我在用子查询围绕查询时遇到麻烦。请继续,如果您会:
样本数据:
CREATE TABLE users (
user_id INT(10) NOT NULL AUTO_INCREMENT,
username VARCHAR(64),
PRIMARY KEY (user_id)
);
CREATE TABLE cars {
car_id INT(10) NOT NULL AUTO_INCREMENT,
user_id INT(10),
caryear YEAR(4),
carmaker VARCHAR(32),
carmodel VARCHAR(32),
PRIMARY KEY (car_id, user_id)
FOREIGN KEY(`user_id`) references users(`user_id`)
) ENGINE=InnoDB;
INSERT INTO users (user_id, name) VALUES (1,'Bob'),(2,'John'),(3,'Sally');
INSERT INTO cars (user_id, caryear, carmaker, carmodel) VALUES
(1,'2004','Audi','A4'),
(1,'2006','Toyota','Camry'),
(1,'2014','Jeep','CJ'),
(2,'1998','Acura','CL'),
(2,'2014','Honda','Accord'),
(3,'2011','Jeep','Rubicon')
好的,因此,如果我想获取用户及其汽车的列表,可以执行以下操作:
SELECT
user_id,
username,
(SELECT GROUP_CONCAT(CONCAT(CarYear,' ',CarMaker,' ',CarModel))
FROM cars
WHERE cars.user_id = users.user_id
) AS Cars
FROM users;
但是,例如,如何获取拥有2004-2014年吉普车的用户列表?
我已经提出了,但是我敢肯定还有一种更优雅的方法:
SELECT
user_id,
username,
(SELECT GROUP_CONCAT(CONCAT(CarYear,' ',CarMaker,' ',CarModel))
FROM cars
WHERE cars.user_id = users.user_id
AND CarMaker = 'Jeep'
AND CarYear BETWEEN 2004 AND 2014
) AS Cars
FROM users
HAVING Cars is not null;
该解决方案的问题在于,它不会显示拥有2004-2014年吉普车的人拥有的其他汽车。
有什么建议吗?
要为拥有吉普车2004-2014的任何用户显示用户及其所有汽车的列表,请执行以下操作:
SELECT
u.user_id,
u.username,
GROUP_CONCAT(CONCAT(c.CarYear,' ',c.CarMaker,' ',c.CarModel)) AS Cars
FROM users u
INNER JOIN cars c USING (user_id)
INNER JOIN (
SELECT DISTINCT user_id FROM cars
WHERE c.CarMaker = 'Jeep'
AND c.CarYear BETWEEN 2004 AND 2014) AS j USING (user_id)
GROUP BY u.user_id, u.username;
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