, 즉 아주 쉽게 기존의 구조를 감안할 때 :
이
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);
그냥 HAVING COUNT(rest_id)에 주어진 숫자 값은 당신이 검색하는 값의 수를 일치하는지 확인 . 다음은 간단한 테스트 케이스 (I 실제로 '중국어', '저녁 식사'와 '주차'가 또 다른 레스토랑, 추가 한 노트입니다 : 또는
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)
,이 같은 관련 테이블을 만들 수는 (그러나 이것은이다 아마 가장 좋은 구조) :
이
---> facility
restaurant ---> restaurant_has_facility ---|
---> facility_type
가
이 쿼리는, 당신은 단지 적절한를 생산하기 위해 하위 쿼리를 거의 동일한 필요가있다 가입 : 여기
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;
은 위의 몇 가지 샘플 SQL입니다 구조 :
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);