Question

我正在开发视频游戏广告资源网站。这是我的数据库表的简化版本以及一些示例数据：

CREATE TABLE `platforms` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(16) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;

INSERT INTO `platforms` VALUES(1, 'Nintendo');
INSERT INTO `platforms` VALUES(2, 'Super Nintendo');
INSERT INTO `platforms` VALUES(3, 'Nintendo 64');

--

CREATE TABLE `games` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `platform_id` int(10) unsigned NOT NULL,
  `name` varchar(255) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;

INSERT INTO `games` VALUES(1, 1, 'Super Mario Bros.');
INSERT INTO `games` VALUES(2, 1, 'Super Mario Bros. 2');
INSERT INTO `games` VALUES(3, 2, 'Super Mario World');
INSERT INTO `games` VALUES(4, 2, 'Super Mario Kart');
INSERT INTO `games` VALUES(5, 3, 'Super Mario 64');
INSERT INTO `games` VALUES(6, 3, 'Mario Kart 64');

--

CREATE TABLE `users` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `username` varchar(64) NOT NULL,
  `password` varchar(64) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

INSERT INTO `users` VALUES(1, 'john_doe', '$2a$10$cQhc4VAXVMEyC1tA.VRoWunpNVi7392adacT/weVBzu6XGI6.Jx/K');
INSERT INTO `users` VALUES(2, 'jane_doe', '$2a$10$Ot2BmlT14hKDxHGIV8jBx.lW76HCWdwuOhNGIYrJO5O7BEtDUWLWu');

--

CREATE TABLE `games_users` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `game_id` int(10) unsigned NOT NULL,
  `user_id` int(10) unsigned NOT NULL,
  `created` datetime NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;

INSERT INTO `games_users` VALUES(1, 1, 1, '2013-04-12 12:18:09');
INSERT INTO `games_users` VALUES(2, 3, 1, '2013-04-12 12:18:42');
INSERT INTO `games_users` VALUES(3, 4, 1, '2013-04-12 12:19:13');
INSERT INTO `games_users` VALUES(4, 2, 2, '2013-04-12 12:19:32');

正如你所看到的，john_doe有1个任天堂游戏（超级马里奥兄弟），2个超级任天堂游戏（超级马里奥世界和超级马里奥赛车）和0个任天堂64游戏。 jane_doe有1个任天堂游戏（超级马里奥兄弟2），0个超级任天堂游戏，0个任天堂64游戏。

我想编写一个特定于用户的查询，该查询将列出所有控制台，并列出用户为每个控制台拥有的游戏数量。

这是john_doe的结果：

platform.id: 1
platform.name: Nintendo
game_count: 1

platform.id: 2
platform.name: Super Nintendo
game_count: 2

platform.id: 3
platform.name: Nintendo 64
game_count: 0

这就是jane_doe的结果：

platform.id: 1
platform.name: Nintendo
game_count: 1

platform.id: 2
platform.name: Super Nintendo
game_count: 0

platform.id: 3
platform.name: Nintendo 64
game_count: 0

我该怎么做？

Answer 1

SELECT p.id, p.name, IFNULL(t.cnt, 0)
FROM platforms p
LEFT JOIN (
  SELECT g.platform_id as 'id', COUNT(g.id) as cnt
  FROM users u
  JOIN games_users gu ON gu.user_id = u.id
  JOIN games g ON g.id = gu.game_id
  WHERE u.username = "jane_doe"
  GROUP BY g.platform_id
  ) t ON t.id = p.id

的 DEMO 的

Answer 2

SELECT p.id, p.name, COUNT(g.id) FROM users u 
LEFT JOIN games_users gu ON u.id = gu.user_id 
LEFT JOIN games g ON g.id = gu.game_id 
LEFT JOIN platforms p ON g.platform_id = p.id 
WHERE u.id = 1 
GROUP BY p.id

http://sqlfiddle.com/#!2/6104d/6

如果你想看0，你必须做一些作弊我想，这就是我实现它的方式。我得到了user_id的总和（因为它们都是相同的）并除以user_id。

SELECT p.id, p.name, IFNULL(SUM(user_id),0)/1 FROM platforms p 
LEFT JOIN games g ON g.platform_id = p.id 
LEFT JOIN games_users gu ON g.id = gu.game_id AND gu.user_id = 1
GROUP BY p.id, p.name


SELECT p.id, p.name, IFNULL(SUM(user_id),0)/2 FROM platforms p 
LEFT JOIN games g ON g.platform_id = p.id 
LEFT JOIN games_users gu ON g.id = gu.game_id AND gu.user_id = 2
GROUP BY p.id, p.name

SQL左连接数

2 个答案: