6

我正在开发一个相对庞大的应用程序，其中包含很多表。我必须编写一个 SQL 查询，简化后涉及 5 个表（请参阅连接的 jpg）。

这个想法如下：人有地址，地址有一个类型（私人，专业等）和一个国家。人们也可以有选择。这些选项（在选项表中以 name_id 说明）可以链接到地址类型。

这个想法是提取所有拥有一个或多个地址的人，这些地址由一个国家指定，并且他们也出现在 [option address] 表中。

例如，假设我们想要地址为 country_id=1 的人。结果集必须排除那些没有与其选项相关联的相同地址类型的人。

嗯......我不确定我是否了解自己:)

但无论如何，这里是创建所有东西的 SQL。

CREATE TABLE `address` (
  `person_id` int(11) NOT NULL,
  `type_id` int(11) NOT NULL,
  `country_id` int(11) NOT NULL,
  UNIQUE KEY `apt` (`person_id`,`type_id`),
  KEY `apid` (`person_id`),
  KEY `atid` (`type_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `address` (`person_id`, `type_id`, `country_id`) VALUES
(1, 1, 1),
(2, 2, 1),
(3, 1, 1),
(3, 2, 2),
(5, 1, 2),
(6, 2, 1),
(7, 1, 1),
(7, 2, 2),
(8, 1, 1),
(9, 2, 1);

CREATE TABLE `address_type` (
  `id` int(11) NOT NULL,
  UNIQUE KEY `tid` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `address_type` (`id`) VALUES
(1),
(2);

CREATE TABLE `option` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name_id` int(11) NOT NULL,
  `person_id` int(11) NOT NULL,
  UNIQUE KEY `oid` (`id`),
  UNIQUE KEY `onp` (`name_id`,`person_id`),
  KEY `opid` (`person_id`),
  KEY `on` (`name_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=9 ;

INSERT INTO `option` (`id`, `name_id`, `person_id`) VALUES
(1, 1, 1),
(2, 1, 2),
(3, 1, 3),
(4, 1, 5),
(5, 1, 6),
(6, 1, 7),
(7, 1, 8),
(8, 1, 9);

CREATE TABLE `option_address_type` (
  `option_id` int(11) NOT NULL,
  `type_id` int(11) NOT NULL,
  UNIQUE KEY `ot` (`option_id`,`type_id`),
  KEY `ooid` (`option_id`),
  KEY `otid` (`type_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `option_address_type` (`option_id`, `type_id`) VALUES
(1, 1),
(2, 2),
(3, 1),
(3, 2),
(4, 2),
(5, 1),
(6, 1),
(7, 1),
(7, 2),
(8, 1),
(8, 2);

CREATE TABLE `person` (
  `id` int(11) NOT NULL,
  UNIQUE KEY `pid` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `person` (`id`) VALUES
(1),
(2),
(3),
(4),
(5),
(6),
(7),
(8),
(9);


ALTER TABLE `address`
  ADD CONSTRAINT `address_ibfk_1` FOREIGN KEY (`person_id`) REFERENCES `person` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
  ADD CONSTRAINT `address_ibfk_2` FOREIGN KEY (`type_id`) REFERENCES `address_type` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;

ALTER TABLE `option`
  ADD CONSTRAINT `option_ibfk_1` FOREIGN KEY (`person_id`) REFERENCES `person` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;

ALTER TABLE `option_address_type`
  ADD CONSTRAINT `option_address_type_ibfk_1` FOREIGN KEY (`option_id`) REFERENCES `option` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
  ADD CONSTRAINT `option_address_type_ibfk_2` FOREIGN KEY (`type_id`) REFERENCES `address_type` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;
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2 回答 2

3

那这个呢：

select person_id
from   address adr
,      `option` opt
,      option_address_type opt_adt
where  adr.country_id = 1
and    opt.person_id = adr.person_id
and    opt_adt.option_id = opt.option_id
and    opt_adt.type_id = adr.type_id

或者

select person_id
from   address adr
inner join   `option` opt
  on     opt.person_id = adr.person_id
  and    adr.country_id = 1
inner join    option_address_type opt_adt
  on      opt_adt.option_id = opt.option_id
  and    opt_adt.type_id = adr.type_id
于 2013-01-31T19:40:04.447 回答
2

我不确定我是否完全理解您想要的结果。当国家为1时，排除地址类型不等于选项地址类型的人？

像这样的东西？

SELECT p.id
FROM Person p
    JOIN address a ON p.Id = a.Person_ID
    JOIN address_type at ON a.Type_ID = at.Id
    JOIN `option` o ON p.Id = o.person_Id
    JOIN option_address_type oat ON o.id = oat.option_id
WHERE a.country_id = 1 AND at.id <> oat.type_id;

还有SQL 小提琴。

祝你好运。

于 2013-01-31T20:03:22.817 回答