一家汽车经销商。鲍勃和爱丽丝试驾凯美瑞，鲍勃和玛丽试驾雅阁，弗雷德和乔不试驾任何东西。经销商想列出两个清单：

谁没有试驾凯美瑞？玛丽、弗雷德、乔
谁没有试驾雅阁？爱丽丝、弗雷德、乔

这是一个 SQL Fiddle 演示：http://sqlfiddle.com/#!15/80d6a

以下是建议的，但它并不完全正确（参见上面的 SQL Fiddle 链接），因为它也给了我试驾其他车型的人：

SELECT c.email
FROM customer AS c
    LEFT JOIN testdrive AS t ON c.id=t.customer_id
    LEFT JOIN vehicle AS v ON t.vehicle_id=v.id AND v.modelname='Camry'
WHERE v.id IS NULL;

【问题讨论】：

标签： sql database join

【解决方案1】：

试试这个：

SQL Fiddle

PostgreSQL 9.3 架构设置：

BEGIN;

-- DROP TABLE IF EXISTS customer;
-- DROP TABLE IF EXISTS vehicle;
-- DROP TABLE IF EXISTS testdrive;

CREATE TABLE customer (
    id SERIAL PRIMARY KEY,
    email TEXT
);

CREATE TABLE vehicle (
    id SERIAL PRIMARY KEY,
    modelname TEXT
);

CREATE TABLE testdrive (
    id SERIAL PRIMARY KEY,
    vehicle_id INTEGER REFERENCES vehicle(id),
    customer_id INTEGER REFERENCES customer(id)
);

INSERT INTO customer (email) VALUES
    ('Bob@foo.org'), ('Alice@bar.net'), ('Mary@baz.com'),
    ('Fred@int.edu'), ('Joe@mut.var');

INSERT INTO vehicle (modelname) VALUES ('Camry'), ('Accord');

INSERT INTO testdrive (vehicle_id, customer_id)
VALUES
    (1, 1), -- Camry, Bob
    (1, 2), -- Camry, Alice
    (2, 1), -- Accord, Bob
    (2, 3); -- Accord, Mary

-- Fred and Joe never test drove anything.
-- Mary didn't test drive the Camry.
-- Alice didn't test drive the Accord.

-- How do I query to find the list of customers who didn't test
-- drive anything, for each vehicle model?

-- Customers who didn't test drive the Camry:
-- Mary, Fred, Joe
SELECT c.email
FROM customer AS c
    LEFT JOIN testdrive AS t ON c.id=t.customer_id
    LEFT JOIN vehicle AS v ON t.vehicle_id=v.id AND v.modelname='Camry'
WHERE v.id IS NULL;

-- Customers who didn't test drive the Accord:
-- Alice, Fred, Joe

COMMIT;

查询 1：

SELECT *
FROM customer c
WHERE NOT EXISTS 
( SELECT * 
  FROM testdrive t 
  INNER JOIN vehicle v 
     on v.id = t.vehicle_id
  WHERE v.modelname = 'Camry' AND t.customer_id = c.id)

Results：

| id |        email |
|----|--------------|
|  3 | Mary@baz.com |
|  4 | Fred@int.edu |
|  5 |  Joe@mut.var |

查询 2：

SELECT *
FROM customer c
WHERE NOT EXISTS 
( SELECT * 
  FROM testdrive t 
  INNER JOIN vehicle v 
     on v.id = t.vehicle_id
  WHERE v.modelname = 'Accord' AND t.customer_id = c.id)

Results：

| id |         email |
|----|---------------|
|  2 | Alice@bar.net |
|  4 |  Fred@int.edu |
|  5 |   Joe@mut.var |

【讨论】：

【解决方案2】：

SELECT c.email
FROM  customer AS c
WHERE NOT c.customer_ID IN (SELECT customer_id FROM testdrive WHERE vehicle_id=@Camry);

【讨论】：