选择匹配MySql中一对多关系右侧的行

Question

我有 4 张桌子。一份用于公司，一份用于产品，一份用于公司地址，一份用于公司董事。

products、director 和 address 表与 company 表是一对多的关系。

所以一家公司可以有很多产品、很多地址和很多董事。

CREATE TABLE IF NOT EXISTS `companies` (
  `company_id` int(11) NOT NULL AUTO_INCREMENT,
  `company_name` varchar(50) NOT NULL,
  PRIMARY KEY (`company_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

CREATE TABLE IF NOT EXISTS `products` (
  `product_id` int(11) NOT NULL AUTO_INCREMENT,
  `company_id` int(11) NOT NULL,
  `product` varchar(50) NOT NULL,
  PRIMARY KEY (`product_id`),
  KEY `company_id` (`company_id`),
  KEY `product` (`product`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;

CREATE TABLE IF NOT EXISTS `directors` (
  `director_id` int(11) NOT NULL AUTO_INCREMENT,
  `company_id` int(11) NOT NULL,
  `surname` varchar(100) NOT NULL,
  `dob` date NOT NULL,
  PRIMARY KEY (`director_id`),
  KEY `company_id` (`company_id`),
  KEY `surname` (`surname`),
  KEY `dob` (`dob`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;

CREATE TABLE IF NOT EXISTS `addresses` (
  `address_id` int(11) NOT NULL AUTO_INCREMENT,
  `company_id` int(1) NOT NULL,
  `postcode` varchar(10) NOT NULL,
  PRIMARY KEY (`address_id`),
  KEY `company_id` (`company_id`),
  KEY `postcode` (`postcode`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=11 ;

INSERT INTO `companies` (`company_id`, `company_name`) VALUES
(1, 'Honda'),
(2, 'Toyota');

INSERT INTO `products` (`product_id`, `company_id`, `product`) VALUES
(1, 1, 'Civic'),
(2, 1, 'Accord'),
(3, 2, 'Corolla'),
(4, 2, 'Prius'),
(5, 1, 'CRV');

INSERT INTO `directors` (`director_id`, `company_id`, `surname`, `dob`)     VALUES
(1, 1, 'Jones', '1990-09-09'),
(2, 1, 'Smith', '1980-08-08'),
(3, 2, 'Lucas', '1970-07-07'),
(4, 1, 'Kelly', '1960-06-06'),
(5, 2, 'Monty', '1950-05-05');

INSERT INTO `addresses` (`address_id`, `company_id`, `postcode`) VALUES
(6, 1, '12345'),
(7, 2, '23456'),
(8, 1, '34567'),
(9, 2, '45678'),
(10, 1, '56789');

我正在尝试编写一个有效的查询（使用 MySql / PDO）来为匹配董事（姓和 dob）和地址（邮政编码）的公司查找产品。

我只想每行列出一个匹配的产品，而不是单独列出每个董事或邮政编码。

到目前为止，我有以下查询，这似乎有效，但它很丑陋，我怀疑在速度和效率方面这是一种荒谬的方法。

SELECT product
FROM products p
LEFT JOIN companies c USING(company_id)
WHERE :lname IN ( 
    SELECT surname 
    FROM directors d 
    WHERE c.company_id = d.company_id )
AND :dob IN ( 
    SELECT dob 
    FROM directors d 
    WHERE c.company_id = d.company_id )
AND :postcode IN ( 
    SELECT postcode 
    FROM addresses a 
    WHERE c.company_id = a.company_id )

提前感谢您的帮助。

Answer 1

至少，directors 上的两个子查询可以通过使用exists 运算符而不是in 重写它们来统一。为了更好地衡量，我用这个运算符重写了整个查询，尽管这不是绝对必要的：

SELECT    product
FROM      products p
LEFT JOIN companies c USING(company_id)
WHERE     EXISTS (SELECT * 
                  FROM directors d 
                  WHERE c.company_id = d.company_id AND 
                        (:lname  = d.lanme OR :dob = d.dob)) AND
           EXISTS (SELECT * 
                   FROM addresses a 
                   WHERE c.company_id = a.company_id AND :postcode = a.postcode)

Answer 2

完全不确定为什么需要子查询？

SELECT p.product FROM products p
INNER JOIN companies c USING(company_id)
INNER JOIN directors d ON d.company_id = c.company_id AND d.surname = 'Jones' AND d.dob = '1990-09-09'
INNER JOIN addresses a ON a.company_id = c.company_id AND a.postcode = '12345'

或者

SELECT p.product FROM products p
INNER JOIN companies c USING(company_id)
INNER JOIN directors d USING(company_id) 
INNER JOIN addresses a USING(company_id)
WHERE d.surname = 'Jones'
AND d.dob = '1990-09-09'
AND a.postcode = '12345'

如果您对这两个查询执行 EXPLAIN，您会发现它们在内部最终是相同的。