MySQL查询引用多对多表

Question

假设下表:

area
-id
-title

properties
-id
-title
-area_id

categories
-id
-title

properties_categories
-property_id
-category_id

以下查询非常接近正确答案，但并不正确。您能告诉我正确的查询是什么吗？

select a.id, a.title, count(p.id) 
FROM area AS a,properties AS p, properties_categories AS pc 
WHERE a.id = p.area_id 
  AND pc.category_id IN (1,2,3) 
  AND pc.property_id = p.id 
GROUP BY a.id;

或者

SELECT A.id, A.title, COUNT(B.id)
FROM area A LEFT JOIN  properties B 
ON A.id=B.area_id 
JOIN properties_categories C 
ON C.property_id=B.id
WHERE C.category_id IN (1,2,3)
GROUP BY A.id, A.title;

如果我编写一个查询，其中以下内容仅从区域和属性表中提取，您将得到正确的结果。

SELECT A.id, A.title, COUNT(B.id) FROM area A LEFT JOIN properties B ON A.id=B.area_id GROUP BY A.id, A.title

这是要试验的数据转储:

CREATE TABLE `area` (
  `id` int(10) UNSIGNED NOT NULL,
  `title` varchar(255) COLLATE utf8_unicode_ci NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;


INSERT INTO `area` (`id`, `title`) VALUES
(2, 'Test1'),
(3, 'Test2'),
(4, 'Test3'),
(5, 'Test4'),
(6, 'Test5');


CREATE TABLE `categories` (
  `id` int(10) UNSIGNED NOT NULL,
  `title` varchar(255) COLLATE utf8_unicode_ci NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;


INSERT INTO `categories` (`id`, `title`) VALUES
(1, 'Category A'),
(2, 'Category B'),
(3, 'Category C'),
(4, 'Category D'),
(5, 'Category E'),
(6, 'Category F');


CREATE TABLE `properties` (
  `id` int(11) NOT NULL,
  `name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
  `area_id` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;


INSERT INTO `properties` (`id`, `name`, `area_id`) VALUES
(2, 'Property A', 2),
(3, 'Property B', 2),
(4, 'Property C', 2),
(5, 'Property D', 2),
(7, 'Property E', 2),
(8, 'Property F', 2),
(10, 'Property G', 2),
(11, 'Property H', 2),
(12, 'Property I', 2),
(13, 'Property J', 2),
(14, 'Property K', 2),
(19, 'Property L', 2),
(20, 'Property M', 4),
(21, 'Property O', 2),
(22, 'Property P', 2),
(23, 'Property Q', 2),
(24, 'Property  R', 2),
(27, 'Property S', NULL),
(29, 'Property T', 2),
(30, 'Property U', 2),
(32, 'Property V', 2),
(33, 'Property W', 2),
(34, 'Property X', 5),
(35, 'Property Y', 5),
(36, 'Property Z', 5),
(37, 'Property A1', 5),
(38, 'Property A2', 3),
(39, 'Property A3', 6);

CREATE TABLE `properties_categories` (
  `category_id` int(11) NOT NULL,
  `property_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

INSERT INTO `properties_categories` (`category_id`, `property_id`) VALUES
(2, 2),
(6, 2),
(2, 3),
(6, 3),
(2, 4),
(6, 4),
(2, 5),
(6, 5),
(2, 7),
(6, 7),
(2, 8),
(6, 8),
(2, 10),
(6, 10),
(2, 11),
(6, 11),
(2, 12),
(6, 12),
(2, 13),
(6, 13),
(2, 14),
(6, 14),
(5, 7),
(5, 3),
(5, 14),
(5, 4),
(5, 12),
(5, 11),
(5, 13),
(5, 5),
(5, 10),
(5, 8),
(2, 20),
(3, 20),
(2, 19),
(6, 19),
(2, 21),
(2, 22),
(2, 23),
(5, 23),
(6, 23),
(1, 22),
(2, 24),
(6, 24),
(2, 29),
(2, 30),
(5, 30),
(2, 33),
(5, 33),
(6, 33),
(2, 32),
(5, 32),
(6, 32),
(1, 34),
(2, 34),
(3, 34),
(4, 34),
(2, 35),
(5, 35),
(2, 36),
(4, 36),
(5, 36),
(2, 37),
(4, 37),
(5, 37),
(2, 38),
(2, 39),
(4, 39),
(5, 39),
(1, 39),
(5, 24),
(4, 38);

Answer 1

试试这个:

SELECT A.id, A.title, COUNT(B.id)
FROM area A LEFT JOIN  properties B 
ON A.id=B.area_id 
JOIN properties_categories C 
ON C.property_id=B.id
WHERE C.category_id IN (1,2,3)
GROUP BY A.id, A.title;

参见MySQL Join Made Easy了解有关在 mysql 中使用联接的见解。