鉴于您现有的结构,这很容易:
SELECT name, place FROM restaurant WHERE id IN (
SELECT rest_id FROM stack
WHERE value IN ('chinese', 'dinner', 'parking')
GROUP BY rest_id
HAVING COUNT(rest_id)=3);
只需确保给定HAVING COUNT(rest_id)
的数值与您正在搜索的值的数量相匹配。这是一个简单的测试用例(请注意,我添加了另一家餐厅,实际上有 'chinese'、'dinner' 和 'parking':
CREATE TABLE `restaurant` (
`id` int(11) NOT NULL auto_increment,
`name` VARCHAR(255),
`place` VARCHAR(255),
PRIMARY KEY (`id`)
) ENGINE=InnoDB;
CREATE TABLE `stack` (
`id` int(11) NOT NULL auto_increment,
`rest_id` int(11) NOT NULL,
`type` VARCHAR(255),
`value` VARCHAR(255),
PRIMARY KEY (`id`)
) ENGINE=InnoDB;
INSERT INTO `restaurant` VALUES
(1, 'rest1', 'ny'),
(2, 'rest2', 'la'),
(3, 'rest3', 'ph'),
(4, 'rest4', 'mlp');
INSERT INTO `stack` VALUES
( 1, 1, 'cuisine', 'chinese'),
( 2, 1, 'serves', 'breakfast'),
( 3, 1, 'facilities', 'party hall'),
( 4, 1, 'serves', 'lunch'),
( 5, 1, 'serves', 'dinner'),
( 6, 1, 'cuisine', 'seafood'),
( 7, 2, 'cuisine', 'Italian'),
( 8, 2, 'serves', 'breakfast'),
( 9, 2, 'facilities', 'parking'),
(10, 2, 'serves', 'lunch'),
(11, 2, 'serves', 'dinner'),
(12, 2, 'cuisine', 'indian'),
(13, 3, 'cuisine', 'chinese'),
(14, 3, 'serves', 'breakfast'),
(15, 3, 'facilities', 'parking'),
(16, 3, 'serves', 'lunch'),
(17, 3, 'serves', 'dinner'),
(18, 3, 'cuisine', 'indian');
SELECT name, place FROM restaurant WHERE id IN (
SELECT rest_id FROM stack
WHERE value IN ('chinese', 'dinner', 'parking')
GROUP BY rest_id
HAVING COUNT(rest_id)=3);
+-------+-------+
| name | place |
+-------+-------+
| rest3 | ph |
+-------+-------+
SELECT name, place FROM restaurant WHERE id IN (
SELECT rest_id FROM stack
WHERE value IN ('chinese', 'dinner')
GROUP BY rest_id
HAVING COUNT(rest_id)=2);
+-------+-------+
| name | place |
+-------+-------+
| rest1 | ny |
| rest3 | ph |
+-------+-------+
SELECT name, place FROM restaurant WHERE id IN (
SELECT rest_id FROM stack
WHERE value IN ('parking', 'hellipad')
GROUP BY rest_id
HAVING COUNT(rest_id)=2);
Empty set (0.00 sec)
或者,您可以像这样创建相关表(但这可能不是最好的结构):
---> facility
restaurant ---> restaurant_has_facility ---|
---> facility_type
查询几乎相同,您只需要您的子查询来生成适当的连接:
SELECT restaurant_name, restaurant_place FROM (
SELECT
r.id AS restaurant_id,
r.name AS restaurant_name,
r.place AS restaurant_place,
ft.name AS facility_name
FROM restaurant AS r
JOIN restaurant_has_facility AS rf ON rf.restaurant_id = r.id
JOIN facility_type AS ft ON ft.id = rf.facility_type_id
ORDER BY r.id, ft.name) AS tmp
WHERE facility_name IN ('chinese', 'dinner', 'parking')
GROUP BY tmp.restaurant_id
HAVING COUNT(tmp.restaurant_id)=3;
以下是上述结构的一些示例 SQL:
CREATE TABLE `restaurant` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT ,
`name` VARCHAR(45) NOT NULL ,
`place` VARCHAR(45) NOT NULL ,
PRIMARY KEY (`id`) )
ENGINE = InnoDB;
CREATE TABLE `facility` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT ,
`name` VARCHAR(45) NOT NULL ,
PRIMARY KEY (`id`) )
ENGINE = InnoDB;
CREATE TABLE IF NOT EXISTS `facility_type` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT ,
`name` VARCHAR(45) NOT NULL ,
PRIMARY KEY (`id`) )
ENGINE = InnoDB;
CREATE TABLE IF NOT EXISTS `restaurant_has_facility` (
`restaurant_id` INT UNSIGNED NOT NULL ,
`facility_id` INT UNSIGNED NOT NULL ,
`facility_type_id` INT UNSIGNED NOT NULL ,
PRIMARY KEY (`restaurant_id`, `facility_id`, `facility_type_id`) ,
INDEX `fk_restaurant_has_facility_restaurant` (`restaurant_id` ASC) ,
CONSTRAINT `fk_restaurant_has_facility_restaurant`
FOREIGN KEY (`restaurant_id` )
REFERENCES `restaurant` (`id` )
ON DELETE CASCADE
ON UPDATE CASCADE)
ENGINE = InnoDB;
INSERT INTO `restaurant` VALUES
(1, 'rest1', 'ny'),
(2, 'rest2', 'la'),
(3, 'rest3', 'ph'),
(4, 'rest4', 'mlp');
INSERT INTO `facility` VALUES
(1, 'cuisine'),
(2, 'serves'),
(3, 'facilities');
INSERT INTO `facility_type` VALUES
(1, 'chinese'),
(2, 'breakfast'),
(3, 'party hall'),
(4, 'lunch'),
(5, 'dinner'),
(6, 'seafood'),
(7, 'Italian'),
(8, 'parking'),
(9, 'indian');
INSERT INTO `restaurant_has_facility` VALUES
(1, 1, 1),
(1, 2, 2),
(1, 3, 3),
(1, 2, 4),
(1, 2, 5),
(1, 1, 6),
(2, 1, 7),
(2, 2, 2),
(2, 3, 8),
(2, 2, 4),
(2, 2, 5),
(2, 1, 9),
(3, 1, 1),
(3, 2, 5),
(3, 3, 8),
(3, 2, 4),
(3, 2, 2),
(3, 1, 9);